Typescript 按参数值过滤对象答案

【问题标题】：Typescript filter object by value of argumentTypescript 按参数值过滤对象
【发布时间】：2023-02-06 06:18:47
【问题描述】：

我有一个受 Prisma 启发的函数，它从对象生成 SQL 查询字符串，然后发出 SQL 查询请求并返回检索到的对象。

这是代码的 typescript playground (original code, see new below) 最小复制。

我目前正在使用泛型来设置预期的输出，但它总是返回完整的对象，即使应用了 select 也是如此。

如果有的话，有没有办法返回根据提供的选择对象过滤的输出类型？我试过使用 keyof (typeof query)["select"] 但这会获取选择类型的键，而不是运行时提供的值。

更新：我在这方面取得了一些进展，我能够让 output2 和 output3 输出正确的类型，但仍然不是 output1。这是一个带有更新代码的新typescript playground链接，我已经更新了帖子中的代码。

游乐场代码：

type ExpectedOutput = {
    aField: string;
    bField: number;
    cField: string;
    dField: number;
    eField: string;
    fField: number;
    gField: string;
}

type ObjectOrTrue<Type> = Type extends Record<string, any>
    ? { [Property in keyof Type]: ObjectOrTrue<Property> }
    : true;

async function myFunction<
    Type extends Record<string, any> = Record<string, unknown>
>(query: {
    select?: Partial<{ [Property in keyof Type]: ObjectOrTrue<Type[Property]> }>;
    from: string;
    where?: Partial<{ [Property in keyof Type]: Type[Property] }>;
    order?: Partial<{ [Property in keyof Type]: "ASC" | "DESC" }>;
    limit?: number;
    offset?: number;
}) {
  const {select} = query;

  // Simulated output of function
  if(select === undefined) {
    console.log('select is undefined');
    console.log({query});
    return {} as Type;
  }
  return {} as {[Property in keyof typeof select]: Type[Property]};
}
async function runAllTests() {
  const output1 = await myFunction<ExpectedOutput>({
    select: {
      aField: true,
      bField: true,
      cField: true,
    },
    from: 'SomeTable',
  });
  /*
  output1 type === ExpectedOutput, but I would like it to be
    {
      aField: string,
      bField: number,
      cField: string,
    }
  */
  const output2 = await myFunction({
    select: {
      aField: true,
      bField: true,
      cField: true,
    },
    from: 'SomeTable',
  });
  /*
  output2 type === {
      aField: unknown,
      bField: unknown,
      cField: unknown,
    } 
    which is what it should be.
  */
  const output3 = await myFunction<ExpectedOutput>({
    from: 'SomeTable',
  });
  /*
  output3 type === ExpectedOutput which is what it should be.
  */
}
runAllTests();

【问题讨论】：

标签： typescript

【解决方案1】：

每当返回类型取决于参数类型时，我通常会求助于额外的泛型。但是，没有办法传递一个泛型并推断出其他泛型，所以我决定将 myFunction 设为工厂作为一种 hack。

如果this proposal 通过，我们将能够恢复工厂。

type ObjectOrTrue<Type> = Type extends Record<string, any>
  ? { [Property in keyof Type]: ObjectOrTrue<Property> }
  : true;

function myFunction<
  Type extends Record<string, any> = Record<string, unknown>
>() {
  return async <
    Select extends Partial<{ [Property in keyof Type]: ObjectOrTrue<Type[Property]> }> | undefined = undefined,
    MyReturnType = Select extends undefined
      ? Type
      : string extends keyof Type
      ? { [Property in keyof Select]: unknown }
      : { [Property in Extract<keyof Type, keyof Select>]: Type[Property] }
  >(query: {
    select?: Select;
    from: string;
    where?: Partial<{ [Property in keyof Type]: Type[Property] }>;
    order?: Partial<{ [Property in keyof Type]: "ASC" | "DESC" }>;
    limit?: number;
    offset?: number;
  }): Promise<MyReturnType> => {
    const { select } = query;

    // Simulated output of function
    if (select === undefined) {
      console.log('select is undefined');
      console.log({ query });
      return {} as never;
    }

    return {} as never;
  }
}

不得不修改runAllTests，我所做的就是在每个调用中添加()：

const output1 = await myFunction<ExpectedOutput>()({
  select: {
    aField: true,
    bField: true,
    cField: true,
  },
  from: 'SomeTable',
});
const output2 = await myFunction()({
  select: {
    aField: true,
    bField: true,
    cField: true,
  },
  from: 'SomeTable',
});
const output3 = await myFunction<ExpectedOutput>()({
  from: 'SomeTable',
});

【讨论】：