Question

可能重复：
How to return the response from an AJAX call from a function?

我正在使用Jquery Ajax来调用服务来更新值。

function ChangePurpose(Vid, PurId) {
        var Success = false;
        $.ajax({
            type: "POST",
            url: "CHService.asmx/SavePurpose",
            dataType: "text",
            data: JSON.stringify({ Vid: Vid, PurpId: PurId }),
            contentType: "application/json; charset=utf-8",
            success: function (data) {
                Success = true;//doesnt goes here
            },
            error: function (textStatus, errorThrown) {
                Success = false;//doesnt goes here
            }

        });
        //done after here
        return Success;
    }

和服务：

[WebMethod]
        public string SavePurpose(int Vid, int PurpId)
        {
            try 
            {
                CHData.UpdatePurpose(Vid, PurpId);
                //List<IDName> abc = new List<IDName>();
                //abc.Add(new IDName { Name=1, value="Success" });
                return "Success";

            }
            catch (Exception ex)
            {
                throw new Exception(ex.Message);
            }
        }

该服务正在从AJAX成功调用。 价值也在改变。但在服务之后，成功或错误功能未被调用，在这种情况下，应该已经调用了成功，但它无效。

我使用了firebug并发现成功或错误函数被跳过并直接转到return Success;

似乎无法找到代码的问题。

提前致谢

更新添加async: false解决了问题

Answer 1

将您的代码更改为

function ChangePurpose(Vid, PurId) {
        var Success = false;
        $.ajax({
            type: "POST",
            url: "CHService.asmx/SavePurpose",
            dataType: "text",
            async:false,
            data: JSON.stringify({ Vid: Vid, PurpId: PurId }),
            contentType: "application/json; charset=utf-8",
            success: function (data) {
                Success = true;//doesnt goes here
            },
            error: function (textStatus, errorThrown) {
                Success = false;//doesnt goes here
            }

        });
        //done after here
        return Success;
    }

您只能从synchronous函数返回值。否则，您必须制作callback。

所以我刚刚将async:false,添加到您的ajax调用

<强>更新

jquery ajax调用默认是异步的。如此成功＆amp;当ajax加载完成时，将调用错误函数。但是你的return语句将在ajax调用开始后执行。

更好的方法

     // callbackfn is the pointer to any function that needs to be called
     function ChangePurpose(Vid, PurId, callbackfn) {
        var Success = false;
        $.ajax({
            type: "POST",
            url: "CHService.asmx/SavePurpose",
            dataType: "text",
            data: JSON.stringify({ Vid: Vid, PurpId: PurId }),
            contentType: "application/json; charset=utf-8",
            success: function (data) {
                callbackfn(data)
            },
            error: function (textStatus, errorThrown) {
                callbackfn("Error getting the data")
            }

        });
     } 


     function Callback(data)
     {
        alert(data);
     }

并将ajax称为

 // Callback is the callback-function that needs to be called when asynchronous call is complete
 ChangePurpose(Vid, PurId, Callback);

Answer 2

尝试将ajax调用封装到函数中，并将async选项设置为false。请注意，自jQuery 1.8以来不推荐使用此选项。

function foo() {
    var myajax = $.ajax({
        type: "POST",
        url: "CHService.asmx/SavePurpose",
        dataType: "text",
        data: JSON.stringify({ Vid: Vid, PurpId: PurId }),
        contentType: "application/json; charset=utf-8",
        async: false, //add this
    });
    return myajax.responseText;
}

您也可以这样做：

$.ajax({
    type: "POST",
    url: "CHService.asmx/SavePurpose",
    dataType: "text",
    data: JSON.stringify({ Vid: Vid, PurpId: PurId }),
    contentType: "application/json; charset=utf-8",
    async: false, //add this
}).done(function ( data ) {
        Success = true;
}).fail(function ( data ) {
       Success = false;
});

您可以阅读有关jqXHR jQuery Object

的更多信息

Jquery Ajax Call，不会调用Success或Error

2 个答案: