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Omit

介绍​

实现内置的Omit<T,K>而不需要使用它.

通过从T中移除K属性来构建一个类型

For example

ts
interface Todo {
  title: string
  description: string
  completed: boolean
}
type TodoPreview = MyOmit<Todo, 'description' | 'title'>
const todo: TodoPreview = {
  completed: false,
}
ts
interface Todo {
  title: string
  description: string
  completed: boolean
}
type TodoPreview = MyOmit<Todo, 'description' | 'title'>
const todo: TodoPreview = {
  completed: false,
}
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起点​

ts
/* _____________ Your Code Here _____________ */
type MyOmit<T, K> = any
 
/* _____________ Test Cases _____________ */
type cases = [
  Expect<Equal<Expected1, MyOmit<Todo, 'description'>>>,
Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.
  Expect<Equal<Expected2, MyOmit<Todo, 'description' | 'completed'>>>
Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.
]
 
interface Todo {
  title: string
  description: string
  completed: boolean
}
 
interface Expected1 {
  title: string
  completed: boolean
}
 
interface Expected2 {
  title: string
}
ts
/* _____________ Your Code Here _____________ */
type MyOmit<T, K> = any
 
/* _____________ Test Cases _____________ */
type cases = [
  Expect<Equal<Expected1, MyOmit<Todo, 'description'>>>,
Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.
  Expect<Equal<Expected2, MyOmit<Todo, 'description' | 'completed'>>>
Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.
]
 
interface Todo {
  title: string
  description: string
  completed: boolean
}
 
interface Expected1 {
  title: string
  completed: boolean
}
 
interface Expected2 {
  title: string
}
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解决方案​

Spoiler warning // Click to reveal answer
ts
// 方法1
type MyOmit<T,K extends keyof T> = {
  [P in keyof T as P extends K ? never : P]: T[P]
}
ts
// 方法1
type MyOmit<T,K extends keyof T> = {
  [P in keyof T as P extends K ? never : P]: T[P]
}
ts
//方法2
/**
 * 方法2会在第三个测试中失败, 在 keyof T 被传递给 Exclude2 的时候。在这个阶段，TypeScript 只是处理了属性名的字符串，而不是保留了完整的属性描述符。
 */
type Exclude2<T,K extends T> = T extends K ? never : T;
 
type MyOmit2<T,K extends keyof T> = {
  [P in Exclude<keyof T, K>]: T[P]
}
 
ts
//方法2
/**
 * 方法2会在第三个测试中失败, 在 keyof T 被传递给 Exclude2 的时候。在这个阶段，TypeScript 只是处理了属性名的字符串，而不是保留了完整的属性描述符。
 */
type Exclude2<T,K extends T> = T extends K ? never : T;
 
type MyOmit2<T,K extends keyof T> = {
  [P in Exclude<keyof T, K>]: T[P]
}
 
ts
//方案2的正确版本
 
type Exclude3<T, K> = T extends K ? never : T;
 
type MyOmit3<T, K extends T> = {
  [P in keyof T as Exclude3<P, K>]: T[P]
}
 
ts
//方案2的正确版本
 
type Exclude3<T, K> = T extends K ? never : T;
 
type MyOmit3<T, K extends T> = {
  [P in keyof T as Exclude3<P, K>]: T[P]
}
 
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