我想在Typescript对象中存储string ->字符串的映射,并强制所有值映射到字符串。例如:
var stuff = {};
stuff["a"] = "foo"; // okay
stuff["b"] = "bar"; // okay
stuff["c"] = false; // ERROR! bool != string
是否有一种方法让我强制值必须是字符串(或任何类型..)?
我想在Typescript对象中存储string ->字符串的映射,并强制所有值映射到字符串。例如:
var stuff = {};
stuff["a"] = "foo"; // okay
stuff["b"] = "bar"; // okay
stuff["c"] = false; // ERROR! bool != string
是否有一种方法让我强制值必须是字符串(或任何类型..)?
当前回答
实际上有一个内置的实用程序记录:
const record: Record<string, string> = {};
record['a'] = 'b';
record[1] = 'c'; // leads to typescript error
record['d'] = 1; // leads to typescript error
其他回答
var stuff: { [key: string]: string; } = {};
stuff['a'] = ''; // ok
stuff['a'] = 4; // error
// ... or, if you're using this a lot and don't want to type so much ...
interface StringMap { [key: string]: string; }
var stuff2: StringMap = { };
// same as above
interface AccountSelectParams {
...
}
const params = { ... };
const tmpParams: { [key in keyof AccountSelectParams]: any } | undefined = {};
for (const key of Object.keys(params)) {
const customKey = (key as keyof typeof params);
if (key in params && params[customKey] && !this.state[customKey]) {
tmpParams[customKey] = params[customKey];
}
}
如果你明白这个概念,请评论
定义接口
interface Settings {
lang: 'en' | 'da';
welcome: boolean;
}
强制键为设置界面的特定键
private setSettings(key: keyof Settings, value: any) {
// Update settings key
}
interface AgeMap {
[name: string]: number
}
const friendsAges: AgeMap = {
"Sandy": 34,
"Joe": 28,
"Sarah": 30,
"Michelle": "fifty", // ERROR! Type 'string' is not assignable to type 'number'.
};
在这里,接口AgeMap强制键为字符串,值为数字。关键字名称可以是任何标识符,应该用于建议接口/类型的语法。
你可以使用类似的语法来强制一个对象对union类型中的每个条目都有一个键:
type DayOfTheWeek = "sunday" | "monday" | "tuesday" | "wednesday" | "thursday" | "friday" | "saturday";
type ChoresMap = { [day in DayOfTheWeek]: string };
const chores: ChoresMap = { // ERROR! Property 'saturday' is missing in type '...'
"sunday": "do the dishes",
"monday": "walk the dog",
"tuesday": "water the plants",
"wednesday": "take out the trash",
"thursday": "clean your room",
"friday": "mow the lawn",
};
当然,您也可以将其设置为泛型类型!
type DayOfTheWeek = "sunday" | "monday" | "tuesday" | "wednesday" | "thursday" | "friday" | "saturday";
type DayOfTheWeekMap<T> = { [day in DayOfTheWeek]: T };
const chores: DayOfTheWeekMap<string> = {
"sunday": "do the dishes",
"monday": "walk the dog",
"tuesday": "water the plants",
"wednesday": "take out the trash",
"thursday": "clean your room",
"friday": "mow the lawn",
"saturday": "relax",
};
const workDays: DayOfTheWeekMap<boolean> = {
"sunday": false,
"monday": true,
"tuesday": true,
"wednesday": true,
"thursday": true,
"friday": true,
"saturday": false,
};
10.10.2018更新: 看看下面@dracstaxi的答案——现在有一个内置的类型Record,它可以帮你完成大部分工作。
1.2.2020更新: 我已经完全从我的回答中删除了预先制作的映射接口。@dracstaxi的回答让他们完全无关紧要。如果您仍然想使用它们,请检查编辑历史记录。
type KeyOf<T> = keyof T;
class SomeClass<T, R> {
onlyTFieldsAllowed = new Map<KeyOf<T>, R>();
}
class A {
myField = 'myField';
}
const some = new SomeClass<A, any>();
some.onlyTFieldsAllowed.set('myField', 'WORKS');
some.onlyTFieldsAllowed.set('noneField', 'Not Allowed!');