How to resove variables in a powershell script block The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) The Ask Question Wizard is Live! Data science time! April 2019 and salary with experiencePass arguments to a scriptblock in powershellSetting Windows PowerShell path variableDetermine installed PowerShell versionTerminating a script in PowerShellHow to run a PowerShell scriptHow to output something in PowerShellPowerShell says “execution of scripts is disabled on this system.”How to pass an argument to a PowerShell script?How do you comment out code in PowerShell?Best way to write to the console in PowerShellHow do I concatenate strings and variables in PowerShell?
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How to resove variables in a powershell script block
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experiencePass arguments to a scriptblock in powershellSetting Windows PowerShell path variableDetermine installed PowerShell versionTerminating a script in PowerShellHow to run a PowerShell scriptHow to output something in PowerShellPowerShell says “execution of scripts is disabled on this system.”How to pass an argument to a PowerShell script?How do you comment out code in PowerShell?Best way to write to the console in PowerShellHow do I concatenate strings and variables in PowerShell?
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Given I have:
$a = "world"
$b = write-host "hello $a"
How to I get the resolved text of the script block, which should be:
write-host "hello world"
UPDATE: additional clarifications
If you just print $b you get the variable and not the resolved value
write-host "hello $a"
If you execute the script block with & $b you get the printed value, not the contents of the script block:
hello world
This question is seeking a string containing the contents of the script block with the evaluated variables, which is:
write-host "hello world"
powershell
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
add a comment |
Given I have:
$a = "world"
$b = write-host "hello $a"
How to I get the resolved text of the script block, which should be:
write-host "hello world"
UPDATE: additional clarifications
If you just print $b you get the variable and not the resolved value
write-host "hello $a"
If you execute the script block with & $b you get the printed value, not the contents of the script block:
hello world
This question is seeking a string containing the contents of the script block with the evaluated variables, which is:
write-host "hello world"
powershell
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...Invoke-Command -ScriptBlock $boutput =hello world
– Lee_Dailey
Mar 8 at 14:43
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
@Lee_Dailey - as you can see from the answer, you can do that using$ExecutionContext.InvokeCommand.ExpandString($b)
– alastairtree
Mar 8 at 16:52
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00
add a comment |
Given I have:
$a = "world"
$b = write-host "hello $a"
How to I get the resolved text of the script block, which should be:
write-host "hello world"
UPDATE: additional clarifications
If you just print $b you get the variable and not the resolved value
write-host "hello $a"
If you execute the script block with & $b you get the printed value, not the contents of the script block:
hello world
This question is seeking a string containing the contents of the script block with the evaluated variables, which is:
write-host "hello world"
powershell
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
Given I have:
$a = "world"
$b = write-host "hello $a"
How to I get the resolved text of the script block, which should be:
write-host "hello world"
UPDATE: additional clarifications
If you just print $b you get the variable and not the resolved value
write-host "hello $a"
If you execute the script block with & $b you get the printed value, not the contents of the script block:
hello world
This question is seeking a string containing the contents of the script block with the evaluated variables, which is:
write-host "hello world"
powershell
powershell
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
edited Mar 8 at 14:55
alastairtree
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
asked Mar 8 at 12:48
alastairtreealastairtree
1,8571631
1,8571631
Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...Invoke-Command -ScriptBlock $boutput =hello world
– Lee_Dailey
Mar 8 at 14:43
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
@Lee_Dailey - as you can see from the answer, you can do that using$ExecutionContext.InvokeCommand.ExpandString($b)
– alastairtree
Mar 8 at 16:52
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00
add a comment |
Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...Invoke-Command -ScriptBlock $boutput =hello world
– Lee_Dailey
Mar 8 at 14:43
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
@Lee_Dailey - as you can see from the answer, you can do that using$ExecutionContext.InvokeCommand.ExpandString($b)
– alastairtree
Mar 8 at 16:52
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00
Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...
Invoke-Command -ScriptBlock $b output = hello world– Lee_Dailey
Mar 8 at 14:43
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...
Invoke-Command -ScriptBlock $b output = hello world– Lee_Dailey
Mar 8 at 14:43
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
@Lee_Dailey - as you can see from the answer, you can do that using
$ExecutionContext.InvokeCommand.ExpandString($b)– alastairtree
Mar 8 at 16:52
@Lee_Dailey - as you can see from the answer, you can do that using
$ExecutionContext.InvokeCommand.ExpandString($b)– alastairtree
Mar 8 at 16:52
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
As in the original question, if your entire scriptblock contents is not a string (but you want it to be) and you need variable substitution within the scriptblock, you can use the following:
$ExecutionContext.InvokeCommand.ExpandString($b)
Calling .InvokeCommand.ExpandString($b) on the current execution context will use the variables in the current scope for substitution.
