javascript if statement not working

Question

I have a random number generator which generates numbers between 5 and 15. I want to be able to change the position of the div 'chest' depending on which number the computer chooses. (all set to 300 because it's being tested.) I do not know why it is not working. (please bear in mind that the div is definitely called chest, and i have a css statement that includes 'left'. Furthermore, the function is definitely being called, and is generating a number.) Any help is appreciated,

Here is my code:

var xChestPosition;
var yChestPosition;
function randomNumberForChest(firstNum, secondNum) {
    xChestPosition = Math.floor(Math.random() * secondNum) + firstNum;
    yChestPosition = Math.floor(Math.random() * secondNum) + firstNum;
    alert(xChestPosition);
    if(xChestPosition==5) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==6) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==7) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==8) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==9) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==10) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==11) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==12) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==13) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==14) {
        $('#chest').css('left','300');
    }
    if(xChestPosition==15) {
        $('#chest').css('left','300');
    }
}

left work only on absolute, relative and fixed positioned elements. — Tushar
– Tushar, Commented Feb 18, 2016 at 3:08
@igetstuckalot by the way what about if( xChestPosition>=5 && xChestPosition<=15){$('#chest').css('left', '300px'); — Satyaki Chatterjee
– Satyaki Chatterjee, Commented Feb 18, 2016 at 4:05

Atticus · Accepted Answer · 2016-02-18 03:08:41Z

1

It's likely that you may need to specify the unit of measurement for the CSS position

$('#chess').css('left', '300px')

answered Feb 18, 2016 at 3:08

Atticus

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Comments

VioLoustic · Accepted Answer · 2016-02-18 03:36:55Z

0

You'd better use:

$('#chest').css({"position": "absolute", "left": "300px"});

rather than use:

$('#chest').css('left','300');

Because, if you want to set left attributes, you must specify its position (absolute, fixed, relative). Your "if" statement can be executed well.

answered Feb 18, 2016 at 3:36

VioLoustic

33 bronze badges

Comments

Jobelle · Accepted Answer · 2016-02-18 04:03:03Z

Try this

var xChestPosition;
    var yChestPosition;
    function randomNumberForChest(firstNum, secondNum) {
        xChestPosition = Math.floor(Math.random() * secondNum) + firstNum;
        yChestPosition = Math.floor(Math.random() * secondNum) + firstNum;
        alert(xChestPosition);
        if (xChestPosition == 5) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 6) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 7) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 8) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 9) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 10) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 11) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 12) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 13) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 14) {
            $('#chest').css('left', '300px');
        }
        if (xChestPosition == 15) {
            $('#chest').css('left', '300px');
        }
    }

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