What Regex would captur everything from ' mark to the end of a line?

Question

Have a text file that denotes remarks with a single '. Some lines have two quotes but need to get everything from the first instance of a ' and the line feed.

I AL01 ' A-LINE '091398 GDK 33394178
402922 0831850 ' '091398 GDK 33394179
I AL02 ' A-LINE '091398 GDK 33394180
400722 0833118 ' '091398 GDK 33394181
I A10A ' A-LINE 102 ' 53198 DJ 33394182
395335 0832203 ' ' 53198 DJ 33394183
I A10B ' A-LINE 102 ' 53198 DJ 3339418

Answer 1

The appropriate regex would be the ' char followed by any number of any chars [including zero chars] ending with an end of string/line token:

'.*$

And if you wanted to capture everything after the ' char but not include it in the output, you would use:

(?<=').*$

This basically says give me all characters that follow the ' char until the end of the line.

Edit: It has been noted that $ is implicit when using .* and therefore not strictly required, therefore the pattern:

'.*

is technically correct, however it is clearer to be specific and avoid confusion for later code maintenance, hence my use of the $. It is my belief that it is always better to declare explicit behaviour than rely on implicit behaviour in situations where clarity could be questioned.

Answer 2

'.*$

Starting with a single quote ('), match any character (.) zero or more times (*) until the end of the line ($).

Answer 3

'.*

I believe you need the option, Multiline.

Answer 4

This will capture everything up to the ' in backreference 1 - and everything after the ' in backreference 2. You may need to escape the apostrophes though depending on language (\')

/^([^']*)'?(.*)$/

Quick modification: if the line doesn't have an ' - backreference 1 should still catch the whole line.

^ - start of string
([^']*) - capture any number of not ' characters
'? - match the ' 0 or 1 time
(.*) - capture any number of characters
$ - end of string