开发者

jQuery AJAX returned result

开发者 https://www.devze.com 2023-01-28 09:19 出处:网络
I\'m having an issue with an ajax call using jQuery.I am posting information to the server, and getting back data as expected.The type of data I am getting back is html.Using firebug, if I console.log

I'm having an issue with an ajax call using jQuery. I am posting information to the server, and getting back data as expected. The type of data I am getting back is html. Using firebug, if I console.log the data, it shows an object with all my开发者_如何学编程 tags. I want to manipulate the form of the returned data, but when I try to console.log the form, I get an empty object. What am I doing wrong? Here is my code:

$.post('add', {'ajax':true}, function(data){  
  var $data = $(data);  
  console.log($data.find('form'));  
});


Look at this example. It works as expected. Maybe your response is not good?


I usually prefer the following syntax for selecting elements out of an HTML response:

$.post('add', {'ajax':true}, function(data){
  var myform = $('form', data); 
  console.log(myform);  
});

The second argument to the $() method is used as the context in which to search.


The problem is with this line:

var $data = $(data);

If you remove it, you will have the html code returned by server as a normal string variable inside data.

0

精彩评论

暂无评论...
验证码 换一张
取 消