PHP regex to extract first part of file name in URL

Question

Possible Duplicate:
How can I convert ereg expressions to preg in PHP?

I'm working on a PHP script.

I have a URL of the form:

http://ecx.images-amazon.com/images/I/5104Xl51zFL._SL175_.jpg

and I simply want to grab the first part of the file name 5104Xl51zFL.

I'm pretty new to regexps but so far I have:

.*images\/I\/(.+?)(\.[^.]*|$)

which according to regextester.com should work but doesn't in my PHP.

Doesn't have to be a regexp if it's not the best solution.

If it's relevant here's my PHP (still debuggy):

function linkExtractor($html)
{
    if(preg_match_all('/<img ([^>]* )?src=[\"\']([^\"\']*\._SL175_\.jpe?g)[\"\']/Ui', $html, $matches, PREG_SET_ORDER)){
        foreach ($matches as $match) {
            $url = $match[2];
            echo "\n\n" .$url . "\nfile name: ";
            if(preg_match_all('.*images\/I\/(.+?)(\.[^.]*|$)', $url, $matched, PREG_SET_ORDER)) {
                foreach($matched as $name) {
                    print_r($matched);
                }
            }
        }
    }
}

score 8 · Accepted Answer · edited Nov 23 '12 at 23:09

8

$url = 'http://ecx.images-amazon.com/images/I/5104Xl51zFL._SL175_.jpg';
$path = parse_url($url, PHP_URL_PATH);
$filename = basename($path);
$partOne = strtok($filename, '.');

edited Nov 23 '12 at 23:09

salathe

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answered Nov 23 '12 at 23:04

Sergey Eremin

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perfect - was looking for a way to do this without regexps anyway – Max Bates Nov 23 '12 at 23:14
wouldn't that give you "com/images/I/5104Xl51zFL" not "5104Xl51zFL"? – Mike Nov 23 '12 at 23:17
Oh I see how I was wrong now. – Mike Nov 23 '12 at 23:22

score 1 · Answer 2 · answered Nov 23 '12 at 23:08

1

This will match everything between /images/I/ and .

$url = 'http://ecx.images-amazon.com/images/I/5104Xl51zFL._SL175_.jpg';
if(preg_match('~/images/I/(.+?)\.~',$url,$match)) {
    $name = $match[1];
}

answered Nov 23 '12 at 23:08

lafor

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Chris Seymour · Answer 3 · 2012-11-23T23:14:59.333

1

One way using regex:

$str='http://ecx.images-amazon.com/images/I/5104Xl51zFL._SL175_.jpg';
preg_match('/.*\/(.*?)\./',$str,$match);
print $match[1];

>>> 5104Xl51zFL

Explanation:

.*    # Match anything
\/    # Up to the last forwardslash (escaped)
(.*?) # Match anything after (captured, lazy)
\.    # Up to the first . (escaped)

edited Nov 23 '12 at 23:14

answered Nov 23 '12 at 23:09

Chris Seymour

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Francis Gagnon · Answer 4 · 2012-11-23T23:14:53.940

1

If you're still looking for a regular expression to extract that part of the filename here is one:

([^/.]++)[^/]+$

This expression will match the full filename. Capture group 1 will contain every character in the filename that is before the first dot.

The ++ makes this regex faster than with only one +. This is because when a piece of text fails to match, the regex will fail more quickly.

edited Nov 23 '12 at 23:14

answered Nov 23 '12 at 23:09

Francis Gagnon

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Koval · Answer 5 · 2012-11-23T23:12:50.313

0

$file = basename("http://ecx.images-amazon.com/images/I/5104Xl51zFL._SL175_.jpg");
echo substr($file, 0, strpos($file, "."));

strpos is much faster than using regular expressions for something that simple

edited Nov 23 '12 at 23:12

answered Nov 23 '12 at 23:07

Koval

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PHP regex to extract first part of file name in URL

5 Answers5