Modify global javascript variable inside a function

Question

I'm trying clone a json object inside a function wanting to access the object outside the function. But when function is done, the object still seems to be undefined. Seems like it's not the same variable?

var jsonUserObj;
$.getJSON("user.json", function(content){
    jsonUserObj = $.parseJSON(JSON.stringify(content));
    console.log(content);
    console.log(jsonUserObj); //looks fine!
});
console.log(jsonUserObj); //undefined

inside the callback function it contains all the data, but it does not remain outside of it. How to make it assessible globally?

Typo: jsonUserObject !== jsonUserObj That's the reason for the undefined. The asynchronous call however, will cause it not to have the new values when you view it immediately following the getJSON() call. — Michael Berkowski
– Michael Berkowski, Commented Oct 23, 2012 at 13:27

kabaros · Accepted Answer · 2012-10-23 13:28:22Z

2

$.getJSON is asynchronous so console.log at the end of your code runs before $.getJSON returns its result.

You should modify the variable inside the callback (where it looks fine) and then use the variable inside that function, this callback is the only place where you can guarantee your variable is set.

You could also use the synchronous version of $.ajax but that's really not recommended (and probably unnecessary).

answered Oct 23, 2012 at 13:28

kabaros

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1 Comment

user1768443 Over a year ago

Hm ok. I can do that, but wanted to avoid it. Thanks anyway

Miguel Sanchez Gonzalez · Accepted Answer · 2012-10-23 13:28:08Z

1

You got a typo:

console.log(jsonUserObject);

It should be

console.log(jsonUserObj);

answered Oct 23, 2012 at 13:28

Miguel Sanchez Gonzalez

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2 Comments

user1768443 Over a year ago

Sorry about the type. Problem still remains though.

Miguel Sanchez Gonzalez Over a year ago

This is because you are using a async function @kabaros explained the behavior of your code

Maess · Accepted Answer · 2012-10-23 13:27:47Z

0

You need to declare var jsonUserObj outside a function.

Also it looks like you have a typeo, is it jsonUserObj or jsonUserObject?

answered Oct 23, 2012 at 13:27

Maess

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5 Comments

Maess Over a year ago

Right, but where is he calling the get? It needs to be called from somewhere, I assume this whole block is contained in another funtion

Michael Berkowski Over a year ago

Doesn't matter though - it is still declared at a higher scope than the getJSON() call, so even if the whole thing is wrapped in a function the scope will still work for this example.

Maess Over a year ago

That doesn't make it global, which is what he states he wants it to be, it makes it scoped to the function.

Michael Berkowski Over a year ago

But you're making the inference that this is all inside a function you can't see and imagine exists.

user1768443 Over a year ago

i have tried to put it in a function and actually call it, but asynch problem still there.

AGB · Accepted Answer · 2012-10-23 13:28:48Z

0

Could it be a question of timing? If that getJSON method is firing asynchronously then it may not have returned it's value by the time you have fired the last line. Does that make sense?

answered Oct 23, 2012 at 13:28

AGB

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Comments

jbabey · Accepted Answer · 2012-10-23 13:33:52Z

0

$.getJSON performs an ajax call, which is asynchronous. The code after it will continue evaluating while it waits for a response. When the response comes back, the program flow will jump back into the success/error/complete handlers of the ajax call.

tldr: Anything you do with data from an ajax call must be in the success handler of that ajax call.

answered Oct 23, 2012 at 13:33

jbabey

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