Reputation: 63588
I thought this would be as easy as:
if(typeof(Array.push) == 'undefined'){
//not defined, prototype a version of the push method
// Firefox never gets here, but IE/Safari/Chrome/etc. do, even though
// the Array object has a push method!
}
And it does work fine in Firefox, but not in IE, Chrome, Safari, Opera, they return all properties/methods of the native Array object as 'undefined' using this test.
The .hasOwnProperty( prop ) method only works on instances... so it doesn't work, but by trial and error I noticed that this works.
//this works in Firefox/IE(6,7,8)/Chrome/Safari/Opera
if(typeof(Array().push) == 'undefined'){
//not defined, prototype a version of the push method
}
Is there anything wrong with using this syntax to determine if a property/method exists on a Native Object / ~"JavaScript Class"~, or is there a better way to do this?
Upvotes: 25
Views: 49700
Reputation: 536409
And it does work fine in Firefox
That's only by coincidence! You can't generally expect a prototype's method to also exist on the constructor function.
if(typeof(Array().push) == 'undefined')
This was nearly right except you forget new
, a perennial JavaScript gotcha. new Array().push
, or [].push
for short, would correctly check an instance had the method you wanted.
Upvotes: 3
Reputation: 631
The proper way to check if a property exists:
if ('property' in objectVar)
Upvotes: 62
Reputation: 134
The .hasOwnProperty
can be accessed on the Array's proptotype, if typeof
is not idiomatic enough.
if (Array.prototype.hasOwnProperty('push')) {
// Native array has push property
}
Upvotes: 8
Reputation: 105898
First of all, typeof is an operator, not a function, so you don't need the parentheses. Secondly, access the object's prototype.
alert( typeof Array.prototype.push );
alert( typeof Array.prototype.foo );
When you execute typeof Array.push
you are testing if the Array object itself has a push method, not if instances of Array have a push method.
Upvotes: 35