Ask for a regular, get the address of the image src

Regular takes the value of SRC in the above code
Image/ad1.gif
Image/ad2.gif
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Regular takes the value of SRC in the above code
Image/ad1.gif
Image/ad2.gif
Image/ad3.gif
Image/ad4.gif
Image/ad5.gif

Get src value str, then Str.match (/image (s*) gif/) [0]

$('img').each(function(index,item){    console.log(item.src.match(/image(\S*)gif/)[0]);})

/src=(?:pattern'|\"|\\\['"]|\\\)?(.*?)(?:pattern'|\"|\\\['"]|\\\)?\sw/

I don't know, it's not right.

//如果你的src中肯定不包含空格的话可以用这个简单的办法，通常是ok的preg_match_all('/

这样得到的内容会包含单引号什么的，对于带有反斜线的单双引号先去掉反斜线，然后trim一下就行了，例如

foreach($_match[1] as str) { $str = str_replace("\\'", "'", $str); $str = str_replace('\\"', '"', $str); echo trim($str, '\'"');}

大致这样，具体的自己调试吧。不太复杂的正则匹配优先考虑关键字符来排除结果，比如本例中的空格及制表符。

http://tool.oschina.net/regex/

 ]*?src=\\? ["]? (.*?) ["]?\S[^>]*?>  

There should be a red envelope here

  $rule = ' # # ';  

You can try it: online regular expression generation tool

