Using two variables to create a jQuery object

Question

Never ran into this situation before and now I am having trouble accomplishing this.

I basically want this:

$(cardMonth).prepend("<option value='' selected='selected'></option>");
$(cardYear).prepend("<option value='' selected='selected'></option>");

To be something like this:

$(cardMonth, cardYear).prepend("<option value='' selected='selected'></option>");

And it doesnt work? Have you tried it? But what are cardMonth and cardYear? Id's or classnames? So its either $('#cardMonth, #cardYear') or $('.cardMonth, .cardYear') — putvande
– putvande, Commented Jul 10, 2013 at 22:27
They are variables. They work fine separated, but trying to use them in the same line does not work. — souporserious
– souporserious, Commented Jul 10, 2013 at 22:28

DevlshOne · Accepted Answer · 2013-07-10 22:33:45Z

2

$(cardMonth).add(cardYear).prepend("<option value='' selected='selected'></option>");

Alternatively:

var els[0] = cardMonth;
var els[1] = cardYear;
$(els).prepend("<option value='' selected='selected'></option>");

answered Jul 10, 2013 at 22:33

DevlshOne

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Comments

Adrian Wragg · Accepted Answer · 2013-07-10 22:35:52Z

1

$.merge($(cardMonth), $(cardYear)).prepend("<option value='' selected='selected'></option>");

answered Jul 10, 2013 at 22:35

Adrian Wragg

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