问题描述
当我同时使用fadeOut
和replaceWith
时,不起作用fadeOut.但是,如果我只使用fadeOut
,它就可以工作.我想一起使用它们的每个拖曳.在我的代码中如何?
When i use of fadeOut
and replaceWith
together, does not work fadeOut. but if i use only of fadeOut
it worked. i want use of each tow they together. how is it in my code?
$.ajax({
type: "POST",
url: url,
data: dataString,
cache: false,
success: function(html){
var $html = $(html);
////////////////////////////////// here ////////////////////////////////
$('.ser_form #paginate input:checkbox:checked').parent().parent().fadeOut("slow");
$('#num_count').replaceWith($html.find('#num_count'));
$('tr#paginate').replaceWith($html.find('tr#paginate'));
$('.pagination').replaceWith($html.find('.pagination'));
////////////////////////////////// here ////////////////////////////////
},
示例::您自己看到的
在上面的示例中,请检查行,然后单击DELETE进行查看.
EXAMPLE: Yourself see
In example above please check row and click on DELETE to see it.
推荐答案
在这里,我尝试模拟您的情况 http://jsfiddle.net/yY2AS/1/
Here i tried to simulate your scenariohttp://jsfiddle.net/yY2AS/1/
无论如何,您的问题对我来说不是很清楚...
your question was not very clear to me, anyway...
在发出"ajax请求"之前
before makeing the `ajax request do
$this = $('.ser_form #paginate input:checkbox:checked').parent().parent();
然后进行ajax调用(我猜ajax调用是为了删除和更新记录)
then make the ajax call(i am guessing that the ajax call is to delete and update record)
在成功回拨中做
success: function(html){
$this.fadeOut("slow");
//rest of the code
这篇关于不要工作"fadeOut",如果有"replaceWith".的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!