JavaScript function that returns AJAX call data [duplicate]
You can't return data returned by an AJAX call unless you want to call it synchronously (and you don't – trust me). But what you can return is a promise of a data returned by an AJAX call and you can do it actually in a very elegant way.
(UPDATE: Please note that currently jQuery Promises are not compatible with the Promises/A+ specification - more info in this answer.)
Basically you can return the return value of your $.ajax(...) call:
function checkUserIdExists(userid){
return $.ajax({
url: 'theurl',
type: 'GET',
cache: false,
data: {
userid: userid
}
});
}
and someone who calls your function can use it like this:
checkUserIdExists(userid).success(function (data) {
// do something with data
});
See this post of mine for a better explanation and demos if you are interested.
you can pass in a callback function:
function checkUserIdExists(userid, callback) {
$.ajax({
...
success: callback
});
}
checkUserIdExists(4, function(data) {
});
With jQuery 1.5, you can use the brand-new $.Deferred
feature, which is meant for exactly this.
// Assign handlers immediately after making the request, // and remember the jqxhr object for this request var jqxhr = $.ajax({ url: "example.php" }) .success(function() { alert("success"); }) .error(function() { alert("error"); }) .complete(function() { alert("complete"); }); // perform other work here ... // Set another completion function for the request above jqxhr.complete(function(){ alert("second complete"); });
Source
As of jQuery 1.8, the "success", "error" and "complete" callback are deprecated. Instead you should be using "done", "fail" and "always".
So you could have:
function checkUserIdExists(userid, callback) {
return $.ajax({
url: 'theurl',
type: 'GET',
cache: false,
data: {
userid: userid
}
})
.done(callback)
.fail(function(jqXHR, textStatus, errorThrown) {
// Handle error
});
}
checkUserIdExists(2, function(data) {
console.log(data); // Do what you want with the data returned
});