CODEWITHSUNDEEP
javascriptjqueryfunctionobject
bento
bento

Reputation: 5026

applying functions to objects in javascript

Excuse the beginner question, but I'm having difficulty wrapping my head around this. Using jquery I can do something like this:

$.each(array, testF); 

function testF(){
    console.log(this.name);  
}

Using $.each I can pass the array to the the function testF, and access the object contained within each row of the array. But when I try to do something like:

array[0].testF()

I get an error stating that the Object has no method 'testF'. I'm wondering if someone can explain what's happening here... why the jQuery one works, but the second doesn't. Again, please excuse the beginner question.

Upvotes: 3

Views: 110

Answers (4)

ThiefMaster
ThiefMaster

Reputation: 318518

  • $.each(array, testF) calls testF(i, elem) for each element in the array with this bound to elem.
  • array[0].testF() tries to call a method testF() that exists on array[0]

The equivalent of what $.each() does is testF.call(array[0], 0, array[0]) or since you do not use the arguments you can simpl do testF.call(array[0])

See MDN for the documentation of .call().

Upvotes: 4

Wampie Driessen
Wampie Driessen

Reputation: 1710

this is because the object array[0] has no function testF, You will need to declare that first. like so:

array[0].testF = function testF()
{
    console.log(this.name);
}

array[0].testF();

Upvotes: 1

Amadan
Amadan

Reputation: 198324

When you write array[0].testF(), this means:

  • Look up the variable array
  • Take the element under the index "0"
  • Within that, take the element under the index "testF"
  • and execute it as a function (also assigning the special variable this to array[0])

However, in most circumstances, array[0] will not have an attribute testF defined. If you were to do this beforehand:

array[0].testF = testF;

then it would work.

Upvotes: 1

Šime Vidas
Šime Vidas

Reputation: 185933

Do it like so:

testF.call( array[0] );

The call function method enables you to set the context (the this value) of the function invocation. So, whenever you want to invoke a function with a custom context, instead of fn();, do fn.call( context );.

Upvotes: 2

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