当我映射和展平数组时，TypeScript 会错误地出错？

Question

我正在使用 GraphQL Codegen 从我的 GraphQL 模式生成 TypeScript 类型。这是我的查询，我使用了一个片段，因此它只有一种易于导出的类型：

query reservationsPage($schoolSettingId: Int!, $day: UnixDate!) {
  roomsForSchoolSettingAll(schoolSettingId: $schoolSettingId) {
    ...reservationsPage
  }
}

fragment reservationsPage on Room {
    id
    name
    children {
      id
      name
      registrationToday(day: $day) {
        id
        checkIn
        checkOut
        importantInformation
        plannedCheckOut
        absent
        pickUpBy {
          id
          name
        }
      }
    }
}

返回的数据如下所示：

const res = [
  {
    id: 1,
    __typename: "Room",
    name: 'Name 1',
    children: []
  },
  {
    id: 2,
    __typename: "Room",
    name: 'Name 2',
    children: [
      {
        id: 3,
        __typename: "ChildProfile",
        name: 'James',
        registration: [
          {
            id: 4,
            __typename: "Registration",
            checkIn: "06:00:00",
            checkOut: "14:00:00",
            absent: null,
            importantInformation: null,
            pickUpBy: null
            plannedCheckOut: null  
          }
        ]
      }
    ]
  }
]

我可以在新变量上使用这种类型，它按预期工作：

const childrenCurrent: ReservationsPageFragment['children'] = roomsForSchoolSettingAll.find(
    (room) => room.id === selectedRoomId,
  )?.children;

然而，当我尝试在另一个变量上使用它时，出现 TypeScript 错误：

const childrenAll: ReservationsPageFragment['children'] = roomsForSchoolSettingAll
  .map((room) => room.children)
  .flat();

TS2322: Type '(({ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }) | null | undefined)[]' is not assignable to type 'Maybe<{ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }>[]'. Type '({ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }) | null | undefined' is not assignable to type 'Maybe<{ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }>'. Type 'undefined' is not assignable to type 'Maybe<{ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }>'.

Answer 1

抱歉，但从你的问题来看，我无法理解你是在寻找错误原因的解释，还是只是寻找一种使其起作用的方法。

在第一种情况下，我不是 GraphQL 高手，所以你可以忽略我的回答，寻找了解 GraphQL 的人的答案。

在第二种情况下，您可以尝试不指定 childrenAll 类型：

const childrenAll = roomsForSchoolSettingAll
  .map((room) => room.children)
  .flat();

或使用类型转换：

const childrenAll: ReservationsPageFragment['children'] = roomsForSchoolSettingAll
  .map((room) => room.children)
  .flat() as unknown as ReservationsPageFragment['children'];

Answer 2

如果您在错误消息文本中进行替换，它会变得更具可读性：

TS2322:
Type '((X) | null | undefined)[]' is not assignable to type 'Maybe<X>[]'.
Type '(X) | null | undefined' is not assignable to type 'Maybe<X>'.
Type 'undefined' is not assignable to type 'Maybe<X>'.

其中 X 代表

{ __typename?: "ChildProfile" | undefined; } & Pick<ChildProfile, "id" | "name"> & { registrationToday?: ({ __typename?: "Registration" | undefined; } & Pick<...> & { ...; }) | null | undefined; }

现在我们可以看到编译器指示类型 Maybe<X> 不能保存值 undefined。
为了满足编译器，你有多种选择

过滤掉 undefined 个值
使用类型转换（如 Daniele Ricci 的回答中所述）
更改 Maybe 类型的定义以允许 undefined 值

如果您选择修改 Maybe 类型，在 GraphQL 代码生成器中有一个如何执行此操作的示例 documentation:

当我映射和展平数组时，TypeScript 会错误地出错？

TypeScript incorrectly errors when I map and flat an array?

typescript

graphql-codegen