Javascript isset() equivalent

Question

In PHP you can do if(isset($array['foo'])) { ... }. In Javascript you often use if(array.foo) { ... } to do the same, but this is not exactly the same statement. The condition will also evaluate to false if array.foo does exists but is null or false (and probably other values as well).

What is the perfect equivalent of PHP's isset in Javascript?

In a broader sense, a general, complete guide on Javascript's handling of variables that don't exist, variables without a value, etc. would be convenient.

Answer 1

I generally use the typeof operator:

if (typeof obj.foo != 'undefined') {
  // ..
}

It will return "undefined" either if the property doesn't exist or it's value is undefined.

_{(See also: Difference between undefined and not being defined.)}

There are other ways to figure out if a property exists on an object, like the hasOwnProperty method:

if (obj.hasOwnProperty('foo')) {
  //..
}

And the in operator:

if ('foo' in obj) {
  //..
}

The difference between the last two, is that the hasOwnProperty method, will check if the property exist physically on the object (the property is not inherited).

The in operator will check on all the properties reachable up in the prototype chain, e.g.:

var obj = { foo: 'bar'};

obj.hasOwnProperty('foo'); // true
obj.hasOwnProperty('toString'); // false
'toString' in obj; // true

As you can see, hasOwnProperty returns false and the in operator returns true when checking the toString method, this method is defined up in the prototype chain, because obj inherits form Object.prototype.

Answer 2

if (!('foo' in obj)) {
  // not set.
}

Answer 3

Should work for ya:

//Function to test if variable is unset/null
function isset(me)
{
 if (me == null || me == '')
  return 0;
 else
  return 1;
}

It does assume that you have at least declared the variable like a good programmer before hand (null if nothing else).