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Filename match with Python regex

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

-1

I have a text file scrapped from my email which contains 1 attachment/mail. The attachment is present under different names with totally different formats, ex.:

filename="John_wheeler 11041997 resume.pdf";
filename="Kujal_newResume(1).pdf"; 
filename=JohnKrasinski_Resume.pdf

My question is: is there any way to find a RegEx pattern that would start searching from filename= and go until the dot character (that separates from file extension)? Getting file extension would be next task, but I can hold that for now. Please help me figure this out.

python regex python-3.x

share|improve this question

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Your question is unclear. Please provide the data/input you have, the code that you have tried, the expected output and the error you got.
  
  – Yusufsn
  Mar 8 at 7:13
  
  

  
  
  
  

add a comment | 

-1

I have a text file scrapped from my email which contains 1 attachment/mail. The attachment is present under different names with totally different formats, ex.:

filename="John_wheeler 11041997 resume.pdf";
filename="Kujal_newResume(1).pdf"; 
filename=JohnKrasinski_Resume.pdf

My question is: is there any way to find a RegEx pattern that would start searching from filename= and go until the dot character (that separates from file extension)? Getting file extension would be next task, but I can hold that for now. Please help me figure this out.

python regex python-3.x

share|improve this question

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Your question is unclear. Please provide the data/input you have, the code that you have tried, the expected output and the error you got.
  
  – Yusufsn
  Mar 8 at 7:13
  
  

  
  
  
  

add a comment | 

I have a text file scrapped from my email which contains 1 attachment/mail. The attachment is present under different names with totally different formats, ex.:

filename="John_wheeler 11041997 resume.pdf";
filename="Kujal_newResume(1).pdf"; 
filename=JohnKrasinski_Resume.pdf

My question is: is there any way to find a RegEx pattern that would start searching from filename= and go until the dot character (that separates from file extension)? Getting file extension would be next task, but I can hold that for now. Please help me figure this out.

python regex python-3.x

share|improve this question

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

filename="John_wheeler 11041997 resume.pdf";
filename="Kujal_newResume(1).pdf"; 
filename=JohnKrasinski_Resume.pdf

share|improve this question

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

share|improve this question

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

edited Mar 8 at 8:13

Michał Turczyn

16.2k132241

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

asked Mar 8 at 7:07

Srinath RadhakrishnanSrinath Radhakrishnan

42

3 Answers
3

active

oldest

votes

1

You could try this pattern: filename="?([^.]+)

It assumes that dot separates filename from extension.

Explanation:

filename="? - match filename= literally and tehn match 0 or 1 apostrophe "

([^.]+) - match one or more characters that is not a dot (match everything until dot) and store it in capturing group

Your desired filename will be stored in capturing group.

Demo

EXTRA: to capture also file extension, you could use such pattern: filename="?([^.]+).([^";]+)

Additional thing here is .([^";]+): matches dot literally with .. Then it matches one or more characters other than " or ; with pattern [^";]+ and stores it in second capturing gropup.

Another demo

share|improve this answer

edited Mar 8 at 8:20

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

add a comment | 

0

How about the following:

(?:filename=)([^.]*).(w*)

This REGEX returns different groups containing the different elements you're interested in.

share|improve this answer

answered Mar 8 at 7:15

tk78tk78

547310

add a comment | 

0

I'm not sure the output you expect. But this may help. RegexDemo

(?<=filename=)["]?(w.*[.].*)(?<=w)["]?

Or if you want to ignore the file extension:

(?<=filename=)["]?(w.*)[.]

share|improve this answer

answered Mar 8 at 7:26

YusufsnYusufsn

574215

add a comment | 

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1

You could try this pattern: filename="?([^.]+)

It assumes that dot separates filename from extension.

Explanation:

filename="? - match filename= literally and tehn match 0 or 1 apostrophe "

([^.]+) - match one or more characters that is not a dot (match everything until dot) and store it in capturing group

Your desired filename will be stored in capturing group.

