$('xxxx').html() 把 DOM 节点转成字符串后,如何快速从字符串中剔除某个节点? - V2EX
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paloalto

$('xxxx').html() 把 DOM 节点转成字符串后,如何快速从字符串中剔除某个节点?

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  •   paloalto Jun 19, 2014 5880 views
    This topic created in 4330 days ago, the information mentioned may be changed or developed.
    现在我的做法是先 .clone(),然后 .remove()。

    var $cOntent= $('#content').clone();
    $content.find('.class-1, .class-2, .class-3').remove();
    var cOntent= $content.html();

    最终的 content 就是 剔除后某节点的字符串了。

    但是这么做有点笨,有没有更好的处理办法?
    7 replies    2014-06-19 10:10:31 +08:00
    lijsh
        1
    lijsh  
       Jun 19, 2014 via Android
    写个函数封装下就不笨了,哈哈
    yyfearth
        2
    yyfearth  
       Jun 19, 2014
    哪里笨了 不就这样么

    如果你只有字符串 那么也就是 (类似)
    var $cOntent= $(content_html_string);
    content.find('xxx').remove();
    var cOntent= content.html()

    如果你想用regex处理string那就更麻烦而且不靠普了
    yyfearth
        3
    yyfearth  
       Jun 19, 2014
    抱歉 第二行应该是 $content.find('.class-1, .class-2, .class-3').remove();
    icanfork
        4
    icanfork  
       Jun 19, 2014
    $(content_html_string).find('class').remove();
    switch
        5
    switch  
       Jun 19, 2014
    可以这样:
    $("#content").clone().find(".class-1, .class-2, .class-3").remove().end();
    switch
        6
    switch  
       Jun 19, 2014
    $("#content").clone().find(".class-1, .class-2, .class-3").remove().end().html()
    justfindu
        7
    justfindu  
       Jun 19, 2014
    你这个方法so 机智 ...
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