打字稿：类型中缺少索引签名 [英] Typescript: Index signature is missing in type

问题描述

我希望 MyInterface.dic 像字典 name：value ，我对它的定义如下：

I want MyInterface.dic to be like a dictionary name: value, I define it as follows:

interface MyInterface {
    dic: { [name: string]: number }
}

现在，我创建一个函数来等待我的类型：

Now I create a function which waits for my type:

function foo(a: MyInterface) {
    ...
}

然后输入：

let o = {
    dic: {
        'a': 3,
        'b': 5
    }
}

我期望 foo（o）是正确的，但是编译器正在下降：

I'm expecting foo(o) to be correct, but the compiler is falling:

foo(o) // Typescript error: Index signature is missing in type { 'a': number, 'b': number }

我知道可能存在强制转换： let o：MyInterface = {...} 其中可以解决问题，但问题是，为什么打字稿无法识别我的打字稿？

I know there is a possible casting: let o: MyInterface = { ... } which do the trick but the question is, why typescript is not recognizing my type?

额外：如果声明 o 内联，则可以正常工作：

Extra: works fine if o is declared inline:

foo({ 
  dic: {
    'a': 3, 
    'b': 5
  }
})

推荐答案

问题是，当推断类型时， o 是：

The problem is that when the type is inferred, then the type of o is:

{ dic: { a: number, b: number } }

与 {dic不同：{[name：string ]：数字}} 。至关重要的是，使用最高签名的人不允许您执行 o.dic ['x'] = 1 之类的操作。有了第二个签名，就可以了。

That's not the same as { dic: { [name: string]: number } }. Critically, with the top signature you're not allowed to do something like o.dic['x'] = 1. With the 2nd signature you are.

它们在运行时是等效类型（实际上，它们是完全相同的值），但是TypeScript的安全性很大一部分来自于这些事实并不相同，并且只有在知道对象明确地被视为一个对象的情况下，才允许您将其视为字典。这就是阻止您意外读取和写入对象上完全不存在的属性的原因。

They are equivalent types at runtime (indeed, they're the exact same value), but a big part of TypeScript's safety comes from the fact that these aren't the same, and that it'll only let you treat an object as a dictionary if it knows it's explicitly intended as one. This is what stops you accidentally reading and writing totally non-existent properties on objects.

解决方案是确保TypeScript知道它打算用作字典。这意味着：

The solution is to ensure TypeScript knows that it's intended as a dictionary. That means:

• 在某处显式提供一种类型，以表明它是字典：

• Explicitly providing a type somewhere that tells it it's a dictionary:

let o：MyInterface

将其设置为字典内联：

Asserting it to be a dictionary inline:

let o = {dic：< {[名称：字符串]：数字}> {'a'：1，'b'：2}}

确保它是TypeScript为您推断的初始类型：

Ensuring it's the initial type that TypeScript infers for you:

foo（{dic：{'a'：1，'b'：2}}）

如果在某些情况下TypeScript认为它是仅具有两个属性的普通对象，然后尝试将其用作一本字典，会很不开心。

If there's a case where TypeScript thinks it's a normal object with just two properties, and then you try to use it later as a dictionary, it'll be unhappy.

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