Regex & PHP - isolate src attribute from img tag

Question

With PHP, how can I isolate the contents of the src attribute from $foo? The end result I\'m looking for would give me just \"http://example.com/img/image.jpg\"

Answer 1

Code

<?php
    $foo = '<img class="foo bar test" title="test image" src="http://example.com/img/image.jpg" alt="test image" width="100" height="100" />';
    $array = array();
    preg_match( '/src="([^"]*)"/i', $foo, $array ) ;
    print_r( $array[1] ) ;

Output

http://example.com/img/image.jpg

Answer 2

lets assume i use

$text ='<img src="blabla.jpg" alt="blabla" />';

in

getTextBetween('src="','"',$text);

the codes will return :

blabla.jpg" alt="blabla" 

which is wrong, we want the codes to return the text between the attribute value quotes i.e attr = "value".

so

  function getTextBetween($start, $end, $text)
            {
                // explode the start string
                $first_strip= end(explode($start,$text,2));

                // explode the end string
                $final_strip = explode($end,$first_strip)[0];
                return $final_strip;
            }

does the trick!.

Try

   getTextBetween('src="','"',$text);

will return:

blabla.jpg

Thanks all the same , because your solution gave me an insight to the final solution .

Answer 3

try this pattern:

'/< \s* img [^\>]* src \s* = \s* [\""\']? ( [^\""\'\s>]* )/'

Answer 4

preg_match solves this problem nicely.

See my answer here: How to extract img src, title and alt from html using php?