Pass jQuery variable to CSS

Question

I'm trying to dynamically calculate the min-width of body based on the number of div.columns that exist on the page, plus a padding of 180px. This is the code I'm using:

function countColumns() {
    var columns =  $('.column').size(); 
    document.write(columns)
}

$('body').css('min-width', (parseInt(columns) * 120) + 180 + 'px') );

Any clue as to why its not working?

columns is local to countColumns(), so you cannot access it from the outside. Also, your call to parseInt() looks unnecessary since size() already returns an int (and you should use length instead, btw). — Frédéric Hamidi
– Frédéric Hamidi, Commented Dec 1, 2011 at 16:43
...and as if that weren´t enough, you never use countColumns() — Stefan
– Stefan, Commented Dec 1, 2011 at 16:50

Rory McCrossan · Accepted Answer · 2011-12-01 16:45:09Z

6

This is due to the scope of the columns variable. You need to either make it global, or assign it to the result of the function, the latter being preferable. Try this:

function countColumns() {
    return $('.column').size(); 
}

$('body').css('min-width', (parseInt(countColumns()) * 120) + 180 + 'px') );

Note, the parseInt is redundant as the size() method will always return an integer.

answered Dec 1, 2011 at 16:45

Rory McCrossan

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1 Comment

colindunn Over a year ago

I used this instead and it worked great $('body').css('min-width', ((countColumns()) * 120) + 180 + 'px'); Thank you!

ShankarSangoli · Accepted Answer · 2011-12-01 16:41:53Z

1

Since columns variable is defined inside the function it is not accessible outside it. You should define it in an outer or global scope. Try this.

var columns;
function countColumns() {
    columns =  $('.column').size(); 
    document.write(columns)
}

$('body').css('min-width', ((parseInt(columns) * 120) + 180) + 'px') );

Alternatively you can even call the function which will return the number of columns. In this way you don't have to maintain the variable. Also in countColumns function I am using lenght property of the jQuery object which gives the same result and also avoids one function call. Try this

   function countColumns() {
        return  $('.column').length; 
    }

    $('body').css('min-width', ((countColumns * 120) + 180) + 'px') );

answered Dec 1, 2011 at 16:41

ShankarSangoli

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3 Comments

JJJ Over a year ago

You also have to call the countColumns() function... on the other hand, there's no reason to use the global variable if the function would just return the value.

Stefan Over a year ago

Also, make sure you´re running the script once the DOM has loaded.

Viruzzo Over a year ago

As a side note: consider LESS for things like this, it's wonderful.

Pastor Bones · Accepted Answer · 2011-12-01 16:43:13Z

0

you're setting the column variable from within the scope of the function, try this instead

var columns
function countColumns() {
    columns =  $('.column').size(); 
    document.write(columns)
}

$('body').css('min-width', (parseInt(columns) * 120) + 180 + 'px') );

answered Dec 1, 2011 at 16:43

Pastor Bones

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Comments

Stefan · Accepted Answer · 2011-12-01 16:51:46Z

0

This should work

$(document).ready(function() {

  var minWidth = ($('.column').length * 120) + 180);
  $('body').css('min-width', (minWidth + 'px') );

});

Never seen anyone use min-width on <body> though :P

edited Dec 1, 2011 at 16:51

answered Dec 1, 2011 at 16:46

Stefan

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