Question

我无法检查我的代码是否错误或此处是否存在语法错误。我花了更多时间考虑在stackoverflow中提问。

$(document).ready(function () {

    AjaxGet = function (url, storeageLocation, mySuccessCallback) {
        var result = $.ajax({
            type: "GET",
            url: "/Regions/GetView",
            async: true,
            contentType: 'application/json',
            dataType: "html",
            success: function (viewData) {
                alert(viewData);
                storeageLocation = viewData; 
                mySuccessCallback(viewData);
            },
            error: function (xhr, ajaxOptions, thrownError) {

            }
        }).responseText;

        return result;
    };
    alert(result);
});

Answer 1

我认为唯一的问题是，您无法从Ajax调用返回值。

尝试更改return result，我认为这样可行。您的代码中没有任何问题。

或者只是尝试将ajax请求类型更改为Synchronous。也许就像

async: false;

Answer 2

Ajax是一种异步技术。因此，您在获得响应之前尝试返回结果。

你可以关闭ajax的async，看起来像这样：

$(document).ready(function () {
    AjaxGet = function (url, storeageLocation, mySuccessCallback) {
        var result = null;

        $.ajax({
            type: "GET",
            url: "/Regions/GetView",
            async: false,
            contentType: 'application/json',
            dataType: "html",
            success: function (viewData, textStatus, jqXHR) {
                alert(viewData);
                storeageLocation = viewData; 

                result = textStatus;

                mySuccessCallback(viewData);
            },
            error: function (xhr, textStatus, thrownError) {
                result = textStatus;
            }
        });

        return result;
    };
    alert(result);
});

Answer 3

        $(document).ready(function () {

            AjaxGet = function (url, mySuccessCallback) {
                var result = $.ajax({
                    type: "GET",
                    url: "/Regions/GetView",
                    async: true,
                    contentType: 'application/json',
                    dataType: "html",
                    success: function (viewData) {
                        mySuccessCallback(viewData);
                    },
                    error: function (xhr, ajaxOptions, thrownError) {

                    }
                }).responseText;

                return result;
            };
            // alert(result);  // will throw error: undefined variable 'result', because you used this variable inside function definition 'AjaxGet', not in this context
            AjaxGet( '/', function(data) { // call the function after defining it to execute and get the results
                alert( data );
            } );
        });

我的代码中的jquery语法错误

3 个答案: