打字稿中的关键字名称 [英] nameof keyword in Typescript

问题描述

正如我所见，没有原生的 nameof-keyword 就像 C# 内置在 TypeScript 中一样.但是，出于与 C# 中存在的相同原因，我希望能够以类型安全的方式引用属性名称.

As I have seen, there is no native nameof-keyword like C# has built into TypeScript . However, for the same reasons this exists in C#, I want to be able to refer to property names in a type safe manner.

当使用 jQuery 插件时，这在 TypeScript 中特别有用(Bootstrap-Tagsinput) 或其他需要配置属性名称的库.

This is especially useful in TypeScript when using jQuery plugins (Bootstrap-Tagsinput) or other libraries where the name of a property needs to be configured.

它可能看起来像:

const name: string = nameof(Console.log);
// 'name' is now equal to "log"

当Console.log被重构和重命名时，name的分配也应该改变.

The assignment of name should change too when Console.log got refactored and renamed.

目前在 TypeScript 中使用此类功能最接近的方法是什么?

推荐答案

正如您已经说过的，从版本 2.8 开始，TypeScript 没有内置功能.但是，有一些方法可以获得相同的结果:

As you have already said, there is no built in functionality on TypeScript as of version 2.8. However, there are ways to get the same result:

ts-nameof 是一个提供与 C# 完全相同的功能的库.有了这个，你可以做到:

ts-nameof is a library that provides the exact functionality as C# does. With this you can do:

nameof(console); // => "console"
nameof(console.log); // => "log"
nameof<MyInterface>(); // => "MyInterface"
nameof<MyNamespace.MyInnerInterface>(); // => "MyInnerInterface"

ts-simple-nameof 提供了另一种选择.它基本上解析一个字符串化的 lambda 来找出属性名称:

ts-simple-nameof offers an alternative. It basically parses a stringified lambda to figure out the property name:

nameof<Comment>(c => c.user); // => "user"
nameof<Comment>(c => c.user.posts); // => "user.posts"

选项 2:定义辅助函数

您可以轻松定义自己的 nameof 来添加类型检查，但是它不会自动重构，因为您仍然需要输入字符串文字:

Option 2: Define a helper function

You can easily define your own nameof that adds the type checking, however it will not refactor automatically as you'll still need to type a string literal:

const nameof = <T>(name: keyof T) => name;

它将返回传递的属性名称，但当属性名称在 T 类型上不存在时，将生成编译时错误.像这样使用它:

It will return the passed property name but will generate a compile time error when the property name does not exist on type T. Use it like so:

interface Person {
    firstName: string;
    lastName: string;
}

const personName1 = nameof<Person>("firstName"); // => "firstName"
const personName2 = nameof<Person>("noName");    // => compile time error

关于此的信用和更多信息

类型 keyof T 现在不仅解析为字符串，而且解析为 string |数量 |符号 (参考).如果您仍然只想解析字符串，请改用此实现:

The type keyof T now not only resolves to a string, but to string | number | symbol (ref). If you still want to resolve strings only, use this implementation instead:

const nameof = <T>(name: Extract<keyof T, string>): string => name;

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