IE浏览器下:
<img style="width: 1024px; height: 806px;" alt="" src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" complete="complete" data-cke-saved-src="/upload/articles/20121204/images/201212040328205506018siw5.jpg"/>FF浏览器下:
<img style="width: 1024px; height: 806px;" src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" complete="complete" data-cke-saved-src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" alt="" />这个是用编辑器上传的图片,现在我需要把图片里面的style全部删除掉,后面的内容保持原样。求高手帮我写个正则!!

解决方案 »

  1.   


    $str='<img style="width: 1024px; height: 806px;" src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" complete="complete" data-cke-saved-src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" alt="" />';
    $aa=preg_replace('/style=".*?"\s+/','',$str);
    echo $aa;
      

  2.   

    $str='<img style="width: 1024px; height: 806px;" alt="" src="/upload/articles/20121204/images/201212040328205506018siw5.jpg" complete="complete" data-cke-saved-src="/upload/articles/20121204/images/201212040328205506018siw5.jpg"/>';echo preg_replace('/(<img\s+[^>]*)style\s*=\s*"[^"]*"/i', '$1', $str);
      

  3.   

    #1注意:只过滤<img内的style,你那样,如果外面有style代码也被过滤掉了。