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Add the X values in Json with respect to Image
Which “href” value should I use for JavaScript links, “#” or “javascript:void(0)”?Add table row in jQueryHow do I format a Microsoft JSON date?Can comments be used in JSON?How can I pretty-print JSON in a shell script?What is the correct JSON content type?Why does Google prepend while(1); to their JSON responses?Why can't Python parse this JSON data?How can I pretty-print JSON using JavaScript?Parse JSON in JavaScript?
9
In a Json, there are multiple image sources as "src" : "image.png", , "src" : "image2.png",
For image.png , right now i am fetching X value as "40" [3rd position in below image]
For image2.png , right now i am fetching X value as "100" [6th position in below image]
Requirement :
Instead of that, i need to add 1st (10) + 3rd (40) + 4th (50) positions values and fetch final X value for "src" : "image.png", as 100 [10+40+50] and
"src" : "image1.png" = 1st (10) + 6th (100) + 7th (105) positions & final value of X is 215....
Codepen : https://codepen.io/kidsdial/pen/zbKaEJ
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);javascript json
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
add a comment |
9
In a Json, there are multiple image sources as "src" : "image.png", , "src" : "image2.png",
For image.png , right now i am fetching X value as "40" [3rd position in below image]
For image2.png , right now i am fetching X value as "100" [6th position in below image]
Requirement :
Instead of that, i need to add 1st (10) + 3rd (40) + 4th (50) positions values and fetch final X value for "src" : "image.png", as 100 [10+40+50] and
"src" : "image1.png" = 1st (10) + 6th (100) + 7th (105) positions & final value of X is 215....
Codepen : https://codepen.io/kidsdial/pen/zbKaEJ
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);javascript json
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
add a comment |
9
9
9
1
In a Json, there are multiple image sources as "src" : "image.png", , "src" : "image2.png",
For image.png , right now i am fetching X value as "40" [3rd position in below image]
For image2.png , right now i am fetching X value as "100" [6th position in below image]
Requirement :
Instead of that, i need to add 1st (10) + 3rd (40) + 4th (50) positions values and fetch final X value for "src" : "image.png", as 100 [10+40+50] and
"src" : "image1.png" = 1st (10) + 6th (100) + 7th (105) positions & final value of X is 215....
Codepen : https://codepen.io/kidsdial/pen/zbKaEJ
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);javascript json
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
In a Json, there are multiple image sources as "src" : "image.png", , "src" : "image2.png",
For image.png , right now i am fetching X value as "40" [3rd position in below image]
For image2.png , right now i am fetching X value as "100" [6th position in below image]
Requirement :
Instead of that, i need to add 1st (10) + 3rd (40) + 4th (50) positions values and fetch final X value for "src" : "image.png", as 100 [10+40+50] and
"src" : "image1.png" = 1st (10) + 6th (100) + 7th (105) positions & final value of X is 215....
Codepen : https://codepen.io/kidsdial/pen/zbKaEJ
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
for (i = 1; i < layer2.length; i++)
var x = layer2[i].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);javascript json
javascript json
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
asked Mar 5 at 5:26
vickey colorsvickey colors
204220
204220
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
7
+50
Using a recursive function you can find all the src and its corresponding summed x values.
Below is a crude example. Refactor as you see fit.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);Here is the CodePen for the same: https://codepen.io/anon/pen/Wmpvre
Hope this helps :)
answered Mar 7 at 6:21
RoysonRoyson
5302514
add a comment |
5
You can do that using recursion. While going though nested obj you could parent x value each item in layers and check if the element in layers have src property add the previous x value to it.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
add a comment |
3
Here is my solution with recursion and mutation. You can cloneDeep the array before if you don't want to mutate it directly.
// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)answered Mar 7 at 7:15
birdbird
997620
add a comment |
3
let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
add a comment |
1
Here is My Solution. Check this out
let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
+50
Using a recursive function you can find all the src and its corresponding summed x values.