Demo

EXTRA: to capture also file extension, you could use such pattern: filename="?([^.]+).([^";]+)

Additional thing here is .([^";]+): matches dot literally with .. Then it matches one or more characters other than " or ; with pattern [^";]+ and stores it in second capturing gropup.

Another demo

share|improve this answer

edited Mar 8 at 8:20

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

add a comment | 

1

You could try this pattern: filename="?([^.]+)

It assumes that dot separates filename from extension.

Explanation:

filename="? - match filename= literally and tehn match 0 or 1 apostrophe "

([^.]+) - match one or more characters that is not a dot (match everything until dot) and store it in capturing group

Your desired filename will be stored in capturing group.

Demo

EXTRA: to capture also file extension, you could use such pattern: filename="?([^.]+).([^";]+)

Additional thing here is .([^";]+): matches dot literally with .. Then it matches one or more characters other than " or ; with pattern [^";]+ and stores it in second capturing gropup.

Another demo

share|improve this answer

edited Mar 8 at 8:20

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

add a comment | 

You could try this pattern: filename="?([^.]+)

It assumes that dot separates filename from extension.

Explanation:

filename="? - match filename= literally and tehn match 0 or 1 apostrophe "

([^.]+) - match one or more characters that is not a dot (match everything until dot) and store it in capturing group

Your desired filename will be stored in capturing group.

Demo

EXTRA: to capture also file extension, you could use such pattern: filename="?([^.]+).([^";]+)

Additional thing here is .([^";]+): matches dot literally with .. Then it matches one or more characters other than " or ; with pattern [^";]+ and stores it in second capturing gropup.

Another demo

share|improve this answer

edited Mar 8 at 8:20

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

share|improve this answer

edited Mar 8 at 8:20

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

answered Mar 8 at 8:12

Michał TurczynMichał Turczyn

16.2k132241

0

How about the following:

(?:filename=)([^.]*).(w*)

This REGEX returns different groups containing the different elements you're interested in.

share|improve this answer

answered Mar 8 at 7:15

tk78tk78

547310

add a comment | 

0

How about the following:

(?:filename=)([^.]*).(w*)

This REGEX returns different groups containing the different elements you're interested in.

share|improve this answer

answered Mar 8 at 7:15

tk78tk78

547310

add a comment | 

How about the following:

(?:filename=)([^.]*).(w*)

This REGEX returns different groups containing the different elements you're interested in.

share|improve this answer

answered Mar 8 at 7:15

tk78tk78

547310

(?:filename=)([^.]*).(w*)

share|improve this answer

answered Mar 8 at 7:15

tk78tk78

547310

answered Mar 8 at 7:15

tk78tk78

547310

answered Mar 8 at 7:15

tk78tk78

547310

0

I'm not sure the output you expect. But this may help. RegexDemo

(?<=filename=)["]?(w.*[.].*)(?<=w)["]?

Or if you want to ignore the file extension:

(?<=filename=)["]?(w.*)[.]

share|improve this answer

answered Mar 8 at 7:26

YusufsnYusufsn

574215

add a comment | 

0

I'm not sure the output you expect. But this may help. RegexDemo

(?<=filename=)["]?(w.*[.].*)(?<=w)["]?

Or if you want to ignore the file extension:

(?<=filename=)["]?(w.*)[.]

share|improve this answer

answered Mar 8 at 7:26

YusufsnYusufsn

574215

add a comment | 

I'm not sure the output you expect. But this may help. RegexDemo

(?<=filename=)["]?(w.*[.].*)(?<=w)["]?

Or if you want to ignore the file extension:

(?<=filename=)["]?(w.*)[.]

share|improve this answer

answered Mar 8 at 7:26

YusufsnYusufsn

574215

(?<=filename=)["]?(w.*[.].*)(?<=w)["]?

(?<=filename=)["]?(w.*)[.]

share|improve this answer

answered Mar 8 at 7:26

YusufsnYusufsn

574215

answered Mar 8 at 7:26

YusufsnYusufsn

574215

answered Mar 8 at 7:26

YusufsnYusufsn

574215