$re = '~(?s).*\b(http://www\.blooming\.com\b.*?\.jpg\b).*~';
$str = '<div>Colorful Floor chair Series</div><div><br /></div><div>Soft
Suede</div><div><br /></div><div>Cute bubble design</div><div><br
/></div><div><p align="center"><p align="center"><img
src="http://gdetail.image-gemkt.com/186/716088198/2010/2/e3b117e2-a7bd-4d.GIF"
/></div><div><p align="center"><p align="center"><img
src="http://www.blooming.com/image/xxxxxxxx.jpg" /></div>';
$subst = "\1";
$result = preg_replace($re, $subst, $str, 1);
echo "The result of the substitution is ".$result;
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for PHP, please visit: http://php.net/manual/en/ref.pcre.php