I know it's a silly question but I am a bit confused with this. For example, if I have an input with an ID: rad1, is there any difference between below lines of code?
var $a = $('#rad1') or
var a = $('#rad1') 
- 2possible duplicate of Why would a JavaScript variable start with a dollar sign?Nikolay Kostov– Nikolay Kostov2015-07-27 11:46:25 +00:00Commented Jul 27, 2015 at 11:46
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6 Answers
No there is no real difference.
It's just a convention that helps you remember that a isn't the DOM element but it's a jQuery object.
var a = document.getElementById('a'); a.innerHTML //fine var $a = $('#a'); $a.html() // fine Ohhh, and by the way, neither a or $a are good variable names ... you should use meaningful variable names not abc characters.
Read the jQuery info tag on this very same site:
Variable Naming Conventions
jQuery wrapped variables are usually named starting with '$' to distinguish them from standard JavaScript objects.
var $this = $(this); 
2 Comments
swdev
I come from java background.. and this answer really enlightened my day.I just know that in JS I can use $ in var name..
zb226
Funny thing is, you can use $ in Java variables as well.
It's only for showing that it's a Jquery variable.
Declaring $a you're showing that your variable is for JQuery objects, it's just a notation. So the most readable thing will be to declare Jquery variable with $ notation
var $obj=$("#obj"); And DOM element without $ notation
var obj = document.getElementById("obj"); 
Comments
There's no difference. It's just a coding convention to help identify that the variable represents a jquery wrapped object.
Comments
No difference its just coding convention , check this
Comments
I think this scenario should illustrate the reason for remembering(off course, assigning '$' ) jquery and simple javascript variables:
<form id='myform' > <input id="name" type="text" value="Peter" /> </form> <script> $(document).ready(function(){ var name = document.getElementById('name'); var $name = $('#name'); console.log(name.value); // javascript's properties are available console.log($name.value); //it is undefined (worth of notice) console.log(name.val()); // error! name.val is not a function ! (jquery function will not be available) console.log($name.val()); // jquery functions are available }); </script> 
Comments
A composition also works fine:
You can also do something like this to show a <div>:
function getCompTable(divId){ var $d = $('#' + divId); $d.show(); } USAGE
getCompTable('compListDiv'); // compListDiv - is a div id="" 
