JQUERY + Return an Array from a Function

Question

I'm struggling to return an array from a function call - code below:

///////////// Generic Functions
function spotJoinPosition() {
    var pos = {  
                offset: $('div#spotJoinSite').offset(),                                
                width: $('div#spotJoinSite').width(),
                height: $('div#spotJoinSite').height()
    }
    return(pos);
}


        var positionData = spotJoinPosition();
        alert(positionData);
        alert(positionData[width]);

When I alert positionData I get [object][object] and then undefined.

Advice?

Andy Ray · Accepted Answer · 2011-09-10 09:31:42Z

6

alert(positionData[width]);

This is alerting a key in positionData, and using the variable width as the key. You haven't defined a variable called width, so it's essentially looking up positionData[undefined]. What you want is positionData.width, or positionData['width'], but there is no reason for quotes here.

Quotes would only be required if you had a key with non alphanumeric characters. positionData['some-key'] works, but positionData.some-key is a syntax error, because variables cannot have - in them.

Also, your code SHOULD be erroring, because width isn't defined anywhere. I'm worried that you have a globally defined width variable somewhere in your code.

answered Sep 10, 2011 at 9:31

Andy Ray

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Comments

lonesomeday · Accepted Answer · 2011-09-10 09:45:39Z

1

That's because positionData is an object (the object you return from spotJoinPosition) and ~~the variable width is undefined~~ the variable width contains a value that is not present on the object.

You want positionData.width or positionData['width'].

See the MDN docs on member operators.

edited Sep 10, 2011 at 9:45

answered Sep 10, 2011 at 9:32

lonesomeday

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1 Comment

Chris Over a year ago

actually, the reason that positionData[width] is outputting undefined is because width is defined somewhere, but the value of width is not a valid key in his positionData object.

Allen Liu · Accepted Answer · 2011-09-10 09:33:18Z

0

Try:

  alert(positionData.offset);
  alert(positionData.width);
  alert(positionData.height);

answered Sep 10, 2011 at 9:33

Allen Liu

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Comments

Chris · Accepted Answer · 2011-09-10 09:45:50Z

0

If you actually want a generic function that returns an array and not an object, you might want to revise to the following:

///////////// Generic Functions
function spotJoinPosition(selector) {
    var el = $(selector),
        result = [el.offset(), el.width(), el.height()];

    return(result);
}

var positionArray = spotJoinPosition("#spotJoinSite");
alert(positionArray[0]); // outputs the position
alert(positionArray[1]); // outputs the width

answered Sep 10, 2011 at 9:45

Chris

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