2011-11-28 74 views
3

刪除元素讓我們說,我想通過Ajax加載該文件:從Ajax響應

<!-- loadme.html --> 

<div class='content'> 
    Hello ! 
    <div class='removeme'>Remove me, please.</div> 
</div> 

我怎樣才能得到只有Hello內容?我試圖倍數的方法來去除.removeme DIV,它總是失敗:

$.ajax({ 
    url: 'loadme.html', 
    success: function(data) { 
     var response = $('<div />').html(data); 

     // First try : 
     var content1 = response.find('.content').html() 
     console.log(content1); // Return : Hello ! <div class="removeme">Remove me, please.</div> 

     // Second Try : 
     var content2 = response.find('.content').remove('.removeme').html() 
     console.log(content2); // Return : Hello ! <div class="removeme">Remove me, please.</div> 

     // Third Try : 
     var content3 = response.find('.content').html(); 
     console.log($(content3).remove('.removeme').html()); // Return : Remove me, please 
    } 
}); 

回答

7

嘗試:

var temp = response.find('.content'); 
temp.children('.removeme').remove(); 
var content4 = temp.html();