CODEWITHSUNDEEP
javascriptjquery
user911
user911

Reputation: 35

How do I debug Jquery Success function

I am using Javascript to get records from the database, everything works the way i want but i cannot show alerts in success handler. When i place a break point to sucess:function(data) it gets hit but this line is not being hit $("#Alert").html(data).show().... Another unusually thing i have noticed is that some time $("#Alert").html(data).show().. gets hit. Is there any way i can debug this? 

        function MethodNAme() {
        ajReq.abort();
        ajReq = $.ajax({
            type: "POST",
            url: "Services/",
            data: ,
            contentType: "application/json; charset=utf-8",
            dataType: "json",
        success: function (data) {
                         getSomething();
            $("#Alert").html(data).show().delay(5000).fadeOut();   
            alert("working");

        }
    }

Upvotes: 0

Views: 286

Answers (2)

Sudhir Bastakoti
Sudhir Bastakoti

Reputation: 100195

your syntax is not correct, you are placing a function getSomething() in middle of $.ajax(),

function MethodNAme() {
    ajReq.abort();
    ajReq = $.ajax({
        type: "POST",
        url: "Services/",
        data: {},
        contentType: "application/json; charset=utf-8",
        dataType: "json",
     //       getSomething(); <-- remove this
        success: function (data) {
            $("#Alert").html(data).show().delay(5000).fadeOut();   
            alert("working");
        }
    });
}

Upvotes: 1

Bilal
Bilal

Reputation: 2673

You can use console.log() to print data in browser console. You can access console window by pressing F12 key and then go to console tab.

Upvotes: 0

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