“未知”类型的参数不可分配给“{}”类型的参数

这是我的代码

 const Res = await fetch(`https://foo0022.firebaseio.com/.json`);
        const ResObj = await Res.json();
        if (!Res.ok || !ResObj) { 
          throw new Error("Page Not Found 404");
        } 
        const ResArr = await Object.values(ResObj)
            .map(v => Object.values(v).flat())//error
            .flat()
            .filter(({ title }) => title.includes(Search))

在该行中，我收到此错误“.map (v => Object.values (v) .flat ())”，我收到此错误 Argument of type 'unknown' is not assignable to parameter of type '{}' .这个问题如何解决？

这里的问题是你需要帮助 TypeScript 理解你正在处理的对象的类型。这fetchAPI 无法提前知道返回对象的形状，因此您必须定义它并断言结果符合它。

看看有什么https://foo0022.firebaseio.com/.json https://foo0022.firebaseio.com/.json，我建议如下：

interface ResObj {
  Mens: {
    Hat: Clothing[];
    Jacket: Clothing[];
    Pants: Clothing[];
    Shoes: Clothing[];
    Suit: Clothing[];
  };
  New: Clothing[];
}
interface Clothing {
  agility: boolean[];
  alt: string;
  color: string[][];
  id: string;
  location?: string; // fix this
  Location?: string; // fix this
  material: string;
  price: string[][];
  prodState: string;
  saiz: string[][];
  shipping: string;
  sold: string;
  src: string[][];
  title: string;
  to: string;
}

当然，这是否准确取决于某种 API 文档。假设这是正确的，你可以更进一步：

  const Res = await fetch(`https://foo0022.firebaseio.com/.json`);
  const ResObj: ResObj | undefined = await Res.json();
  if (!Res.ok || !ResObj) {
    throw new Error("Page Not Found 404");
  }

Now ResObj将被称为类型ResObj你就可以开始操纵它了。一个问题是标准库的类型Object.values() https://github.com/microsoft/TypeScript/blob/v3.5.1/lib/lib.es2017.object.d.ts#L22-L32 and Array.prototype.flat() https://github.com/microsoft/TypeScript/blob/v3.5.1/lib/lib.es2019.array.d.ts#L152-L223不要反映你正在和他们一起做什么。我们可以为它们构建一些自定义类型...但在这种情况下，我将用类型匹配的新函数包装它们：

  // return an array of all object values...
  // if the object is already an array, the output is the same type.
  // otherwise it's the union of all the known property types
  function vals<T extends object>(
    arr: T
  ): Array<T extends Array<infer U> ? U : T[keyof T]> {
    return Object.values(arr); // need es2017 lib for this
  }

  // Flatten an array by one level... 
  function flat<T>(
    arr: Array<T>
  ): Array<Extract<T, any[]>[number] | Exclude<T, any[]>> {
    return arr.flat(); // need esnext lib for this
  }

如果您以前从未使用过 TypeScript，这些函数类型可能会令人困惑，特别是因为它们依赖于条件类型 https://github.com/Microsoft/TypeScript/wiki/What%27s-new-in-TypeScript#conditional-types梳理出数组属性。

然后我们可以像这样重写你的代码：

  const ResArr = flat(vals(ResObj).map(v => flat(vals(v)))).filter(
    ({ title }) => title.includes(Search)
  );

并且没有错误，并且编译器理解ResArr是一个数组Clothing对象。

链接到代码 https://typescript-play.js.org/#code/JYOwLgpgTgZghgYwgAgEoQM4HkBGArZAbwChlkBZCEDALiNLOQAk4w6BhAGwHswALUAHMA2gF0A3AzIApRAGsIbZF14CQIiVOQAFOOFrKe-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

好的，希望有帮助；祝你好运！