Regular expression to find the arguments in a function call

Question

I need to extract the arguments from a function Say, Function_1(arg1,agr2,....argn).

Is it possible to strip the arguments alone with out brackets.

I tried using $[\w,]*$ expression

Argument, as in passed to a function call, OR is it parameter of a function? — nhahtdh, Jun 26 '12 at 07:18

score 2 · Accepted Answer · answered Jun 26 '12 at 07:15

2

If you want to "catch" the matches you need to put braces around them:

 \(([\w,]*)\)

This way you will find everything arg1,arg2 in $1.

answered Jun 26 '12 at 07:15

arnep

Yeah.. much simpler , u will get it in the first group. – gout Jun 26 '12 at 07:17

score 1 · Answer 2 · answered Jun 26 '12 at 07:13

1

use this Regular Expression (?<=[\w]*$)[\w,]*(?=$) to solve the purpose

answered Jun 26 '12 at 07:13

gout

What language have you tested this on? – nhahtdh Jun 26 '12 at 07:19
1

The answer is using variable-width look-behind, which is only supported in C#. http://stackoverflow.com/questions/3159524/why-doesnt-finite-repetition-in-lookbehind-work-in-some-flavorsp – nhahtdh Jun 26 '12 at 07:23

2 Answers2