how to select image src using PHP

Question

So i have some images with several forms like this :

<a href="" class="link-img" alt="">
  <img editable="true" style="display: block; cursor: default;" class="main-image" 
    width="538" height="auto" src="src" alt="large image">
</a>

and like this :

<a href="" class="link link-img">
 <img src="src" style="width: 100%; display: block; cursor: pointer;" editable="true" 
  class="main-image imageLink" width="" height="auto" alt="">
</a>

so my code to select the src image is :

$c = preg_replace('/<a href="(.+)" class="link link-img" alt="(.+)"><img src="(.+)"><\/a>/i'
,'<% link url="$1" caption="<img style=max-width:500px; src=$8 >" html="true" %>',$c);

i tried that several times but the code doesn't worked, so please if someone has any idea i will be very appreciative.

To extract data from HTML, use an HTML parser and not some regexp. — Sirko
– Sirko, Commented Jan 16, 2015 at 16:43
Good thing PHPDomDocument isn't for use on the client side then — castis
– castis, Commented Jan 16, 2015 at 16:53

A l w a y s S u n n y · Accepted Answer · 2015-01-16 17:02:22Z

2

Try this way to grab src from image src="([^"]+)"

enter image description here

EDIT : see regex here https://www.regex101.com/r/yF8tJ1/1

CODE EXAMPLE:

$re = "/src=\"([^\"]+)\"/"; 
$str = "<a href=\"\" class=\"link-img\" alt=\"\">\n  <img editable=\"true\" style=\"display: block; cursor: default;\" class=\"main-image\" \n    width=\"538\" height=\"auto\" src=\"src\" alt=\"large image\">\n</a>\n\n<a href=\"\" class=\"link link-img\">\n <img src=\"src\" style=\"width: 100%; display: block; cursor: pointer;\" editable=\"true\" \n  class=\"main-image imageLink\" width=\"\" height=\"auto\" alt=\"\">\n</a>"; 

preg_match_all($re, $str, $matches);

edited Jan 16, 2015 at 17:02

answered Jan 16, 2015 at 16:55

A l w a y s S u n n y

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4 Comments

Ismael Miguel Over a year ago

I suggest a different regex. /src=((['"]?)(?:[^\2]|\2+)*\2)/g. This one matches broken HTML attributes as close as I can get to the behavior of HTML/XML parsers.

Ismael Miguel Over a year ago

Actually, use /src=((['"]?)(?:[^\2]|\2+)+?\2/gi. This one works properly. An input like <img src="test"> will match src="test", <img src="test"" > will match src="test"" and <img src='test""/><span>My new 'test'</span><img src='55'> will match src='test""/><span>My new ' and src='55.

Ismael Miguel Over a year ago

Sorry for flooding this! I've made a new regex: src=([^"'>]+|(['"]?)(?:[^\2]|\2+)+?\2). This one works now with <img src=55>, which matches src=55.

Ismael Miguel Over a year ago

Just trying to make the impossible. Some people would get really angry with such regex. But hey, it works!

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