I've this function
function getTags(level){
$.getJSON("php/get-tags.php", { "parent": level }, function(json) {
return json;
});
}
I'm calling this function as
$(function(){
var tags = getTags('0');
});
The problems is, in the function getTags() the return json is like
{"tags":["Mathematics","Science","Arts","Engineering","Law","Design"]}
but at var tags = getTags('0'), catching the return value, it gets an undefined value.
Is the way I'm returning the value incorrect?
-
1possible duplicate of JavaScript asynchronous return value / assignment with jQueryFelix Kling– Felix Kling2012年01月11日 21:10:04 +00:00Commented Jan 11, 2012 at 21:10
5 Answers 5
Like many others already correctly described, the ajax request runs asynchronous by default. So you need to deal with it in a proper way. What you can do, is to return the jXHR object which is also composed with jQuery promise maker. That could looke like
function getTags(level){
return $.getJSON("php/get-tags.php", { "parent": level });
}
and then deal with it like
$(function(){
getTags('0').done(function(json) {
// do something with json
});
});
Comments
getJSON is asynchronous, the function that called it will have finished by the time the HTTP response gets back and triggers the callback.
You cannot return from it.
Use the callback to do whatever you want to do with the data.
Comments
You are trying to call an asynchronous function in a synchronous fashion.
When you call getTags, the it triggers a call to your PHP page, if javascript was blocking, then your page would hang until your server responded with the JSON. You need to re-think your logic and trigger a callback.
function getTags(level){
$.getJSON("php/get-tags.php", { "parent": level }, function(json) {
//do something with the JSON here.
});
}
Comments
You cannot return from an AJAX call. It's asynchronous. You need to have all logic dealing with the returned data in the callback.
Comments
If you need it to work like that, your getTagsfunction must return a value. It does not at the moment. You could achieve this by using $.ajax and setting async to false.