Using a variable in a regex pattern in PHP

Question

I have this regex:

'/^files\/(.+\..+)$/';

But I want to replace the "files/" section with a runtime variable.

I tried this:

$filePath = "files/";
'/^'.$filePath.'(.+\..+)$/';

which didn't work, as well as this:

$filePath = preg_quote("files/");
'/^'.$filePath.'(.+\..+)$/';

but I still get an error in this loop on the preg_match line, saying that "(" is an unknown modifier.

foreach (glob(FILE_PATH."*.*") as $filename) {
    preg_match($pattern, $filename, $matches);
    echo "<option class='file' value='".$matches[1]."'>".$matches[1]."</option>";
}

Any help would be appreciated, thanks.

EDIT

This works...

$filePath = 'files\/';

...but in actual usage, I'm trying to use a constant declared in another file.

So:

define('FILE_PATH', 'files/');

...and then trying to use that constant. It can't have the escape character embedded because some other parts of the application need it without the escape character.

Does $filePath contain slashes (/)? That conflicts with the regex delimiters you use. You should always regex-escape content you inject into expressions. — arkascha
– arkascha, Commented Nov 25, 2012 at 14:46
@arkascha he does (that's what preg_quote is for). he is just missing a parameter ;) — Martin Ender
– Martin Ender, Commented Nov 25, 2012 at 14:48

Martin Ender · Accepted Answer · 2012-11-25 14:46:32Z

3

You need to hand preg_quote the delimiter:

$filePath = preg_quote("files/", '/');

Since you can use a lot of characters as delimiters, preg_quote cannot know, that you are going to use /, so it does not escape it by default. That's what the second (optional) parameter is for.

Otherwise, your second approach (the one using preg_quote) is the way to go.

answered Nov 25, 2012 at 14:46

Martin Ender

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1 Comment

ARW Over a year ago

Perfect! Don't know how I missed that, thank you very much! Will accept the answer as soon as it lets me!

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