Using Regular Expressions to parse functions from a string in PHP

Question

Is there a logical way to strip out function names and their arguments using Regular Expressions in PHP? Currently I have it splitting line by line so that I am able to have each function on each line for easier markup.

So:

doSomeFunction(arg, moreargs, stuff);
breakSomething();
ini_set(include_path, /home/htdocs/);

becomes

array([0] => 'doSomeFunction(arg, moreargs, stuff);', [1] => 'breakSomething();' and so on...

The real problem I am having is getting the function name using regex to have syntax like:

$functionfromregex($args);

The functions are constructed using PHP's "Variable Function Name" ability, so it is possible. Arguments are taken in as arrays since there is no reasonable way to glue them togather the way that PHP needs to parse them as. The real downside is that I am not too familiar with Regular Expressions, any help will be greatly appreciated, and if you need any other information feel free to ask.

This is one of those cases where you're parsing structured text and a regular expression is not the way to go. Regular expressions aren't a magic wand that you wave at every problem that happens to involve text. — Andy Lester
– Andy Lester, Commented Sep 5, 2013 at 12:37

Gumbo · Accepted Answer · 2010-03-17 16:57:59Z

5

If you want to parse PHP, you can use token_get_all to get the tokens of the given code.

answered Mar 17, 2010 at 16:57

Gumbo

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1 Comment

JonnyLitt Over a year ago

That is brilliant and simple. Cheers mate.

St. John Johnson · Accepted Answer · 2010-03-17 17:22:51Z

1

You could also look into the Reflection class.

answered Mar 17, 2010 at 17:22

St. John Johnson

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Comments

Andy · Accepted Answer · 2013-09-05 09:57:27Z

0

So I'm not sure in php, I'm doing something similar in java and the following code strips it in a similar way:

[^(&&[^)&&[^,]]] applied to foo(param1, param2, param3) will return ["foo", "param1", "param2", "param3"].

The problem I'm having is that this will string out all '(' and ')', so there is no way to tell if your param is actually a variable or a method.

answered Sep 5, 2013 at 9:57

Andy

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