How to check if a JavaScript variable is NOT undefined? [duplicate]

Question

Things I’ve tried that don’t seem to work:

if(lastName != "undefined")

if(lastName != undefined)

if(undefined != lastName)

if(lastName != undefined) This didn't work? Are you getting a ReferenceError in the console? If so, instead of avoiding the error by using typeof (as Crockford followers will suggest), you should be declaring your variables properly. — user1106925
– user1106925, Commented Apr 17, 2012 at 13:59
if "if(typeof lastName !== "undefined")" is not working for you, you may want to check your code for other problems — joelmdev
– joelmdev, Commented Apr 17, 2012 at 14:02
Any of the last three will work if you're coding properly. Your issue is elsewhere. — user1106925
– user1106925, Commented Apr 17, 2012 at 14:06
Why not simply inverse working condition ?? if(!(lastName === undefined)) It is true for null and false for undefined. — Jan
– Jan, Commented Nov 13, 2019 at 12:49

sbeliv01 · Accepted Answer · 2012-04-17 13:58:39Z

535

var lastname = "Hi";

if(typeof lastname !== "undefined")
{
  alert("Hi. Variable is defined.");
}

answered Apr 17, 2012 at 13:58

sbeliv01

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5 Comments

Jordan Over a year ago

Can't we just check it with if(lastname)... stackoverflow.com/questions/5113374/…

Allan Jardine Over a year ago

@Jordan No - because there are other values lastname could take which would also match that condition - e.g. 0, null, '' (empty string). These are called "falsy" values.

Dale Francis Over a year ago

@almcaffee a falsey value is not the same as undefined

Naveed Ali Over a year ago

more secure code is: if (lastname && typeof lastname !== "undefined"){ alert("Hi. Variable is defined."); } checking if 'lastname 'and is also not 'undefined'

Martin Fric Over a year ago

@NaveedAli no it is not. If lastname is false it wont work. And here we need only undefined

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