Mongodb 通过 ids 数组按顺序查找

发布于 2025-01-11 19:35:04 字数 808 浏览 0 评论 0原文

我正在制作一个约会应用程序，其中算法通过排序返回匹配用户 uuid 的数组。所以我想按该 uuid 数组按顺序查找所有用户。

我尝试这样做

//init algorithm
const algo = new Match(currentUser, users);
//calculate score
algo.calcScore();
//sort it
algo.sort();
//get array of uids of matching users
const matches = algo.matching_users;

//find users by that array of uid
const matching_users = await User
  .find({
    uid: {
      $in: matches
    }
  })
  .skip(total_items - items)
  .limit(items)
  .lean()
  .exec()

这会返回用户，但不按顺序返回。例如，让 uids 数组为：["c", "d", "a", "b"] 上面的查询返回以下位置的用户对象：[{...uid: "a"},{...uid: "b"},{...uid: "c"},{...uid : "d"}, ]

预期输出：[{...uid: "c"},{...uid: "d"},{...uid: "a"} ,{...uid: "b"}, ]

原文

I am making a dating app where an algorithm returns an array of matching user's uuid by sorting it.
So I want to find all the users by that array of uuid in order.

I tried doing this

//init algorithm
const algo = new Match(currentUser, users);
//calculate score
algo.calcScore();
//sort it
algo.sort();
//get array of uids of matching users
const matches = algo.matching_users;

//find users by that array of uid
const matching_users = await User
  .find({
    uid: {
      $in: matches
    }
  })
  .skip(total_items - items)
  .limit(items)
  .lean()
  .exec()

This returns the users but not in order.
For example let array of uids be: ["c", "d", "a", "b"]
the above query returns user objects in: [{...uid: "a"},{...uid: "b"},{...uid: "c"},{...uid: "d"}, ]

Expected output: [{...uid: "c"},{...uid: "d"},{...uid: "a"},{...uid: "b"}, ]

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