PHP提取数据库内容中的图片地址并循环输出

/*
1 (?s) 代表 pattern.dotall,也就是匹配换行，允许 img里出现在多行
2 .*?代表非贪婪匹配任意字符，直到后面的条件出现
3 ?: 代表这个匹配但不被捕获，也就是不在结果出现 [\.gif|\.jpg] 是或者的意思
*/
$pattern="/<img.*?src=[\'|\"](.*?(?:[\.gif|\.jpg]))[\'|\"].*?[\/]?>/";
$str='<p style="padding: 0px; margin-top: 0px; margin-bottom: 0px; line-height: 200%;"><img border="0" src="upfiles/2009/07/1246430143_4.jpg" alt=""/></p><p style="padding: 0px; margin-top: 0px; margin-bottom: 0px; line-height: 200%;"><img border="0" src="upfiles/2009/07/1246430143_3.jpg" alt=""/></p><p style="padding: 0px; margin-top: 0px; margin-bottom: 0px; line-height: 200%;"><img border="0" src="upfiles/2009/07/1246430143_1.jpg" alt=""/></p>';
preg_match_all($pattern,$str,$match);
print_r($match);
/*
array
(
[0] => array
(
[0] => <img border="0" src="upfiles/2009/07/1246430143_4.jpg" alt=""/>
[1] => <img border="0" src="upfiles/2009/07/1246430143_3.jpg" alt=""/>
[2] => <img border="0" src="upfiles/2009/07/1246430143_1.jpg" alt=""/>
)
[1] => array
(
[0] => upfiles/2009/07/1246430143_4.jpg
[1] => upfiles/2009/07/1246430143_3.jpg
[2] => upfiles/2009/07/1246430143_1.jpg
)
)
*/