promise 内的多个 resolve() 调用返回异步 function

Question

我有以下 testAsync function 在我调用resolve()后应该返回已解决的 promise 。

async function wait(time) {
    await new Promise((resolve, reject) => { setTimeout(resolve, time) });
}

async function testAsync(testParam) {
    return new Promise(async (resolve, reject) => {
        try {
            await wait(2000); // do some async work
            if (testParam === 1) {
                console.log("test param is equal to 1");
                resolve("first IF");
            }
            if (testParam < 10) {
                console.log("test param is smaller than 10");
                resolve("second IF");
            }
        } catch (error) {
            reject(error);
        }
    });
}

(async () => {
    let result = await testAsync(1);
    console.log(result);
})();

控制台 output：

test param is equal to 1
test param is smaller than 10
first IF

我的问题是，如果我以一种包含更多非排他性的resolve()调用的方式构建代码，则代码会解析它遇到的第一个调用，但不会停止并返回该位置。 为什么会继续？ resolve()不是松散地等于return吗？ 我应该在每次resolve()调用后添加return吗？

Answer 1

return new Promise(async (resolve, reject) => { is antipattern. Adding async in front of a function will already return a Promise. new Promise has to be only used to convert a none Promise base async function into one that uses a Promise .

resolve() 不是松散地等于返回吗？

它只是在某种意义上相等，它设置了 promise 的“返回”值，但它不会中断执行流程。 所以是的，如果您想防止执行后续代码，您需要在解析后return 。

但正如我已经说过的，显示的构造无论如何都是反模式，并且testAsync应该如下所示：

async function testAsync(testParam) {
  await wait(2000); // do some async work

  if (testParam === 1) {
    console.log("test param is equal to 1");
    return "first IF";
  }

  if (testParam < 10) {
    console.log("test param is smaller than 10");
    return "second IF";
  }
}

wait更可能是这样写的：

function wait(time) {
    return new Promise((resolve, reject) => { setTimeout(resolve, time) });
}