Jquery 从Ajax响应中删除元素
假设我想通过AJAX加载此文件:Jquery 从Ajax响应中删除元素,jquery,ajax,wrapper,Jquery,Ajax,Wrapper,假设我想通过AJAX加载此文件: <!-- loadme.html --> <div class='content'> Hello ! <div class='removeme'>Remove me, please.</div> </div> 尝试: $.ajax({ url: 'loadme.html', success: function(data) { var response
<!-- loadme.html -->
<div class='content'>
Hello !
<div class='removeme'>Remove me, please.</div>
</div>
尝试:
$.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
}
});
var temp = response.find('.content');
temp.children('.removeme').remove();
var content4 = temp.html();