[英]Typescript template literal for object key mapping
I'm trying to create a generic type that would map the keys using template literals.我正在尝试创建一个通用类型,该类型将使用模板文字 map 键。 In general i just want all the keys listed in the generic type to be present in the output type, but modified slightly.
一般来说,我只希望通用类型中列出的所有键都出现在 output 类型中,但稍作修改。 Something along the lines of:
类似的东西:
type TFoobar = "foo" | "bar";
const foobar: TFlagDict<TFoobar> = {
"foo_flag": true,
"bar_flag": true
};
I have tried implementing it like so:我试过像这样实现它:
type TFlagDict<TProperties extends string> = {
[key in TProperties]: {[k in `${key}_flag`]: boolean}
}[TProperties]
And while it does have a proper typing, it makes the properties optional (at least one is required, but there is nothing that enforces they are all present)虽然它确实有一个正确的类型,但它使属性成为可选的(至少一个是必需的,但没有任何东西强制它们都存在)
const foo: TFlagDict<TFoobar> = {
"foo_flag": true
}; //valid, shouldnt be
const bar: TFlagDict<TFoobar> = {
"bar_flag": true
}; //valid, shouldnt be
const foobar: TFlagDict<TFoobar> = {
"foo_flag": true,
"bar_flag": true
}; //valid
You were close:你很接近:
type TFoobar = "foo" | "bar";
const foobar: TFlagDict<TFoobar> = {
"foo_flag": true,
"bar_flag": true
};
type TFlagDict<TProperties extends string> = {
[key in TProperties as `${key}_flag`]: boolean // key remapping
}
type O = TFlagDict<TFoobar>
const foo: TFlagDict<TFoobar> = {
"foo_flag": true
}; //error
const bar: TFlagDict<TFoobar> = {
"bar_flag": true
}; //error
const foobar2: TFlagDict<TFoobar> = {
"foo_flag": true,
"bar_flag": true
}; //valid
You are allowed to use as
inside iterator.您可以使用
as
内部迭代器。 See docs for more context有关更多上下文,请参阅文档
You can use TFooBar
instead of key
to enforce such validation:您可以使用
TFooBar
而不是key
来强制执行此类验证:
type TFlagDict<TProperties extends string> = {
[key in TProperties]: {[k in `${TFoobar}_flag`]: boolean}
}[TProperties]
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