How to get a specific word using regular expression if it contains the part of the given word from the comma separated string?

Asked Jun 04 '20 at 20:29

Active Jun 04 '20 at 20:31

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I have a string = "file_20200604170659.DAT,file_20200604170659.TRG,file_20200604170659.CSV"

I need to write a Regular Expression to get the file that contains the given output type.

For Example: If I ask for CVS, I just want to get file_20200604170659.CSV from the above string.

edited Jun 04 '20 at 20:31

Wiktor Stribiżew

asked Jun 04 '20 at 20:29

Pad Sharma

If your language's regex supports *lookbehinds* and you wish to match `'CSV'` files, you might write `(?<=^|,)file_\d{14}\.CSV\b`. `(?<=^|,)` is a *positive lookbehind* (`?<=`) that asserts the match start at the beginning of a line (`^`) or (`|`) be immediately preceded by a comma. `\b` is a word boundary, which prevent `'CSV'` to be followed by a word character (e.g., `'CSVV'`). The match would be `'file_'` followed by 14 digits, followed by `'.CSV'`. This would match `'file_20200604170659.CSV'`. [Demo](https://regex101.com/r/XwpbLX/2/) – Cary Swoveland Jun 04 '20 at 21:34

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