Question

在一个快递项目中，我有2个映射，它们都通过一个操纵up实例运行，并且都返回数组。当前，我正在使用Promise.all在两个映射上等待完成，但是它只返回第一个数组的值，而不是第二个数组的值。我该如何做才能得到两个映射变量的结果？

const games = JSON.parse(JSON.stringify(req.body.games));

const queue = new PQueue({
  concurrency: 2
});

const f = games.map((g) => queue.add(async () => firstSearch(g.game, g.categories)));
const s = games.map((g) => queue.add(async () => secondSearch(g.game, g.categories)));

return Promise.all(f, s)
  .then(function(g) {
    console.log(g); //only returns `f` result, not the `s`
  });

Answer 1

Promise.all接受Promises数组作为参数。您需要将两个数组作为单个数组参数传递

return Promise.all(f.concat(s))
  .then(function(g) {
    console.log(g); //only returns `f` result, not the `s`
  });

Answer 2

不需要使用PQueue，bluebird已开箱即用地支持此操作：

(async () => {
  const games = JSON.parse(JSON.stringify(req.body.games));
  let params = { concurrency: 2};
  let r1 = await Promise.map(games, g => firstSearch(g.game, g.categories), params);
  let r2 = await Promise.map(games, g => secondSearch(g.game, g.categories), params);
  console.log(r1, r2);
 })();

或更正确，但包含更多代码（因此最后一次搜索不会等待）：

(async () => {
  const games = JSON.parse(JSON.stringify(req.body.games));
  let params = { concurrency: 2};
  let fns = [
    ...games.map(g => () => firstSearch(g.game, g.categories)),
    ...games.map(g => () => secondSearch(g.game, g.categories)),
  ];
  let results = await Promise.map(fns, fn => fn(), params);
  console.log(results);
 })();

等待多种地图方法以使用promise返回，然后获取所有地图的返回值

2 个答案: