Regex to extract file extension from URL

Question

I am looking for a regex that would match .js in the following URI:

 /foo/bar/file.js?cache_key=123

I'm writing a function that tries to identify what kind of file is being passed in as a parameter. In this case, the file ends with the extension .js and is a javascript file. I'm working with PHP and preg_match so I'm assuming this is a PCRE compatible regular expression. Ultimately I'll build on this expression and be able to check for multiple file types that are being passed in as a URI that isn't just limited to js, but perhaps css, images, etc.

i alarmed alien · Accepted Answer · 2014-10-30 23:38:50Z

6

You can use a combination of pathinfo and a regular expression. pathinfo will give you the extension plus the ?cache_key=123, and you can then remove the ?cache_key=123 with a regex that matches the ? and everything after it:

$url = '/foo/bar/file.js?cache_key=123';

echo preg_replace("#\?.*#", "", pathinfo($url, PATHINFO_EXTENSION)) . "\n";

Output:

js

Input:

$url = 'my_style.css?cache_key=123';

Output:

css

Obviously, if you need the ., it's trivial to add it to the file extension string.

ETA: if you do want a regex solution, this will do the trick:

function parseurl($url) {
    # takes the last dot it can find and grabs the text after it
    echo preg_replace("#(.+)?\.(\w+)(\?.+)?#", "$2", $url) . "\n";
}

parseurl('my_style.css');
parseurl('my_style.css?cache=123');
parseurl('/foo/bar/file.js?cache_key=123');
parseurl('/my.dir.name/has/dots/boo.html?cache=123');

Output:

css
css
js
html

edited Oct 30, 2014 at 23:38

answered Oct 30, 2014 at 23:18

i alarmed alien

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Comments

Croises · Accepted Answer · 2014-10-30 23:19:18Z

1

Use:

.+\.(js|css|etc)[?]?

extension in $matches[1]

Or you can just use

.+\.(js|css|etc)\?

if the final ?cache... is always used

edited Oct 30, 2014 at 23:19

answered Oct 30, 2014 at 23:14

Croises

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Comments

Jens A. Koch · Accepted Answer · 2014-10-30 23:35:51Z

0

DEMO

Code

$input_line = '/foo/bar/file.js?cache_key=123';

// lets grab the part part between filename and ?
preg_match("/\w+\/\w+\/\w+(.*)\?/", $input_line, $output_array);

var_dump($matches);

echo $matches[0];

Output

Array
(
   [0] => foo/bar/file.js?
   [1] => .js
)

.js

If you know the extensions beforehand (whitelist approach), you might switch from matching everything (.*) to matching specific extensions /.*\.(js|jpg|jpeg|png|gif)/

preg_match("/.*\.(js|jpg|jpeg|png|gif)/", $input_line, $matches);
echo $matches[1]; // js

edited Oct 30, 2014 at 23:35

answered Oct 30, 2014 at 23:25

Jens A. Koch

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3 Comments

i alarmed alien Over a year ago

This solution is rather dependent upon there being a set number of directories...

Jens A. Koch Over a year ago

true! structure is /word/word/word(match)? . maybe it's sufficient. who knows. there is always pathinfo().

i alarmed alien Over a year ago

You could make the number of directories flexible using (\/\w+)* -- matches 0 or more /\w+

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