jQuery — Variable not recognised but manual input of exact same value is?

Question

Best explained with code:

$(".myParent .myChild:nth-child(3n)").css('border-top-color','#ffffff');

√ Works

myVar = "3n";
$(".myParent .myChild:nth-child(myVar)").css('border-top-color','#ffffff');

X Doesn't work

This is obviously jQuery programming 101... but seriously, why on earth won't that work?! I'm passing on the same thing!

I tried it as > myVar = 3n (no string), obviously that shouldn't work, and it didn't.

mcgrailm · Accepted Answer · 2011-03-28 12:42:44Z

5

you var has to be concatenated

 var myVar = "3n";

 $(".myParent .myChild:nth-child("+myVar+")").css('border-top-color','#ffffff');

answered Mar 28, 2011 at 12:42

mcgrailm

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2 Comments

Ah OK... so it's looking at the css as a string already so I have to go out the string to pick up the variable. Thanks a million.

@RGBK Yes but I wouldn't call it CSS to the right of the dot is CSS to the left is your jQuery selector 8 ^ )