The $ dollar sign

Question

I have something simple like this:

$(selector).append("somestuff");

But since I'm going to reuse the selector I cache it with:

var $selector = $(selector);

So I end up with:

$selector.append("somestuff");

My question is, should I be doing that, or should I be doing:

var selector = $(selector);
selector.append("somestuff");

Trying either one, both works. Which method is correct, and why? Is the $ in $selector unnecessary because the jquery object has already been declared in $(selector)?

Edit

Thanks for the answers. It seems very simple and quite clear. Still, there seems to be disagreement over whether or not I should use $ in the variable. It would be nice for everyone to vote up an answer. :)

Your question indicates you're a little confused about how objects behave. You might want to review a bit of object oriented programming basics. — Spencer Ruport, Jul 24 '09 at 21:28

RichieHindle · Answer 1 · 2011-09-29T15:18:23.040

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$ is just a name - names in JavaScript can contain dollar signs, and can consist of just a dollar sign.

Whether you use a dollar sign in your name isn't relevant to jQuery - there's nothing special about the dollar sign, except that jQuery defines a function called $.

edited Sep 29 '11 at 15:18

answered Jul 24 '09 at 21:24

RichieHindle

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I've seen JQuery code with `$j`... is that a JQuery thing or just a custom variable name? – CodyBugstein Jul 23 '14 at 20:18
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@Imray: It's common practice to prefix variables which refer to jQuery objects with a dollar sign, eg. `var $highlights = $('div.highlight');`. – RichieHindle Jul 23 '14 at 21:29
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Cool thanks! I never knew that and would've been hard to figure out. – CodyBugstein Jul 23 '14 at 21:37

score 33 · Accepted Answer · answered Jul 24 '09 at 21:24

33

I've seen it done both ways. All you are doing is creating a variable with the name '$selector', so they are functionally equivalent. The nice thing about it is that it does make them easy to pick out as jQuery objects.

answered Jul 24 '09 at 21:24

Mara Morton

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score 21 · Answer 3 · edited Dec 27 '11 at 15:27

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The cash sign is just an alias for the jQuery function. Starting the variable name with $ has no effect on that variable.

It is customary though, especially in jQuery plugin development, to use variables already wrapped in jQuery with a cash sign, so you know you can call jQuery methods, without having to wrap them.

edited Dec 27 '11 at 15:27

Lightness Races in Orbit

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answered Jul 24 '09 at 21:24

Ben Crouse

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+1. $ prefix to your variable indicates that it's a jQuery object and you can call all the jQuery methods without thinking whether you need to pass the variable to the jQuery method. – SolutionYogi Jul 24 '09 at 21:28

score 11 · Answer 4 · answered Jul 24 '09 at 21:26

11

RichieHindle is correct. To expand:

Javascript variables allow the '$' character. So for example, you could have the following:

var $i = 1;
var i = 1;

Both i and $i have the same value, and both are perfectly legitimate variable names.

jQuery assigns itself (the jQuery object) to '$' because traditionally, that's what Javascript frameworks have done for selectors. There's no inherent meaning to '$' beyond what jQuery gives it.

answered Jul 24 '09 at 21:26

thedz

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I think JavaScript frameworks chose `$` for this because most people were not really aware it could be used in variable names so it rarely clashed when introducing frameworks into existing code. – hippietrail Jan 06 '12 at 11:13

score 9 · Answer 5 · edited Jul 02 '11 at 14:09

9

I like to prefix all my jQuery object names with $ so that I know it's actually a jQuery object, not a DOM reference.

It's a good naming convention.

edited Jul 02 '11 at 14:09

Lucas Jones

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answered Jul 24 '09 at 21:25

mkoryak

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score 3 · Answer 6 · edited Dec 27 '11 at 15:28

3

In this case, $selector is an example of Hungarian notation. It is there so that you know that $selector is a jQuery object.

If you make it your custom to always have jQuery objects start with $ if you suddenly use one wrong it will look wrong which will help you find a problem in your code.

The language couldn't care less. They could have just as well named the jQuery function J or anything else; the $ is not a special symbol.

edited Dec 27 '11 at 15:28

Lightness Races in Orbit

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answered Mar 25 '11 at 09:24

Zachary K

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Of course, strictly speaking, it doesn't let you "know" that `$selector` is a jQuery object; you could just as easily get it wrong and write `var $selector = 0;`. So it's no guarantee. – Lightness Races in Orbit Dec 27 '11 at 15:29
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OK to be fair it lets you know that $selector *SHOULD* be a jquery object – Zachary K Dec 27 '11 at 19:22

score 2 · Answer 7 · answered Jul 24 '09 at 21:24

2

Yeah.

jQuery always returns a jQuery object. It's what allows chaining.

You've already defined the variable so you get a jQuery object back so $ is unnecessary

answered Jul 24 '09 at 21:24

easement

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score -1 · Answer 8 · answered Jul 24 '09 at 21:25

-1

I don't believe that you need to add the jQuery selector to jQuery objects (since it's already a part of the object). We don't add the selector and haven't run into any problems.

answered Jul 24 '09 at 21:25

Grant Heaslip

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It's not "adding the selector". – Lightness Races in Orbit Dec 27 '11 at 15:27

score -2 · Answer 9 · answered Jul 24 '09 at 21:26

-2

"$" is a function in jQuery. So when you call $(selector), you're actually calling the function $ with selector as the argument.

Generally, don't use the "$" as part of a variable name for javascript. You will only confuse yourself.

answered Jul 24 '09 at 21:26

vh.

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$ is not a function, just an alias. Also, using $ as part of a variable name is a good practice to identify that that variable is a jQuery variable. – ricardohdz May 14 '13 at 18:28

The $ dollar sign

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