Below is a crude example. Refactor as you see fit.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);Here is the CodePen for the same: https://codepen.io/anon/pen/Wmpvre
Hope this helps :)
answered Mar 7 at 6:21
RoysonRoyson
5302514
add a comment |
7
+50
Using a recursive function you can find all the src and its corresponding summed x values.
Below is a crude example. Refactor as you see fit.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);Here is the CodePen for the same: https://codepen.io/anon/pen/Wmpvre
Hope this helps :)
answered Mar 7 at 6:21
RoysonRoyson
5302514
add a comment |
7
+50
7
+50
7
+50
Using a recursive function you can find all the src and its corresponding summed x values.
Below is a crude example. Refactor as you see fit.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);Here is the CodePen for the same: https://codepen.io/anon/pen/Wmpvre
Hope this helps :)
answered Mar 7 at 6:21
RoysonRoyson
5302514
Using a recursive function you can find all the src and its corresponding summed x values.
Below is a crude example. Refactor as you see fit.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);Here is the CodePen for the same: https://codepen.io/anon/pen/Wmpvre
Hope this helps :)
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function getAllSrc(layers)
let arr = [];
layers.forEach(layer =>
if(layer.src)
arr.push(src: layer.src, x: layer.x);
else if(layer.layers)
let newArr = getAllSrc(layer.layers);
if(newArr.length > 0)
newArr.forEach((src, x) =>
arr.push(src, x: (layer.x + x));
);
);
return arr;
function getData(data)
let arr = getAllSrc(data.layers);
for (let src, x of arr)
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 6:21
RoysonRoyson
5302514
answered Mar 7 at 6:21
RoysonRoyson
5302514
answered Mar 7 at 6:21
RoysonRoyson
5302514
answered Mar 7 at 6:21
RoysonRoyson
5302514
5302514
add a comment |
add a comment |
5
You can do that using recursion. While going though nested obj you could parent x value each item in layers and check if the element in layers have src property add the previous x value to it.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
add a comment |
5
You can do that using recursion. While going though nested obj you could parent x value each item in layers and check if the element in layers have src property add the previous x value to it.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
add a comment |
5
5
5
You can do that using recursion. While going though nested obj you could parent x value each item in layers and check if the element in layers have src property add the previous x value to it.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
You can do that using recursion. While going though nested obj you could parent x value each item in layers and check if the element in layers have src property add the previous x value to it.
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
function changeData(obj,x=0)
if(obj.src) obj.x += x;
if(obj.layers)
for(const item of obj.layers)
changeData(item,x+obj.x
changeData(jsonData);
function showData(obj)
if(obj.src) document.body.innerHTML += `src:$obj.src<br>
x:$obj.x<br>`;
if(obj.layers)
for(let i of obj.layers) showData(i)
showData(jsonData);
console.log(jsonData);answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
edited Mar 8 at 15:44
answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
answered Mar 7 at 6:25
Maheer AliMaheer Ali
6,953518
6,953518
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
add a comment |
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
Thanks again for your help, As the other answer gave answer before 3 minutes, i accepted that answer....
– vickey colors
Mar 11 at 5:10
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
@vickeycolors Time doesn't matter in giving answer. The thing which matters is short and efficient and I think mine code is better. And if you was not to accept my answer you shouldn't have asked me that much help.well if you think his code is better so no problem
– Maheer Ali
Mar 11 at 5:26
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
I agree, i will rethink about it and i will take the final decision, thanks.....
– vickey colors
Mar 11 at 5:27
add a comment |
3
Here is my solution with recursion and mutation. You can cloneDeep the array before if you don't want to mutate it directly.
// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)answered Mar 7 at 7:15
birdbird
997620
add a comment |
3
Here is my solution with recursion and mutation. You can cloneDeep the array before if you don't want to mutate it directly.
// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)answered Mar 7 at 7:15
birdbird
997620
add a comment |
3
3
3
Here is my solution with recursion and mutation. You can cloneDeep the array before if you don't want to mutate it directly.
// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)answered Mar 7 at 7:15
birdbird
997620
Here is my solution with recursion and mutation. You can cloneDeep the array before if you don't want to mutate it directly.
// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)// Code
function recursion(obj, currentX = 0)
if(obj.src) obj.x = currentX
else if(obj.layers)
obj.layers.forEach(subObj => recursion(subObj, (currentX + subObj.x)))
// Data
let jsonData =
"layers" : [
"x" : 10,
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40,
"layers" : [
"x" : 50,
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
// Use
recursion(jsonData)
console.log(jsonData)answered Mar 7 at 7:15
birdbird
997620
answered Mar 7 at 7:15
birdbird
997620
answered Mar 7 at 7:15
birdbird
997620
answered Mar 7 at 7:15
birdbird
997620
997620
add a comment |
add a comment |
3
let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
add a comment |
3
let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
add a comment |
3
3
3
let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);let jsonData =
"layers": [
"x": 10,
"layers": [
"x": 20,
"y": 30,
"name": "L2a"
,
"x": 40,
"layers": [
"x": 50,
"src": "image.png",
"y": 60,
"name": "L2b-1"
,
"x": 70,
"y": 80,
"name": "L2b-2"
],
"y": 90,
"name": "user_image_1"
,
"x": 100,
"layers": [
"x": 105,
"src": "image2.png",
"y": 110,
"name": "L2C-1"
,
"x": 120,
"y": 130,
"name": "L2C-2"
],
"y": 140,
"name": "L2"
],
"y": 150,
"name": "L1"
]
;
function getData(data)
var dataObj = ;
let layer1 = data.layers;
let layer2 = layer1[0].layers;
let y = layer1[0].x;
for (i = 1; i < layer2.length; i++)
var x = y;
x += layer2[i].x;
x += layer2[i].layers[0].x;
var src = layer2[i].layers[0].src;
document.write("src :" + src);
document.write("<br>");
document.write("x:" + x);
document.write("<br>");
getData(jsonData);answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
answered Mar 7 at 13:28
Syed Mehtab HassanSyed Mehtab Hassan
388115
388115
add a comment |
add a comment |
1
Here is My Solution. Check this out
let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
add a comment |
1
Here is My Solution. Check this out
let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
add a comment |
1
1
1
Here is My Solution. Check this out
let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
Here is My Solution. Check this out
let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));let jsonData =
"layers" : [
"x" : 10, //
"layers" : [
"x" : 20,
"y" : 30,
"name" : "L2a"
,
"x" : 40, //
"layers" : [
"x" : 50, //
"src" : "image.png",
"y" : 60,
"name" : "L2b-1"
,
"x" : 70,
"y" : 80,
"name" : "L2b-2"
],
"y" : 90,
"name" : "user_image_1"
,
"x" : 100,
"layers" : [
"x" : 105,
"src" : "image2.png",
"y" : 110,
"name" : "L2C-1"
,
"x" : 120,
"y" : 130,
"name" : "L2C-2"
],
"y" : 140,
"name" : "L2"
],
"y" : 150,
"name" : "L1"
]
;
var dataObj = [];
var x, src;
function getData(data)
let layer1 = data.layers;
for(var i = 0; i < layer1.length; i++)
if(x === 0)
x = layer1[i].x;
continue;
x = layer1[i].x;
if(typeof layer1[i].layers !== 'undefined')
createOutput(layer1[i].layers);
return dataObj;
function createOutput(dataArray)
if(typeof dataArray === 'undefined')
dataObj.push(x: x, src: src);
x = 0;
getData(jsonData);
return;
dataArray.forEach(element => typeof element.src !== 'undefined')
x += element.x;
src = element.src;
createOutput(element.layers);
)
console.log(JSON.stringify(getData(jsonData)));answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
answered Mar 15 at 4:42
Darshana PathumDarshana Pathum
984
984
add a comment |
add a comment |
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