如何处理lodash.omitBy返回值的类型？

发布于 2025-01-12 22:34:58 字数 1473 浏览 1 评论 0原文

omitFalsyFields 泛型函数将删除值为 falsy 的键（''、undefined、null 等。 .) 的一个对象。

import _, { Dictionary } from 'lodash';

export function omitFalsyFields<T extends Dictionary<any>>(params: T) {
  return _.omitBy(params, (k) => !k);
}

interface CreateUserParams {
  name?: string;
  email: string;
}

const params: CreateUserParams = omitFalsyFields({ name: '', email: '[email protected]' })

我期望它的返回值的类型是CreateUserParams类型。 params 的值为 { email: '[email ;protected]' }，因此类型是兼容的使用CreateUserParams。

但是，出现 TS 类型错误：

“Dictionary”类型中缺少属性“email”，但“CreateUserParams”类型中需要属性“email”

TypeScript 游乐场

原文

omitFalsyFields generic function will remove the key which value is falsy('', undefined, null, etc...) of an object.

import _, { Dictionary } from 'lodash';

export function omitFalsyFields<T extends Dictionary<any>>(params: T) {
  return _.omitBy(params, (k) => !k);
}

interface CreateUserParams {
  name?: string;
  email: string;
}

const params: CreateUserParams = omitFalsyFields({ name: '', email: '[email protected]' })

I expect the type of its returned value is CreateUserParams type. The value of params is { email: '[email protected]' }, so the type is compatible with CreateUserParams.

But, got TS type error:

Property 'email' is missing in type 'Dictionary' but required in type 'CreateUserParams'

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黑色毁心梦 2025-01-19 22:34:58

如果您想使此函数通用，您可以执行以下操作：

export function omitFalsyFields<T>(params: T): Dictionary<T> {
  return _.omitBy(params, (k: any) => !k);
}

现在您可以传递任何类型，并且它应该应用于输入和输出

const params = omitFalsyFields<CreateUserParams>({
  name: "",
  email: "[email protected]"
})

但是对此需要注意的是 omitBy 返回一个 Dictionary，即使执行 Dictionary 它也总是会强制转换为 any，因为有时很难在运行时推断类型（如果不是不可能的话），因此如果您想分配一个风俗返回类型或将通用 T 类型指定为输出，请执行以下操作：

// note that T is an alias for you generic type
export function omitFalsyFields<T>(params: T) {
  // use the "as" keyword to override the returned type from omitBy.
  // With partial you change all properties of T to optional,
 // as omitBy removes falsy properties it makes sense to apply this utility
  return _.omitBy(params, (k: any) => !k) as Partial<T>;
}

const params = omitFalsyFields...
console.log(params.name) // undefined
console.log(params.email) // [email protected]

这样做，特定类型也会传递到输出，请查看完整示例。

If you want to make this function generic, you can do the following:

export function omitFalsyFields<T>(params: T): Dictionary<T> {
  return _.omitBy(params, (k: any) => !k);
}

Now you can pass any type and it should be applied to the input and output

const params = omitFalsyFields<CreateUserParams>({
  name: "",
  email: "[email protected]"
})

But one caveat about this is that omitBy returns a Dictionary<any>, even doing Dictionary<T> it will always casts to any because it's very hard to infer types at runtime if not impossible sometimes, therefore if you want to assign a custom return type or assign the generic T type as output, do the following:

// note that T is an alias for you generic type
export function omitFalsyFields<T>(params: T) {
  // use the "as" keyword to override the returned type from omitBy.
  // With partial you change all properties of T to optional,
 // as omitBy removes falsy properties it makes sense to apply this utility
  return _.omitBy(params, (k: any) => !k) as Partial<T>;
}

const params = omitFalsyFields...
console.log(params.name) // undefined
console.log(params.email) // [email protected]

Doing so, the specific type will be passed to the output as well, take a look at the complete example.

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