这个javascript对象数组如何工作？

Question

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?

Answer 1

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"];

Answer 2

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..

Answer 3

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";

Answer 4

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. Eg:

var name = "test";
topics[name] = 123; // topics["test"] now exists

Answer 5

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

Answer 6

如果您的代码没有导致ReferenceError ，您可能已经在名为name全局作用域中创建了一个变量，并创建了name的内容的新属性。