通过将字段添加到 Omit 类型并结合泛型来创建原始类型

Question

在打字稿中，您可以通过添加使用Omit 删除字段来重新创建类型：

interface Animal {
  name: string
}

interface Dog extends Animal {
  goodboy: true
}

const makeDog = (name: string, attributes: Omit<Dog, 'name'>): Dog => ({
  ...attributes,
  name
})

但是，当您尝试使用泛型类型执行此操作时，它会失败并出现错误：

// Type 'Omit<T, "name"> & { name: string; }' is not assignable to type 'T'.
// 'Omit<T, "name"> & { name: string; }' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint 'Animal'.
const makeAnimal = <T extends Animal>(name: string, attributes: Omit<T, 'name'>): T => ({
  ...attributes,
  name
})

有没有办法创建makeAnimal 函数？

Typescript playground

Answer 1

这样做的原因是一般情况下不安全

interface Animal {
  name: string
}

interface Dog extends Animal {
  goodboy: true
}

interface Cat extends Animal {
  name: 'cat'
}

const makeAnimal = <T extends Animal>(name: string, attributes: Omit<T, 'name'>): T => ({
  ...attributes,
  name
})

// This would make a Cat named 'Mittens', while the interface says a Cat is always named 'cat'
const cat = makeAnimal<Cat>('Mittens', {})

您需要从基本类型中显式选择，以便在调用站点上将其合并：

const makeAnimal = <T extends Animal>(name: string, attributes: Omit<T, 'name'>): Omit<T, 'name'> & Pick<Animal, 'name'> => ({
  ...attributes,
  name
})

// OK
const fluffy: Dog = makeAnimal<Dog>('Fluffy', { goodboy: true })

// valid type error:
// Type 'Omit<Cat, "name"> & Pick<Animal, "name">' is not assignable to type 'Cat'.
// Types of property 'name' are incompatible.
// Type 'string' is not assignable to type '"cat"'.
const cat: Cat = makeAnimal<Cat>('Mittens', {})