javascript - Javascript 中的任务运行器

Question

我有一个任务列表，只有在解决每个任务的所有依赖关系后才需要执行所有这些任务。我正在努力找出一种方法来在最佳时间完成所有任务的运行。

// Each node is a async job, illustrated by setTimeout.
// A and C can run at the same time.
// D, needs to wait for A and C to be done.
// E needs to wait for A and D to be done.

function runTasks(tasks) {
 // run tasks
}

// Sample of tasks
var tasks = {
  'a': {
    job: function (finish) {
      setTimeout(function () {
        console.log('a done');
        finish();
      }, 500);
    },
  },
  'c': {
    job: function (finish) {
      setTimeout(function () {
        console.log('c done');
        finish();
      }, 200);
    },
    dependencies: [],
  },
  'd': {
    job: function (finish) {
      setTimeout(function () {
        console.log('d done');
        finish();
      }, 100);
    },
    dependencies: ['a','c'],
  },
  'e': {
    job: function (finish) {
      setTimeout(function () {
        console.log('e done');
        finish();
      }, 200);
    },
    dependencies: ['a', 'd'],
  },
};

Answer 1

您可以启动可以启动的并行 promise 。

1. 首先启动所有没有依赖关系的独立任务
2. 完成独立任务后存储在 map /集合中
3. 然后使用整套重复步骤过滤一组新任务。

const wait = (ms, task) =>
  new Promise((resolve) =>
    setTimeout(() => {
      console.log(`${task} done`);
      resolve();
    }, ms)
  );

const tasks = {
  a: {
    job: () => wait(500, "A"),
    dependencies: [],
  },
  c: {
    job: () => wait(500, "C"),
    dependencies: [],
  },
  d: {
    job: () => wait(500, "D"),
    dependencies: ["a", "c"],
  },
  e: {
    job: () => wait(500, "E"),
    dependencies: ["a", "d"],
  },
};

const run = async () => {
  const completed = new Set();
  const keys = Object.keys(tasks);
  const runTasks = async (pendingTasks) => {
    const promises = pendingTasks.map((key) => tasks[key].job());
    await Promise.all(promises);
    pendingTasks.forEach((key) => completed.add(key));
  };

  const runner = async () => {
    const pendingTasks = keys.filter((key) => {
      if (!completed.has(key)) {
        const { dependencies } = tasks[key];
        return dependencies.every((dependency) => completed.has(dependency));
      }
      return false;
    });
    if (pendingTasks.length !== 0) {
      await runTasks(pendingTasks);
      return runner();
    }
  };

  await runner();
};

run();