Regex: How can I get the parameter from function?

Question

I am working on getting parameter from a function string. For example, my function is named "translate". So all I need to do is to get everything inbetween "translate(" and ")". Is there a way that i can use regex to do that? So far I have something like:

"/translate\((?<keyName>*)\)/i"

Unfortunately it's not working. Can anybody help me with this one?

Thanks a lot!

And what's the overall goal? What do you intend to accomplish with the parameter name? — outis
– outis, Commented Mar 9, 2010 at 22:29
What about function names in side comments: /* comment translate() */? And in string literals: str = " string translate() "? — Bart Kiers
– Bart Kiers, Commented Mar 9, 2010 at 22:34

tiftik · Accepted Answer · 2010-03-09 22:29:58Z

1

translate\((.*?)\) will match the whole function call and capture the parameter into backreference 1. So you can replace it with \1 if you want to extract the parameter.

If you want to match only the parameter and the regex engine you're using supports look-behinds/look-aheads, you can use this one: (?<=translate\().*?(?=\))

answered Mar 9, 2010 at 22:29

tiftik

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tiftik Over a year ago

Note that these solutions obviously don't support escaping ). So they won't match or capture translate("<*)))><") as intended.

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