CODEWITHSUNDEEP
jquerysyntax
Kurkula
Kurkula

Reputation: 6762

jquery syntax error in my code

I am unable to check if my code is wrong or there is any syntax error here. I spent more time so thought of asking in stackoverflow.

$(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);
});

Upvotes: -1

Views: 100

Answers (3)

Prakash GPz
Prakash GPz

Reputation: 1523

        $(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 );
            } );
        });

Upvotes: 1

kot-6eremot
kot-6eremot

Reputation: 71

Ajax is an async technology. So, you trying to return result before getting the response.

You can turn off async for ajax and that will look like so:

$(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);
});

Upvotes: 1

Afzaal Ahmad Zeeshan
Afzaal Ahmad Zeeshan

Reputation: 15860

The only problem here that I think is, that you cannot return a value from an Ajax call.

Try changing the return result and I think that would work. There isn't any trouble in your code.

Or just try changing the ajax request type to Synchronous. Maybe it would be like

async: false;

Upvotes: 1

Related Questions