jQuery Jquery检查ajax帖子是否成功
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Jquery checking success of ajax post
提问by zsharp
how do i define the success and failure function of an ajax $.post?
我如何定义 ajax $.post 的成功和失败功能？
回答by Jere.Jones
The documentation is here: http://docs.jquery.com/Ajax/jQuery.ajax
文档在这里：http: //docs.jquery.com/Ajax/jQuery.ajax
But, to summarize, the ajax call takes a bunch of options. the ones you are looking for are error and success.
但是，总而言之，ajax 调用需要很多选项。您正在寻找的是错误和成功。
You would call it like this:
你会这样称呼它：
$.ajax({
url: 'mypage.html',
success: function(){
alert('success');
},
error: function(){
alert('failure');
}
});
I have shown the success and error function taking no arguments, but they can receive arguments.
我已经展示了不带参数的成功和错误函数，但它们可以接收参数。
The error function can take three arguments: XMLHttpRequest, textStatus, and errorThrown.
error 函数可以接受三个参数：XMLHttpRequest、textStatus 和 errorThrown。
The success function can take two arguments: data and textStatus. The page you requested will be in the data argument.
成功函数可以接受两个参数：data 和 textStatus。您请求的页面将在数据参数中。
回答by Matthew Crumley
If you need a failure function, you can't use the $.get or $.post functions; you will need to call the $.ajax function directly. You pass an options object that can have "success" and "error" callbacks.
如果需要失败函数，则不能使用 $.get 或 $.post 函数；您需要直接调用 $.ajax 函数。您传递一个可以具有“成功”和“错误”回调的选项对象。
Instead of this:
取而代之的是：
$.post("/post/url.php", parameters, successFunction);
you would use this:
你会用这个：
$.ajax({
url: "/post/url.php",
type: "POST",
data: parameters,
success: successFunction,
error: errorFunction
});
There are lots of other options available too. The documentationlists all the options available.
还有很多其他选项可用。该文档列出了所有可用的选项。
回答by Li3ro
using jQuery 1.8 and above, should use the following:
使用 jQuery 1.8 及更高版本，应使用以下内容：
var request = $.ajax({
type: 'POST',
url: 'mmm.php',
data: { abc: "abcdefghijklmnopqrstuvwxyz" } })
.done(function(data) { alert("success"+data.slice(0, 100)); })
.fail(function() { alert("error"); })
.always(function() { alert("complete"); });
check out the docs as @hitautodestruct stated.
查看@hitautodestruct 所述的文档。
回答by yashpal
I was wondering, why they didnt provide in jquery itself, so i made a few changes in jquery file ,,, here are the changed code block:
我想知道，为什么他们没有在 jquery 本身中提供，所以我在 jquery 文件中做了一些更改，，这里是更改后的代码块：
original Code block:
原始代码块：
post: function( url, data, callback, type ) {
// shift arguments if data argument was omited
if ( jQuery.isFunction( data ) ) {
type = type || callback;
callback = data;
data = {};
}
return jQuery.ajax({
type: "POST",
url: url,
data: data,
success: callback,
dataType: type
});
Changed Code block:
更改代码块：
post: function (url, data, callback, failcallback, type) {
if (type === undefined || type === null) {
if (!jQuery.isFunction(failcallback)) {
type=failcallback
}
else if (!jQuery.isFunction(callback)) {
type = callback
}
}
if (jQuery.isFunction(data) && jQuery.isFunction(callback)) {
failcallback = callback;
}
// shift arguments if data argument was omited
if (jQuery.isFunction(data)) {
type = type || callback;
callback = data;
data = {};
}
return jQuery.ajax({
type: "POST",
url: url,
data: data,
success: callback,
error:failcallback,
dataType: type
});
},
This should help the one trying to catch error on $.Post in jquery.
这应该有助于试图在 jquery 中捕获 $.Post 上的错误的人。
Updated: Or there is another way to do this is :
更新：或者还有另一种方法来做到这一点：
$.post(url,{},function(res){
//To do write if call is successful
}).error(function(p1,p2,p3){
//To do Write if call is failed
});
回答by Veenkar
This style is also possible:
这种风格也是可能的：
$.get("mypage.html")
.done(function(result){
alert("done. read "+result.length+" characters.");
})
.fail(function(jqXHR, textStatus, errorThrown){
alert("fail. status: "+textStatus);
})
