Question

每次最终作为未定义的函数返回函数的响应!!

var result = Checkusers();(result is undefined) 

function CheckUser() {

    var EmpName = $("#txtName").val();

    $.ajax({        
        type: "POST",
        url: location.pathname + "/UserExist",
        data: "{Name:'" + EmpName + "'}",
        contentType: "application/json; charset=utf-8",
        datatype: "jsondata",
        async: "true",
        success: function (response) {
            console.log(response.d);
            var obj = eval('(' + response.d + ')');
            return obj;
        },
        error: function (response) {
            alert(response.status + ' ' + response.statusText);         
        }
    });
}

我将此功能称为

var Result = CheckUser();
if(Result== false){ 
    //do something
} else{
    //do something
}

过去一天我一直在努力！我在一节中读到，这是因为“Ajax是异步的”。。但我怎么能处理呢？

Answer 1

您最好将回调函数传递给CheckUser

function CheckUser(callback) {

    var EmpName = $("#txtName").val();

    $.ajax({        
        type: "POST",
        url: location.pathname + "/UserExist",
        data: "{Name:'" + EmpName + "'}",
        contentType: "application/json; charset=utf-8",
        datatype: "jsondata",
        async: "true",
        success: function (response) {
            console.log(response.d);
            var obj = eval('(' + response.d + ')');
            callback(obj);
        },
        error: function (response) {
            alert(response.status + ' ' + response.statusText);
            callback(null);         
        }
    });
}

然后你会这样调用这个函数

CheckUser(function (res) {
    if (res === null) {
       //false
    } else {
       //true
    }
});

Answer 2

错字

Checkusers -with s

CheckUser（） - 没有s

Answer 3

更改

datatype:'jsondata'

到

 dataType: "json"

由于ajax是异步的，你不能像这样返回数据。你需要在ajax的成功处理程序上编写东西

$.ajax({
type: "POST",
url: location.pathname + "/UserExist",
data: "{Name:'" + EmpName + "'}",
contentType: "application/json; charset=utf-8",
dataType: "json",
async: "true",
success: function (response) {
   //do your stuff here

},
error: function (response) {
    alert(response.status + ' ' + response.statusText);
}
});

如果要从ajax返回数据，则应使用async : false作为

function CheckUser() {
var EmpName = $("#txtName").val();
var CheckUser;
$.ajax({
    type: "POST",
    url: location.pathname + "/UserExist",
    data: "{Name:'" + EmpName + "'}",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    async: false,
    // ...
    success: function (jsonData) {
        CheckUser = jsonData
    }
});
return CheckUser
}

但它不是一个好方法，它会冻结浏览器

Answer 4

函数CheckUser（）没有任何return语句，这就是为什么你的结果总是未定义的。

将回调传递给您的函数，并在您的ajax调用成功或错误时调用它。

类似的东西：

function CheckUser(callback) {

var EmpName = $("#txtName").val();

$.ajax({        
    type: "POST",
    url: location.pathname + "/UserExist",
    data: "{Name:'" + EmpName + "'}",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    async: "true",
    success: function (response) {
        console.log(response.d);
        var obj = eval('(' + response.d + ')');
        callback(null, obj);
    },
    error: function (response) {
        alert(response.status + ' ' + response.statusText);
        callback(response);         
    }
});
}

然后：

var Result = CheckUser(function ( err, data) {
    // check if no err then process whatever data format you have
});

jquery ajax函数未定义

4 个答案: