Jquery each cannot call methods

Question

So far i have the code below:

$('.pagination').each(function(){
    var paginationWidth = $(this).width();
    var pixelOffset = '-' + paginationWidth + 'px';
    console.log(paginationWidth.css('margin-left', pixelOffset ));
});

Console log shows "Object 57 has no method 'css'", the number being the width. Why is this the case on an each element? If its not an array why can't .css() be called upon each of said elements?

var paginationWidth = $(this).width(); which is not a jquery object (just a number). Try $(this).css('margin-left', pixelOffset ) — Khanh TO
– Khanh TO, Commented Dec 30, 2013 at 13:46

Chris · Accepted Answer · 2013-12-30 13:48:34Z

2

The issue here is nothing to do with the each loop. The .width() jQuery method returns a value, and you're then trying to set a CSS property of that value. The $(this) inside your each loop is still the object.

Your last line wants to be this instead.

$(this).css('margin-left', pixelOffset )

answered Dec 30, 2013 at 13:48

Chris

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Pranav C Balan · Accepted Answer · 2013-12-30 13:48:40Z

1

Try this,

$('.pagination').each(function(){
    var page=$(this);
    var paginationWidth = page.width();
    var pixelOffset = '-' + paginationWidth + 'px';
    console.log(page.css('margin-left', pixelOffset ));
});

$(this).width() is not returns an object.It returns an integer number which representing it's width.

answered Dec 30, 2013 at 13:48

Pranav C Balan

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Stokedout · Accepted Answer · 2013-12-30 13:48:45Z

1

I think what you want to do is $(this)... literally if you excuse the joke... See below.

('.pagination').each(function(){
   var paginationWidth = $(this).width();
   var pixelOffset = '-' + paginationWidth + 'px';
   console.log($(this).css('margin-left', pixelOffset ));
);

answered Dec 30, 2013 at 13:48

Stokedout

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kundan Karn · Accepted Answer · 2013-12-30 13:53:12Z

0

jQuery works like

$(selector).function/event(.....

but paginationWidth is not a selector. selectors are objects to be selected.

Better to use

$('any-selector-to apply-margin-left').css('margin-left', pixelOffset );

answered Dec 30, 2013 at 13:53

kundan Karn

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