PHP Regex preg_match arguments from a specific function name

Question

Let's say I have

$funcName = 'FooBar';
$myFuncString = $funcName.'("Arg1, Arg2")';

How could I preg_match() so it returns "Arg1, Arg2" (with doubles quotes and without $funcName?

I have tried:

$pattern = '/(?:'.$funcName.'|\".*?)\"/';
preg_match($pattern, myFuncString, $matches);

It returns "Arg1, Arg2" even if $funcName does not match. Any suggestions?

So you want to match $funcName in the string? Your current regex has an or there. You also are using a non-capture group but I think you do want to capture part of this.. — chris85
– chris85, Commented Feb 26, 2016 at 2:28
Your code actually worked you just had a typo, eval.in/525841. myFuncString != $myFuncString. — chris85
– chris85, Commented Feb 26, 2016 at 2:39

Aziz Saleh · Accepted Answer · 2016-02-26 02:32:54Z

1

This should work:

$funcName = 'FooBar';
$myFuncString = $funcName.'("Arg1, Arg2")';

//$funcName = 'NooBar';
$pattern = '/'.$funcName.'\("(.*)"\)/';
$matches = preg_match($pattern, $myFuncString, $res);

echo $res[1];

I had to add $res to the preg_match which has the results and correct the pattern.

Demo: http://sandbox.onlinephpfunctions.com/code/6de9d1236b82339199a2c64e914b5ca5b80e4e77

edited Feb 26, 2016 at 2:32

answered Feb 26, 2016 at 2:22

Aziz Saleh

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3 Comments

Wistar Over a year ago

Thank you. It returns [[0] => FooBar("Arg1, Arg2") [1] => Arg1, Arg2], how can only get the arguments with doubles quotes ('"Arg1, Arg2"') and not the function name?

chris85 Over a year ago

You should make the .* not greedy. If there were ") later on this line it would match until the last occurrence.

Aziz Saleh Over a year ago

@Wistar You echo $res[1]. @chris85 that is a good suggestion, my code assumes that that there aren't anymore (). The only time I see that is in minified/encoded php files.

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