node.js - 如何在使用promise时分别运行多个函数？

Question

我正在尝试按顺序运行 4 个函数，我已经尝试了下面的代码，但在运行第一个和第二个函数后，函数如何卡住了

代码示例:

const Bluebird = require('bluebird');

///// also tried changing all Bluebird below to Promise -> didn't work

//const Promise = require('bluebird');

const promisesFunc = async (array) => {
    let interval = 1000;
    const delayPromise1 = (data, delayDuration) => {
      return new Promise((resolve) => {
        setTimeout(() => {
            /// do some code here requires .map function may take 10s or more
            resolve();
        }, delayDuration)
      });
    };

    const delayPromise2 = (data, delayDuration) => {
      return new Promise((resolve) => {
        setTimeout(() => {
            /// do some code here requires .map function may take 10s or more
            resolve();
        }, delayDuration)
      });
    };

    const delayPromise3 = (data, delayDuration) => {
      return new Promise((resolve) => {
        setTimeout(() => {
            /// do some code here requires .map function may take 10s or more
            resolve();
        }, delayDuration)
      });
    };

    const delayPromise4 = (data, delayDuration) => {
      return new Promise((resolve) => {
        setTimeout(() => {
            /// do some code here requires .map function may take 10s or more
            resolve();
        }, delayDuration)
      });
    };

    try {
    /////////////// first attempt //////////////
    await Bluebird.map(array, (data, index) => delayPromise1(data, index * interval))
    await Bluebird.map(array, (data, index) => delayPromise2(data, index * interval))
    await Bluebird.map(array, (data, index) => delayPromise3(data, index * interval))
    await Bluebird.map(array, (data, index) => delayPromise4(data, index * interval))
    console.log('done ***************************************************');
    setTimeout(() => {
      console.log('response was sent');
      res.status(200).json('done')
    }, 1000);

    /////////////// second attempt ////////////

    const promises = Bluebird.map(array, (data, index) => delayPromise1(data, index * interval))
      .then(() => Bluebird.map(array, (data, index) => delayPromise2(data, index * interval)))
      .then(() => Bluebird.map(array, (data, index) => delayPromise3(data, index * interval)))
      .then(() => Bluebird.map(array, (data, index) => delayPromise4(data, index * interval)))
      .then(() => {
        setTimeout(() => {
          console.log('response was sent');
          res.status(200).json('done')
        }, 1000);
      })
      .catch(err => console.error(err))

      await Promise.all([promises]);

      ///////////// third attempt ////////////////////

      const promises1 = array.map((data, index) => delayPromise1(data, index * interval));
      const promises2 = array.map((data, index) => delayPromise2(data, index * interval));
      const promises3 = array.map((data, index) => delayPromise3(data, index * interval));
      const promises4 = array.map((data, index) => delayPromise4(data, index * interval));
      await Promise.all([promises1, promises2, promises3, promises4]);
      setTimeout(function(){
        console.log('response was sent');
        res.status(200).json('done')
      }, 1000);


    } catch (e) {
      console.error(e);
    }
}
promisesFunc(array)

注意在第一次和第二次尝试中，delayPromise1和delayPromise2函数成功运行，但随后停止并且不继续，其他解决方案分别不起作用

知道为什么会发生这种情况吗？是否有更好的解决方案来保证这些函数分别运行。

PS函数结构需要像这样。并且需要分别按顺序运行。

映射到数据数组中，然后在delayPromise函数中对其进行一些处理，然后resolve()。在某些情况下，解决 promise 不必等待代码执行完成。

Answer 1

你的第三次尝试是正确的选择。但是，您可以在 Promise.all 解析并且所有异步操作完成后发送响应。

const promises1 = array.map((data, index) => delayPromise1(data, index * interval));
const promises2 = array.map((data, index) => delayPromise2(data, index * interval));
const promises3 = array.map((data, index) => delayPromise3(data, index * interval));
const promises4 = array.map((data, index) => delayPromise4(data, index * interval));

await Promise.all([promises1, promises2, promises3, promises4]).then(() => {
  console.log('response was sent');
  return res.status(200).json('done')
})