The following is one way to create a scripblock and retrieve its contents:
$a = "world"
$b = [ScriptBlock]::create("write-host hello $a")
$b
write-host hello world
You can use your scriptblock notation as well to accomplish the same thing, but you need to use the & call operator:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
A feature to using the method above is that if you change the value of $a at any time and then call the scriptblock again, the output will be updated like so:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
$a = "hi"
& $b
write-host hello hi
The GetNewClosure() method can be used to create a clone of the scriptblock above to take a theoretical snapshot of the scriptblock's current evaluation. It will be immune to the changing of the $a value later the code:
$b = "write-host hello $a".GetNewClosure()
& $b
write-host hello world
$a = "new world"
& $b
write-host hello world
The notation denotes a scriptblock object as you probably already know. That can be passed into Invoke-Command, which opens up other options. You can also create parameters inside of the scriptblock that can be passed in later. See about_Script_Blocks for more information.
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just printswrite-host "hello $a"and the question asks forwrite-host "hello world"
– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
Found a short answer. Just run this$executioncontext.invokecommand.expandstring($b).
– AdminOfThings
Mar 8 at 15:51
|
show 1 more comment
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1 Answer
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As in the original question, if your entire scriptblock contents is not a string (but you want it to be) and you need variable substitution within the scriptblock, you can use the following:
$ExecutionContext.InvokeCommand.ExpandString($b)
Calling .InvokeCommand.ExpandString($b) on the current execution context will use the variables in the current scope for substitution.
The following is one way to create a scripblock and retrieve its contents:
$a = "world"
$b = [ScriptBlock]::create("write-host hello $a")
$b
write-host hello world
You can use your scriptblock notation as well to accomplish the same thing, but you need to use the & call operator:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
A feature to using the method above is that if you change the value of $a at any time and then call the scriptblock again, the output will be updated like so:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
$a = "hi"
& $b
write-host hello hi
The GetNewClosure() method can be used to create a clone of the scriptblock above to take a theoretical snapshot of the scriptblock's current evaluation. It will be immune to the changing of the $a value later the code:
$b = "write-host hello $a".GetNewClosure()
& $b
write-host hello world
$a = "new world"
& $b
write-host hello world
The notation denotes a scriptblock object as you probably already know. That can be passed into Invoke-Command, which opens up other options. You can also create parameters inside of the scriptblock that can be passed in later. See about_Script_Blocks for more information.
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just printswrite-host "hello $a"and the question asks forwrite-host "hello world"
– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
Found a short answer. Just run this$executioncontext.invokecommand.expandstring($b).
– AdminOfThings
Mar 8 at 15:51
|
show 1 more comment
As in the original question, if your entire scriptblock contents is not a string (but you want it to be) and you need variable substitution within the scriptblock, you can use the following:
$ExecutionContext.InvokeCommand.ExpandString($b)
Calling .InvokeCommand.ExpandString($b) on the current execution context will use the variables in the current scope for substitution.
The following is one way to create a scripblock and retrieve its contents:
$a = "world"
$b = [ScriptBlock]::create("write-host hello $a")
$b
write-host hello world
You can use your scriptblock notation as well to accomplish the same thing, but you need to use the & call operator:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
A feature to using the method above is that if you change the value of $a at any time and then call the scriptblock again, the output will be updated like so:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
$a = "hi"
& $b
write-host hello hi
The GetNewClosure() method can be used to create a clone of the scriptblock above to take a theoretical snapshot of the scriptblock's current evaluation. It will be immune to the changing of the $a value later the code:
$b = "write-host hello $a".GetNewClosure()
& $b
write-host hello world
$a = "new world"
& $b
write-host hello world
The notation denotes a scriptblock object as you probably already know. That can be passed into Invoke-Command, which opens up other options. You can also create parameters inside of the scriptblock that can be passed in later. See about_Script_Blocks for more information.
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just printswrite-host "hello $a"and the question asks forwrite-host "hello world"
– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
Found a short answer. Just run this$executioncontext.invokecommand.expandstring($b).
– AdminOfThings
Mar 8 at 15:51
|
show 1 more comment
As in the original question, if your entire scriptblock contents is not a string (but you want it to be) and you need variable substitution within the scriptblock, you can use the following:
$ExecutionContext.InvokeCommand.ExpandString($b)
Calling .InvokeCommand.ExpandString($b) on the current execution context will use the variables in the current scope for substitution.
