How to extract image names with PHP or regex?

Question

I have the following data in MySQL.

<p><img src="../../../../assets/images/frontpage/image1.png" 
alt="" width="790" height="356" /></p>

Now I want to get image2.png with PHP or regex. The extension can be gif or jpg. And a length of image name can be any length.

Worth noting is whether you want to be able to match this on multiple img-tags or just one. — rzetterberg, Jul 03 '11 at 17:04
http://stackoverflow.com/questions/3650125/how-to-parse-html-with-php — derobert, Jul 03 '11 at 17:08
possible duplicate of [Regular expression in PHP to return array with all images from html, eg: all src="images/header.jpg" instances](http://stackoverflow.com/questions/5721025/regular-expression-in-php-to-return-array-with-all-images-from-html-eg-all-src) — mario, Jul 03 '11 at 17:08
And, of course, http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags/1732454#1732454 — derobert, Jul 03 '11 at 17:11
"I want to get image2.png with PHP or regex" doesn't really tell us what you're trying to do. "Get" image2.png how? Your example has "image1.png". — Lightness Races in Orbit, Jul 03 '11 at 20:44

Justin Aquadro · Accepted Answer · 2011-07-03T17:11:17.700

3

This would match a path in an img tag and capture the file in the first interior capturing group.

<?php
if (preg_match('%<img\s.*?src=".*?/?([^/]+?(\.gif|\.png|\.jpg))"%s', $subject, $regs)) {
    $image = $regs[1];
} else {
    $image = "";
}
?>

edited Jul 03 '11 at 17:11

answered Jul 03 '11 at 17:01

Justin Aquadro

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This outputs, ../../../assets/images/frontpage/image1.png – shin Jul 03 '11 at 17:07
I needed to change a capture-all to a group. I think it should work now. – Justin Aquadro Jul 03 '11 at 17:12
I added the following to the above code and it works. $path = explode('/', $image); $image_name = end($path); return $image_name; – shin Jul 03 '11 at 17:16

EdoDodo · Answer 2 · 2011-07-03T17:05:35.557

0

This code should do what you need:

<?php
$regex = '@src[ ]*=[ ]*"[a-z/.]*/(.*?\.(?:png|gif|jpg))@i"';
$match = array();
if (preg_match($regex, $html, $match)) {
    $imglocation = $match[1];
} else {
    die('Failed to find image name.');
}

edited Jul 03 '11 at 17:05

answered Jul 03 '11 at 17:00

EdoDodo

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$html = '
'; $regex = 'src[ ]*=[ ]*"[a-zA-Z/.]*/(.*?\.(?:png|gif|jpg))"'; $match = array(); if (preg_match($regex, $html, $match)) { $imglocation = $match[1]; } else { die('Failed to find image name.'); }. This gives the following error. Warning: preg_match() [function.preg-match]: Delimiter must not be alphanumeric or backslash in C:\xampp\htdocs\test\test1.php on line 19 Failed to find image name. – shin Jul 03 '11 at 17:02
1

Whoops, I forgot to include a delimiter. Added now. – EdoDodo Jul 03 '11 at 17:05

score 0 · Answer 3 · answered Jul 03 '11 at 17:57

0

Try this:

$str = '<p><img src="../../../../assets/images/frontpage/image1.png" 
alt="" width="790" height="356" /></p>'; 
$imageType = end(explode("."));

answered Jul 03 '11 at 17:57

The Mask

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score -2 · Answer 4 · edited Jul 03 '11 at 20:29

-2

Use this expression: var expr=/[a-z]+\.(gif|jpg)$/;

edited Jul 03 '11 at 20:29

Jon

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answered Jul 03 '11 at 17:02

Ankur

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this will return an array so you can get image1.jpg getting by 0th index of array :) – Ankur Jul 03 '11 at 17:11
@Tomalak it looks like Javascript to me. – James Khoury Jul 04 '11 at 02:10

How to extract image names with PHP or regex?

4 Answers4