Undefined when trying to access array created from JSON data

Question

I have been playing with JSON/PHP/JS today and I am having problems..

My PHP works fine which gets a row of data from my SQL table then encodes as JSON. My JS/Jquery file loads the data from the PHP file fine using $.getJSON.

However, when i try and use the data later in the page like so: gender = user['player'].gender, I get user.player is undefined.

This is my code that is relevant:

function getUserInfo() {
    var url = "./php/getUserInfo.php";
    $.getJSON(url, function(data) {
        $.each(data.members, function(i, dat) {

            user['player'] = {
                gender: dat.gender,
                fname: dat.first_name,
                lname: dat.last_name,
                username: dat.username,
            };
        });
    });
}

user = {};
getUserInfo();

//Displays an object, which has the correct information I want.
console.log(user);
var gender = user['player'].gender;
console.log(gender);

the last line of code gives me the error that user.playe is not defined. but it should display male Help would be helpful, I have tried many things to fix this but cant seem to do so.

Sushanth -- · Accepted Answer · 2012-11-27 16:35:44Z

4

$.getJSON is Asynchronous.

As soon as the request is sent , it goes to the next line and does not wait for the request to be completed.

So when you try to access the variable , it is still not available..

Try putting that piece of code after the $.each and it should work.

answered Nov 27, 2012 at 16:35

Sushanth --

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5 Comments

nmyster Over a year ago

how would i go about fixing this then.

nmyster Over a year ago

well the returned data is only one rows worth, would removing the .each loop help and I just add the data to the first element.

Sushanth -- Over a year ago

To check which actually works .. Try putting two console.log statements , one inside the callback , and one outside.. Check what happens

nmyster Over a year ago

console logs female in the callback but undefined when out of the function.

Sushanth -- Over a year ago

Which is called first , the undefined is logged first ... SO basically you need to move that piece of code inside the callback

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