我应该将.done()和.fail()用于新的jQuery AJAX代码，而不是成功和错误

问题描述

我这样编码:

$.ajax({ cache: false,
    url: "/Admin/Contents/GetData",
    data: { accountID: AccountID },
    success: function (data) {
        $('#CityID').html(data);
    },
    error: function (ajaxContext) {
        alert(ajaxContext.responseText)
    }
});

但是，当我在最后查看jQuery .ajax()文档时，似乎建议我应该在下面进行这样的编码，或者至少建议添加.done()和.fail():

But when I look at the jQuery .ajax() documentation at the end it seems to suggest I should be coding like this below or at least it suggests adding a .done() and a .fail():

var request = $.ajax({ cache: false,
    url: "/Admin/Contents/GetData",
    data: { accountID: AccountID }
});

request.done(function (data) {
    xxx;
});
request.fail(function (jqXHR, textStatus) {
    xxx;
});

更新

如果我这样编写代码，将其分为三个，还是有一些优势?

If I code like this is it the same or is there some advantage to breaking it into three ?

$.ajax({ cache: false,
    url: "/Admin/Contents/GetData",
    data: { accountID: AccountID }
}).done(function (data) {
    xxx;
}).fail(function (jqXHR, textStatus) {
    xxx;
});

推荐答案

如user2246674所述，使用success和error作为ajax函数的参数是有效的.

As stated by user2246674, using success and error as parameter of the ajax function is valid.

要与先前的答案保持一致，请阅读doc:

To be consistent with precedent answer, reading the doc :

jqXHR.success()，jqXHR.error()和jqXHR.complete()回调将在jQuery 1.8中弃用.要准备将其最终删除的代码，请改用jqXHR.done()，jqXHR.fail()和jqXHR.always().

The jqXHR.success(), jqXHR.error(), and jqXHR.complete() callbacks will be deprecated in jQuery 1.8. To prepare your code for their eventual removal, use jqXHR.done(), jqXHR.fail(), and jqXHR.always() instead.

如果使用回调操作功能(例如，使用方法链接)，请使用.done()，.fail()和.always()代替success()，error()和complete().

If you are using the callback-manipulation function (using method-chaining for example), use .done(), .fail() and .always() instead of success(), error() and complete().

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