Getting error in jquery ajax "function not defined"

Question

I am returning some values from jquery ajax. I am getting the result but with this I am also getting a error. Please first of all take a look at code

function get_CommentCount(handleData) {
    $.ajax({
        url: 'Profile/get_CommentCount',
        type: "post",
        dataType: 'json',
        success: function(data) {
            handleData(data);
        }
    });
}

calling this function like this

get_CommentCount(function(output) {
    console.log('output', output)
});

Its give me error that TypeError: handleData is not a function. Please anybody tell me why I am getting this error. I have gone through the stackoverflow questions but I am not finding any solution for me. May be you find it duplicate but I post this question after surfing the stackoverflow.

Thanks

And where you have defined handleData. I wonder you have problems with scope. Put a breakpoint into your success function and check 'this' value — Tistkle
– Tistkle, Commented Jun 13, 2016 at 7:43
@AzadChouhan your code seems to work fine: jsfiddle.net/k87ncaje. Could you show an example of the issue? — Rory McCrossan
– Rory McCrossan, Commented Jun 13, 2016 at 7:51

Kirankumar Dafda · Accepted Answer · 2016-06-13 07:48:22Z

1

In the success function of your ajax you have wrote handle(data), and it is a function that you havent wrote any where may be, so this error is showing you that TypeError: handleData is not a function so you have define a function like

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

and this will works for you.

answered Jun 13, 2016 at 7:48

Kirankumar Dafda

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3 Comments

Satpal Over a year ago

OP is passing it as callback(parameter), its pretty valid code

Azamantes Over a year ago

He is passing the function as a callback.

Kirankumar Dafda Over a year ago

Then, the error is getting only after function call

Cheitan · Accepted Answer · 2016-06-13 07:46:46Z

1

I think the problem here is that you are passing an anonymous function to get_CommentCount. Try with a normal function, it should work better.

answered Jun 13, 2016 at 7:46

Cheitan

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1 Comment

Rory McCrossan Over a year ago

The function is being passed as a parameter - it's completely valid.

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