up vote 3 down vote favorite

I am trying to understand something very basic. If I have an object like this:

var topics = {}

And I do this:

topics[name] = ["chapter 1", "chapter 2", "chapter 3"];

When I log this object, I don't see the name attribute. What have I done exactly? Have I created a key called name with the value of an array?

Of course I know I can do that by just doing

topics.name =  ["chapter 1", "chapter 2", "chapter 3"];

But then what is this doing?

topics[name] = ["chapter 1", "chapter 2", "chapter 3"];

Could someone please clarify?

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73% accept rate
This doesn't answer you question but I believe you really want to ask the difference between topics[name] = ["chapter 1", "chapter 2", "chapter 3"]; and topics["name"] = ["chapter 1", "chapter 2", "chapter 3"];? – Matt Fellows Oct 21 '11 at 13:26
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6 Answers

up vote 3 down vote accepted

When you use the [] notation it expects an expression in between, that translates to a string.

Using name not enclosed in quotes 'name' it assumes you are using a variable called name.

Which in your case is undefined.

The correct usage would be

topics["name"] = ["chapter 1", "chapter 2", "chapter 3"];

If you want to use a variable you can do things like

var prop = 'name';
var topics = {};

topics[prop] = ["chapter 1", "chapter 2", "chapter 3"];

this will create the name property on the object.. useful for dynamic/automatic creation/filling of objects..

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Does that mean I can assign a function with expressions that evaluate to a string (like a topicName) or a number (like a topicID)? – Amit Erandole Oct 21 '11 at 13:53
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@Amit, yes you can do topics[func()] = '...'; if func() returns something that can be converted to string.. (if that is what you are asking..) – Gaby aka G. Petrioli Oct 21 '11 at 14:34
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up vote 3 down vote

Your are creating a property on the object with a name based on the value of the name variable. If you want to create a property called name in that way you need to do:

topics["name"] = ["chapter 1", "chapter 2", "chapter 3"];
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up vote 3 down vote

That should generate an error, unless you have variable name defined as a string or a number.

These three are equivalent:

var name = "key";
topics[name] = "value";

topics["key"] = "value";

topics.key = "value";
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up vote 1 down vote

If your code did not result in a ReferenceError, you may have already had a variable in global scope named name, and created a new property of name's contents.

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up vote 2 down vote

There are generally three ways to distinguish:

topics.name
topics["name"]
topics[name]

The first two are equivalent. So .xxx represents the literal key xxx, as does ["xxx"].

The last one will use whatever the name variable contains. E.g.:

var name = "test";
topics[name] = 123; // topics["test"] now exists
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up vote 2 down vote

The resulting structure can be seen in the following screen capture. As stated by mck89, you probably want to use the "name" syntax

enter image description here

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thanks for the comparative screenshots. That made things a lot clear – Amit Erandole Oct 21 '11 at 13:53
Don't use JPEG for screenshots. it makes them all fuzzy. – Dan D. Jan 18 at 19:41
@DanD. sorry about that... It usually outputs as a png... I wonder why it didn't that time. – Joseph Marikle Jan 18 at 20:29
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