Run Javascript Jquery function on two different classes

Question

i have those classes:

.slider
.sliderlogo

and i have those pieces of codes:

var sc=$(".slider img").size();
$(".slider #"+count).show("slide",{direction:'right'},500);
$(".slider #"+count).delay(5500).hide("slide",{direction:'left'},500);

what's the right way to add the other "slidelogo" class to the JS so it will run on the other class as well?

generaly its like:

(".slider #1,.sliderlogo #1")

bit its not working for the above code lines. thanks in advance.

ID must be unique in the entire document.

David Hedlund
– David Hedlund

2013-11-13 14:30:20 +00:00
Commented Nov 13, 2013 at 14:30 — David Hedlund
– David Hedlund, Commented Nov 13, 2013 at 14:30

xdumaine · Accepted Answer · 2013-11-13 14:34:23Z

1

It looks like you might have issues with your markup, as IDs must be unique on the document and cannot begin with a number. If you must use the Ids, try building them as something like 'slider' + count and 'sliderLogo' + count. so that each is unique.

answered Nov 13, 2013 at 14:34

xdumaine

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2 Comments

Netanel Dadon Over a year ago

yeah i forgot that i numbered 2 groups of images in the same id's. however,ID's can start with numbers, since HTML5 arrived.

xdumaine Over a year ago

@NetanelDadon Technically that is true, but as with all HTML5 features, you should be aware of them in case you have to support older browsers.

Skatox · Accepted Answer · 2013-11-13 14:34:30Z

1

The ID should be unique in the entire document, so i recommend you to use classes instead, something like .item-1, .item-2 and then change the code to:

var sc=$(".slider img").size();
var selector = ".slider .item-" + count + ", .sliderlogo .item-" + count;
$(selector).show("slide",{direction:'right'},500);
$(selector).delay(5500).hide("slide",{direction:'left'},500);

answered Nov 13, 2013 at 14:34

Skatox

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Comments

Basic · Accepted Answer · 2013-11-13 14:37:35Z

1

As David says, the id must be unique, so try changing them to

.slider #slider1

and

.sliderlogo #sliderlogo1

etc.

then use:

(".slider #slider"+count+",.sliderlogo #sliderlogo"+count)

edited Nov 13, 2013 at 14:37

Basic

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answered Nov 13, 2013 at 14:34

Pete Thorne

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