preg_replace_callback returns duplicate values

Question

I'm trying to understand the preg_replace_callback function. Inside my callback, I am getting the matched string twice instead of the just one time it appears in the subject string. You can go here and copy/paste this code into the tester and see that it returns two values when I would expect one.

preg_replace_callback(
    '"\b(http(s)?://\S+)"', 
    function($match){           
        var_dump($match);
        die();
    },
    'http://a'
);

The output looks like this:

array(2) { [0]=> string(8) "http://a" [1]=> string(8) "http://a" }

The documentation mentions returning an array if the subject is an array, or a string otherwise. What's happening?

AbraCadaver · Accepted Answer · 2015-05-13 17:01:04Z

3

You have a full pattern match \b(http(s)?://\S+) in $match[0] and a match for the parenthesized capture group (http(s)?://\S+) in $match[1].

In this case just use $match[0] something like:

$result = preg_replace_callback(
    '"\b(http(s)?://\S+)"', 
    function($match){           
        return $match[0] . '/something.php';
    },
    'http://a'
);

Replaces http://a with http://a/something.php.

edited May 13, 2015 at 17:01

answered May 13, 2015 at 16:55

AbraCadaver

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3 Comments

smilebomb Over a year ago

Why is it that s is not also returned as a parenthesized capture group?

Jonathan Kuhn Over a year ago

your subject string doesn't an s and you have a ? making the group optional. If you moved the ? inside of the parenthesis, the group would always exist, but be blank or s depending on whether it is there.

smilebomb Over a year ago

Oh, I'm swapping between my own code and the example code here. My mistake.

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