Need regular expression to replace the part of image source

I want to replace below image source http://www.fishingwithdon.com/wp-content/uploads/2006/06/drop_off_area.jpg

with http://www.test.com/test/uploads/2006/06/drop_off_area.jpg

Means I want to replace "http://www.fishingwithdon.com/wp-content/" with "http://www.test.com/test/"

But string I want to replace will not be same each time.

Any help will be appreciate开发者_运维技巧d.

Thanks,

Umesh Kulkarni

---

preg_replace ('/^(http:).*(\/)$/', 'http://www.test.com/test/', $src)

That should do the trick

---

$str = preg_replace(
    '~\bhttp://www\.fishingwithdon\.com/wp-content/~',
    'http://www.test.com/test/',
    $str
);

http://codepad.org/xTvLGXC8

But wouldn't a simple str_replace be sufficient in your case?

---

$html  = "<img src=\"http://www.fishingwithdon.com/wp-content/uploads/2006/06/drop_off_area.jpg\"/>";
preg_match_all('#<img[^>]*>#i', $html, $match);
if(count($match) > 0)
{
    foreach($match[0] as $image)
    { 
        $new = preg_replace('/http:\/\/www\.(.+)\.com\//', 'http://www.test.com/test/', $image);
        $html = str_replace($image, $new, $html);
    }
}
echo $html;

Can Modify accordingly to adjust for different url situations ie (non-www, non-.com, https vs http)

---

Are you sure that you need a regex for that? Wouldn't str_replace be sufficient? It even says:

If you don't need fancy replacing rules (like regular expressions), you should always use this function instead of preg_replace().

Example:

$replace_needle = 'http://www.fishingwithdon.com/wp-content/';
$replacement = 'http://www.test.com/test/';
$replace_haystack = 'http://www.fishingwithdon.com/wp-content/uploads/2006/06/drop_off_area.jpg';
$result = str_replace($replace_needle, $replacement, $replace_haystack);