The following is one way to create a scripblock and retrieve its contents:
$a = "world"
$b = [ScriptBlock]::create("write-host hello $a")
$b
write-host hello world
You can use your scriptblock notation as well to accomplish the same thing, but you need to use the & call operator:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
A feature to using the method above is that if you change the value of $a at any time and then call the scriptblock again, the output will be updated like so:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
$a = "hi"
& $b
write-host hello hi
The GetNewClosure() method can be used to create a clone of the scriptblock above to take a theoretical snapshot of the scriptblock's current evaluation. It will be immune to the changing of the $a value later the code:
$b = "write-host hello $a".GetNewClosure()
& $b
write-host hello world
$a = "new world"
& $b
write-host hello world
The notation denotes a scriptblock object as you probably already know. That can be passed into Invoke-Command, which opens up other options. You can also create parameters inside of the scriptblock that can be passed in later. See about_Script_Blocks for more information.
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
As in the original question, if your entire scriptblock contents is not a string (but you want it to be) and you need variable substitution within the scriptblock, you can use the following:
$ExecutionContext.InvokeCommand.ExpandString($b)
Calling .InvokeCommand.ExpandString($b) on the current execution context will use the variables in the current scope for substitution.
The following is one way to create a scripblock and retrieve its contents:
$a = "world"
$b = [ScriptBlock]::create("write-host hello $a")
$b
write-host hello world
You can use your scriptblock notation as well to accomplish the same thing, but you need to use the & call operator:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
A feature to using the method above is that if you change the value of $a at any time and then call the scriptblock again, the output will be updated like so:
$a = "world"
$b = "write-host hello $a"
& $b
write-host hello world
$a = "hi"
& $b
write-host hello hi
The GetNewClosure() method can be used to create a clone of the scriptblock above to take a theoretical snapshot of the scriptblock's current evaluation. It will be immune to the changing of the $a value later the code:
$b = "write-host hello $a".GetNewClosure()
& $b
write-host hello world
$a = "new world"
& $b
write-host hello world
The notation denotes a scriptblock object as you probably already know. That can be passed into Invoke-Command, which opens up other options. You can also create parameters inside of the scriptblock that can be passed in later. See about_Script_Blocks for more information.
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
edited Mar 8 at 16:59
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
answered Mar 8 at 13:19
AdminOfThingsAdminOfThings
1,9671212
1,9671212
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just printswrite-host "hello $a"and the question asks forwrite-host "hello world"
– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
Found a short answer. Just run this$executioncontext.invokecommand.expandstring($b).
– AdminOfThings
Mar 8 at 15:51
|
show 1 more comment
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just printswrite-host "hello $a"and the question asks forwrite-host "hello world"
– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
Found a short answer. Just run this$executioncontext.invokecommand.expandstring($b).
– AdminOfThings
Mar 8 at 15:51
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
Sorry, yeah meant $a not $b, although I think the question still stands. Updated the question
– alastairtree
Mar 8 at 13:40
This is not correct and misses the crus of the question - printing $b just prints
write-host "hello $a" and the question asks for write-host "hello world"– alastairtree
Mar 8 at 14:51
This is not correct and misses the crus of the question - printing $b just prints
write-host "hello $a" and the question asks for write-host "hello world"– alastairtree
Mar 8 at 14:51
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
Did you read through the post and try the different scenarios? You have to have quotes around everything inside of the scriptblock to print the entire text, which is there in my examples.
– AdminOfThings
Mar 8 at 14:56
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
I understand but you have changed the question to replace the contents of the script block with a string. Instead the question assumes you already have a script block, and want to print its contents without executing it.
– alastairtree
Mar 8 at 15:01
4
4
Found a short answer. Just run this
$executioncontext.invokecommand.expandstring($b).– AdminOfThings
Mar 8 at 15:51
Found a short answer. Just run this
$executioncontext.invokecommand.expandstring($b).– AdminOfThings
Mar 8 at 15:51
|
show 1 more comment
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Possible duplicate of Pass arguments to a scriptblock in powershell
– arco444
Mar 8 at 13:14
you have a scriptblock and you need to invoke/run the scriptblock. [grin] simply calling the $Var that holds the scriptblock will give the literal content, not run it. you can run it thus ...
Invoke-Command -ScriptBlock $boutput =hello world– Lee_Dailey
Mar 8 at 14:43
I dont believe this is a duplicate because i am not executing the script block - i want a string of its syntax with evaluated variables
– alastairtree
Mar 8 at 14:57
@Lee_Dailey - as you can see from the answer, you can do that using
$ExecutionContext.InvokeCommand.ExpandString($b)– alastairtree
Mar 8 at 16:52
@alastairtree - ha! i learned something new! [grin] i will delete this comment soon - and immediately delete my wrong comment to avoid confusing folks.
– Lee_Dailey
Mar 8 at 17:00