Question

102

javascript function return not working

rated 0 times [ 103] [ 1] / answers: 1 / hits: 32272 / 15 Years ago, fri, november 13, 2009, 12:00:00

I have a problem returning a variable in my function, the below script works fine:

function sessionStatus(){

    $(document).ready(function(){

        $.getJSON(scriptRoot+sessionStatus.php,function(status){

            alert(status);

        });

    });

}



sessionStatus();

Bet when I try the following I get a message box with the message undefined:

function sessionStatus(){

    $(document).ready(function(){

        $.getJSON(scriptRoot+sessionStatus.php,function(status){

            return status;

        });

    });

}

alert(sessionStatus());

This is really bugging me, I just can't seem to see what I've done wrong.

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cayden

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answered 15 Years ago dusty · Accepted Answer

There are two things you should know:

1: the JSON thing is asynchronous, so the function call to sessionStatus could already be done when the JSON is still being fetched. The following would work:

function sessionStatus(callback){

    $(document).ready(function(){

        $.getJSON(scriptRoot + sessionStatus.php, function(status){

                callback(status);

        });

    });

}

sessionStatus(function(s){alert(s);});

or rather:

function sessionStatus(callback){

    $(document).ready(function(){

        $.getJSON(scriptRoot + sessionStatus.php, callback);

    });

}

sessionStatus(function(s){alert(s);});

2: even when it would be synchronous, you are only giving a return value from the inner function, so the sessionStatus would return nothing. Check out this code (not related to your JSON thing):

function do() {

    var x = 0;

    (function(){

       x = 2;

    })();

    return x;

}

or:

function do() {

    var x = (function(){

       return 2;

    })();

    return x;

}

Both return 2. Hopefully this explains a bit.