javascript - Typescript 中的 reduce() 函数实例化空对象

Question

我正在尝试将我的 Javascript 函数转换为它的 Typescript 版本，但我无法摆脱读取的带下划线的代码 acc[curr.name] (在 VSCode 中)。将鼠标悬停在上面时说:

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{}'.No index signature with a parameter of type 'string' was found on type '{}'.

我按键名计算出现次数。

我应该在哪里定义类型，为什么空对象或带数字的对象不起作用。我不确定如何解决它，因为我尝试了多种方法，例如:

data.reduce<{}>()和 data.reduce<{property: number}>()

代码:

const data = [
  { name: "name1", },
  { name: "name1", },
  { name: "name1", },
  { name: "name1", },
  { name: "name2", },
  { name: "name2", },
  { name: "name2", },
  { name: "name1", },
];

// count data by key name
const resultData = data.reduce((acc, curr) => {
  return acc[curr.name] ? ++acc[curr.name] : (acc[curr.name] = 1), acc;
}, {});

运行结果:

const resultData = {
  name1: 4,
  name2: 3,
  name3: 1,
};

Answer 1

设置索引签名，如 data.reduce<{ [index: string]: number }>

引用:https://www.typescriptlang.org/docs/handbook/2/objects.html#index-signatures

const resultData = data.reduce<{ [index: string]: number }>((acc, curr) => {
  return acc[curr.name] ? ++acc[curr.name] : (acc[curr.name] = 1), acc;
}, {});

或者您可以使用 as具有初始值的关键字。

const resultData = data.reduce((acc, curr) => {
  return acc[curr.name] ? ++acc[curr.name] : (acc[curr.name] = 1), acc;
}, {} as { [index: string]: number });