jQuery $.ajax success must be a callback function?

I've created a wrapper function for jQuery's $.ajax() method so I can pass different dataTypes and post variables - like so:

function doPost(dType, postData, uri){

    $.ajax({
        url: SITE_URL + uri,
        dataType: dType,
        data: postData,
        success: function(data){
          return data;
        });
}

The problem I'm having is gettin开发者_Go百科g the data (which is always JSON) back out. I've tried setting a var ret before the $.ajax() function call and setting it as ret = data in the success function. Am I being stupid about this? If I don't set a success function, will the $.ajax simply return the data? Or is it just success: return data? Or does success require a callback function to handle the data, which could just be return data?

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Well, you are inside a function - take advantage of variable scope ;-)

function doPost(dType, postData, uri) { 
    var result;
    $.ajax({ 
        url: SITE_URL + uri, 
        dataType: dType, 
        data: postData, 
        async: false,
        success: function(data){ 
          result = data;
        }); 
    return result;
} 

This actually works, but I guess the async part is obligatory then... Otherwise the call of $.ajax would return immediately, and result would still be undefined - you would always get undefined as result of your function call.

However, when you make the $.ajax call synchronous, it blocks until the data is received, and you can return the data as result of your own function.

But you have to be clear that when using this method, no other code will be executed until the ajax load finishes!

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When you call $.ajax() it creates the request and then moves on to the next thing. It doesn't sit around and wait for the return and block the next lines of code. That means that unless the success callback function can't return the value of data for the doPost function.

If you really want to have a wrapper function then you could do something like this:

function doPost(dType, postData, uri, success){
    $.ajax({
        url: SITE_URL + uri,
        dataType: dType,
        data: postData,
        success: success
    });
}

And define your success handler when you call doPost, but the wrapper function really isn't doing much for you at the point.