为什么 TS 不让我索引到兼容的对象类型？

Question

我有以下类型定义。

type Key<Row> = {
  [P in keyof Row]: Row[P] extends string ? P : never
}[keyof Row];

type ID<Row> = Row[Key<Row>];

type Selected<Row> = {
  selected: boolean;
} & Row;

Key<Row> 返回Row 的所有属性键，其值为字符串。 See this answer 了解它的构造方式。

ID<Row> 是Row 的所有字符串值的并集。它始终是string，但如果字符串值是字符串常量，则可以很好地缩小范围，例如在

interface X {
  prop: "abc";
}

ID<X> 始终是"abc"。

Selected<Row> 是 Row 加上属性 selected: boolean 的交集类型。

以上都定义好了，我不明白为什么下面的函数定义会抛出错误：

const getId = <Row>(row: Selected<Row>, key: Key<Row>): ID<Selected<Row>> =>
  row[key];

TS 给我的类型错误粘贴在下面，但我不知道为什么 TS 不喜欢我的代码。

据我所见，row[key] 应该始终有效，因为即使 row 是 Selected<Row>，Key<Row> 只包含一组更窄的键，那么键入具有更多属性的对象有什么问题我们实际使用的密钥？

Here is a link to the TS playground with this code showing the error

---

Type 'Selected<Row>[{ [P in keyof Row]: Row[P] extends string ? P : never; }[keyof Row]]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)] & Row[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<...>["selected"] extends string ? "selected" : never)]'.
  Type 'Selected<Row>[Row[keyof Row] extends string ? keyof Row : never]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)] & Row[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<...>["selected"] extends string ? "selected" : never)]'.
    Type 'Selected<Row>[keyof Row]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)] & Row[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<...>["selected"] extends string ? "selected" : never)]'.
      Type 'Row[string] | Row[number] | Row[symbol]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)] & Row[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<...>["selected"] extends string ? "selected" : never)]'.
        Type 'Row[string]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)] & Row[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<...>["selected"] extends string ? "selected" : never)]'.
          Type 'Row[string]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)]'.
            Type 'Row' is not assignable to type '{ selected: boolean; }'.
              Type 'Row[string]' is not assignable to type 'boolean'.
                Type 'Selected<Row>[keyof Row]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)]'.
                  Type 'keyof Row' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                    Type 'keyof Row' is not assignable to type 'Selected<Row>[keyof Row] extends string ? keyof Row : never'.
                      Type 'Selected<Row>[keyof Row]' is not assignable to type 'boolean'.
                        Type 'Selected<Row>[Row[keyof Row] extends string ? keyof Row : never]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)]'.
                          Type 'Row[keyof Row] extends string ? keyof Row : never' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                            Type 'Row[keyof Row] extends string ? keyof Row : never' is not assignable to type 'Selected<Row>["selected"] extends string ? "selected" : never'.
                              Type 'Selected<Row>[Row[keyof Row] extends string ? keyof Row : never]' is not assignable to type 'boolean'.
                                Type 'Selected<Row>[{ [P in keyof Row]: Row[P] extends string ? P : never; }[keyof Row]]' is not assignable to type '{ selected: boolean; }[(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)]'.
                                  Type 'Row[keyof Row] extends string ? keyof Row : never' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                                    Type 'keyof Row' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                                      Type 'string | number | symbol' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                                        Type 'string' is not assignable to type '(Selected<Row>[keyof Row] extends string ? keyof Row : never) & (Selected<Row>["selected"] extends string ? "selected" : never)'.
                                          Type 'string' is not assignable to type 'Selected<Row>[keyof Row] extends string ? keyof Row : never'.
                                            Type 'keyof Row' is not assignable to type 'Selected<Row>[keyof Row] extends string ? keyof Row : never'.
                                              Type 'string | number | symbol' is not assignable to type 'Selected<Row>[keyof Row] extends string ? keyof Row : never'.
                                                Type 'string' is not assignable to type 'Selected<Row>[keyof Row] extends string ? keyof Row : never'.
                                                  Type 'Row[keyof Row] extends string ? keyof Row : never' is not assignable to type 'Selected<Row>["selected"] extends string ? "selected" : never'.
                                                    Type 'Selected<Row>[{ [P in keyof Row]: Row[P] extends string ? P : never; }[keyof Row]]' is not assignable to type 'boolean'.
                                                      Type 'Selected<Row>[Row[keyof Row] extends string ? keyof Row : never]' is not assignable to type 'boolean'.
                                                        Type 'Selected<Row>[keyof Row]' is not assignable to type 'boolean'.
                                                          Type 'Row[string] | Row[number] | Row[symbol]' is not assignable to type 'boolean'.
                                                            Type 'Row[string]' is not assignable to type 'boolean'.

Answer 1

编译器没有你那么聪明，尤其是在推理依赖于未解析的泛型参数的条件类型时（如NonSelected<Row>[P] extends string ? P : never）。如果您已经检查了逻辑并确定您正在做的事情是安全的，那么明智的type assertion 是有保证的：

const getId = <Row>(row: Selected<Row>, key: Key<Row>) =>
  row[key] as unknown as ID<Selected<Row>>;

或者，你可以给编译器一些它可以推理的东西，例如由K类型的键索引的O类型的对象将产生@987654326类型的值@：

const getId = <Row>(row: Selected<Row>, key: Key<Selected<Row>>): ID<Selected<Row>> =>
  row[key];

其中任何一个都应该安抚编译器，并且您的示例继续按预期运行。

希望对您有所帮助。祝你好运！