How to execute statement in javascript if variable defined [duplicate]

Question

Possible Duplicate:
How to check a not defined variable in javascript
Determining if a javascript object has a given property

In my beforeSend function i have this

$.myloader.show();

But some scripts dont have that element so my this line gives me error

$.myloader.show(); gives me error that $.myloader does not exists

How can i do something like

if($.myloader exists)
$.myloader.show();

Answer 1

The most generic and solid solution is :

if (typeof $.myloader != 'undefined') {

If you're sure your variable can't hold anything else than a function or undefined, you might use

if ($.myloader) {

But do this only when you're sure of the possible values because this test also match false, 0 and ''.

Answer 2

You can do it as folllow...

if(typeof $.myloader != 'undefined')
{    // your code here.  }; 

Answer 3

Undefined values are 'falsey', and evaluate to false in an if statement, so all you need is:

if( $.myloader )
    $.myloader.show();

Answer 4

Given that jQuery ($) is an object:

if ($.hasOwnProperty('myloader'))
{
    //should work
    //try: $.hasOwnProperty('fn'), it'll return true, too
}

if the myloader object isn't directly bound to the jQuery object, but via one of its prototypes:

if ('myloader' in $)
{
    //works, like:
    if ('hasOwnProperty' in $)
    {//is true, but it's not a property/method of the object itself:
        console.log($.hasOwnProperty('hasOwnProperty'));//logs false
    }
}

You can check for the methods you want to use in the same way:

if ('myloader' in $ && 'show' in $.myloader)
{
    $.myloader.show();
}

Answer 5

You can do this even more shortly:

$.myloader.show() ? console.log('yep') : console.log('nope');

Or:

$.myloader.show() || console.log('fail');

Answer 6

If you want it short, you can also do it like this:

$.myloader && $.myloader.show();

The first operand (before the &&) is called a guard. However some people argue that this is not readable and safe enough, so you might want to use the if (typeof ...) solution.