获取 TypeError: $.ajax(...).done 不是函数 [Ajax, Jquery]
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Getting TypeError: $.ajax(...).done is not a function [Ajax, Jquery ]
提问by Shailendra Kumar
In my Jquery, i am using Ajax and getting below error msg.
在我的 Jquery 中，我使用 Ajax 并收到以下错误消息。
TypeError: $.ajax(...).done is not a function
[Break On This Error] ).success(function(response) {
I tired using success instead of done. but still getting same msg.
我厌倦了使用成功而不是完成。但仍然收到相同的消息。
TypeError: $.ajax(...).success is not a function
[Break On This Error] ).success(function(response) {
sample piece of code is mentioned below:
下面提到了示例代码：
$(document).ready(function () {
alert('in get');
$.ajax({
data: {
'contentId': contentId,
'USER_ID': USER_ID,
'actionType': 'GETRATING',
'portletGuid': portletGuid
},
type: 'GET',
url: ajaxRatingServlet,
cache: false
}).success(function (response) {
getUserPreference(response);
});
回答by Konsole
Replace your successwith doneor use success inside ajax function.
在 ajax 函数中替换您success的done或使用成功。
An alternative construct to the success callback option, the .done()method replaces the deprecated jqXHR.success() method.
成功回调选项的替代构造，.done()方法替换了已弃用的 jqXHR.success() 方法。
EG
例如
$(document).ready(function () {
$.ajax({
data: {
'contentId': contentId,
'USER_ID': USER_ID,
'actionType': 'GETRATING',
'portletGuid': portletGuid
},
type: 'GET',
url: ajaxRatingServlet,
cache: false
}).done(function (response) {
console.log(response);
});
//or use success inside ajax as other answered
$(document).ready(function() {
alert('in get');
$.ajax({
data: { 'contentId':contentId, 'USER_ID':USER_ID, 'actionType':'GETRATING', 'portletGuid':portletGuid },
type:'GET',
url:ajaxRatingServlet,
cache:false,
success: function(response) {
getUserPreference(response);
}
});
});
回答by Mujtaba Haider
try using success function inside ajax function ,
尝试在 ajax 函数中使用成功函数，
$(document).ready(function() {
alert('in get');
$.ajax({
data: { 'contentId':contentId, 'USER_ID':USER_ID, 'actionType':'GETRATING', 'portletGuid':portletGuid },
type:'GET',
url:ajaxRatingServlet,
cache:false,
success: function(response) {
getUserPreference(response);
}
});
});
回答by Shakti Patel
You can used this Demo
你可以使用这个Demo
There are some extra function which help you
有一些额外的功能可以帮助你
$(
function(){
// Get a reference to the content div (into which we will load content).
var jContent = $( "#content" );
// Hook up link click events to load content.
$( "a" ).click(
function( objEvent ){
var jLink = $( this );
// Clear status list.
$( "#ajax-status" ).empty();
// Launch AJAX request.
$.ajax(
{
// The link we are accessing.
url: jLink.attr( "href" ),
// The type of request.
type: "get",
// The type of data that is getting returned.
dataType: "html",
error: function(){
ShowStatus( "AJAX - error()" );
// Load the content in to the page.
jContent.html( "<p>Page Not Found!!</p>" );
},
beforeSend: function(){
ShowStatus( "AJAX - beforeSend()" );
},
complete: function(){
ShowStatus( "AJAX - complete()" );
},
success: function( strData ){
ShowStatus( "AJAX - success()" );
// Load the content in to the page.
jContent.html( strData );
}
}
);
// Prevent default click.
return( false );
}
);
}
);
回答by Aaron Digulla
successand erroraren't attributes of the object returned by the call to $.ajax(). Instead, you must pass them as configs in the call:
success并且error不是调用返回的对象的属性$.ajax()。相反，您必须在调用中将它们作为配置传递：
$.ajax({..., success: function(data){}})
