tonga tonga - 3 months ago 16
Javascript Question

jQuery ajax success callback function definition

I want to use jQuery ajax to retrieve data from a server.

I want to put the success callback function definition outside the

.ajax()
block like the following. So do I need to declare the variable
dataFromServer
like the following so that I will be able to use the returned data from the success callback?

I've seen most people define the success callback inside the
.ajax()
block. So is the following code correct if I want to define the success callback outside?

var dataFromServer; //declare the variable first

function getData() {
$.ajax({
url : 'example.com',
type: 'GET',
success : handleData(dataFromServer)
})
}

function handleData(data) {
alert(data);
//do some stuff
}

Answer

Just use:

function getData() {
    $.ajax({
        url : 'example.com',
        type: 'GET',
        success : handleData
    })
}

The success property requires only a reference to a function, and passes the data as parameter to this function.

You can access your handleData function like this because of the way handleData is declared. JavaScript will parse your code for function declarations before running it, so you'll be able to use the function in code that's before the actual declaration. This is known as hoisting.

This doesn't count for functions declared like this, though:

var myfunction = function(){}

Those are only available when the interpreter passed them.

See this question for more information about the 2 ways of declaring functions