javascript 调用异步时使用 reduce 进行映射

Question

我需要在映射函数回调中调用异步函数 f。我也想在之后应用 reduce 但无法使其工作:

https://www.typescriptlang.org/play?#code/MYewdgzgLgBAhjAvDA2gRgDQwEwF0DcAUHBAJ5jAwBmArhVAJbjUAUAlDAN4C+hANgFNYUAdCTwAdAFs4ABxaEYSpSXKUWADywMOiAHxdFy43ADucBrCrsix44NgA3OH3EbbdlYIBOUFs742IyUASG8hGm8wGACPJV42CXCAExpgARYWcIgaPigsAUEpATAoXQNs3NgAahhCgWLSoKA

  const a = [1, 2];
  async function f() {}
  let test = a.map(
       async (x, i) => {
          await f();
          let val = x;
          alert(val)
        return val;
      }
  ).reduce((result, element) => result + element)

Answer 1

使用Promise.all 等待异步映射，然后return reduce 的结果。请注意，在此示例中，test 也将 return 一个承诺，因此您需要 await 其结果才能记录它。

const a = [1, 2];
async function f() {}

async function test() {
  const mapping = a.map(async (x, i) => {
    await f();
    return x;
  });
  const result = await Promise.all(mapping);
  return result.reduce((acc, c) => acc + c, 0);
}

async function main() {
  console.log(await test())
}

main();

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