javascript - 如何在 Mongoose 中只获得唯一值？

Question

我只需要从集合中获取唯一值。例如:

const userID = `user1`;
const users = await Chat
  .find({'$or': [{to: userID}, {from: userID}]})
  .select(`-_id to from`)
  .lean();

// users will be like:
[
 {from: `user1`, to: `user2`},
 {from: `user1`, to: `user2`},
 {from: `user1`, to: `user2`},
 {from: `user2`, to: `user1`},
 {from: `user3`, to: `user1`},
 // ... 10089 more items
];

// and I want this in result 
const result = [`user2`, `user3`]; // exclude current user too

我知道我可以使用 JS 做到这一点。我可以从用户创建一个数组并运行 new Set() ，但它会很慢。 Mongoose 可以代替我这样做吗？

Answer 1

您可以尝试聚合查询，

$match您的情况

$group通过 null 并构造 from 的唯一数组用户和 to用户

$setUnion获得独一无二的users来自 from和 to数组

$filter迭代上述联合数组的循环并删除当前用户

const userID = mongoose.Types.ObjectId(`user1`);
const users = await Chat.aggregate([
  { $match: { $or: [{ to: userID }, { from: userID }] } },
  {
    $group: {
      _id: null,
      from: { $addToSet: "$from" },
      to: { $addToSet: "$to" }
    }
  },
  {
    $project: {
      _id: 0,
      users: {
        $filter: {
          input: { $setUnion: ["$from", "$to"] },
          cond: { $ne: ["$$this", userID] }
        }
      }
    }
  }
]).exec();

console.log(users[0].users);

Playground