Second parameter in preg_replace_callback()

问题

I have a problem with the function preg_replace_callback() in PHP. I want to call a function which requires two parameters.

private function parse_variable_array($a, $b)
{
    return $a * $b;
}

On the internet I found this piece of code:

preg_replace_callback("/regexcode/", call_user_func_array(array($this, "foo"), array($foo, $bar)), $subject);

But in the function foo I cannot use the matches array that is usual with a preg_replace_callback

I hope you can help me!

回答1:

The callback is called as is, you cannot pass additional parameters to it. You can make a simple wrapper function though. For PHP 5.3+, that's easily done with anonymous functions:

preg_replace_callback(..., function ($match) {
    return parse_variable_array($match, 42);
}, ...);

For older PHP versions, make a regular function that you pass as usual as the callback.

来源：https://stackoverflow.com/questions/8735565/second-parameter-in-preg-replace-callback