Question

我对mongo聚合还很陌生，所以我的鼻子在mongo文档中试图解决它。越来越近了！我正在对数据集运行一些查询，以从保存数据文档中获取当前结果。但是我得到的响应是集合中第一个结果的重复。

const result = await MemoryCard.aggregate([
    {
      $lookup: {
        from: 'users',
        localField: 'owner',
        foreignField: '_id',
        as: 'owner',
      },
    },
    {'$unwind': '$owner'},
    {$unwind: '$progress'},
    {$match: {
      'progress.createdAt': {
        $gte: d,
        $lte: new Date(),
      },
      'progress.file_one': {$eq: true},
      'progress.file_two': {$eq: true},
    }},
    {$unwind: '$profiles'},
    {
      '$project': {
        'owner.email': 1,
        'owner.username': 1,
        'progress': 1,
        'profiles': 1,
      },
    },
  ]);

结果我回来了，您可以知道第一位是重复的：

[
    {
        "_id": "5ca8a0bf6a45e10ad41a7fb3",
        "owner": {
            "email": "kevin2@s*******.com",
            "username": "kevin"
        },
        "profiles": {
            "data": [
                100,
                100,
                84,
                91,
                100
            ],
            "tools": [
                "5c7d4c9971338741c09c6c68",
                "5c7d4c9971338741c09c6c6b",
                "5c7d4c9971338741c09c6c74",
                "5c7d4c9971338741c09c6c73",
                "5c7d4c9971338741c09c6c75"
            ],
            "timestamp": "2019-04-06T12:51:07.151Z",
            "_id": "5ca8a0bf6a45e10ad41a7fb4",
            "profile_id": "5c864fd3f98eeb1afc9cc809",
            "createdAt": "2019-04-06T12:51:11.471Z",
            "updatedAt": "2019-04-06T12:51:11.471Z"
        },
        "progress": {
            "_id": "5caf4675999c0c14d0cda13e",
            "tool": "5c7d4c9971338741c09c6c66",
            "createdAt": "2019-04-11T13:51:49.352Z",
            "updatedAt": "2019-04-11T13:55:17.527Z",
            "file_two": true,
            "file_one": true
        }
    },
    {
        "_id": "5ca8a0bf6a45e10ad41a7fb3",
        "owner": {
            "email": "kevin2@s********.com",
            "username": "kevin"
        },
        "profiles": {
            "data": [
                100,
                100,
                84,
                91,
                100
            ],
            "tools": [
                "5c7d4c9971338741c09c6c68",
                "5c7d4c9971338741c09c6c6b",
                "5c7d4c9971338741c09c6c74",
                "5c7d4c9971338741c09c6c73",
                "5c7d4c9971338741c09c6c75"
            ],
            "timestamp": "2019-04-06T12:51:07.151Z",
            "_id": "5ca8a0bf6a45e10ad41a7fb4",
            "profile_id": "5c864fd3f98eeb1afc9cc809",
            "createdAt": "2019-01-06T12:51:11.471Z",
            "updatedAt": "2019-01-06T12:51:11.471Z"
        },
        "progress": {
            "_id": "5caf4675999c0c14d0cda13e",
            "tool": "5c7d4c9971338741c09c6c66",
            "createdAt": "2019-04-11T13:51:49.352Z",
            "updatedAt": "2019-04-11T13:55:17.527Z",
            "file_two": true,
            "file_one": true
        }
    }
]```

Where as i'm expecting multiple users.

Answer 1

因此，我通过意识到重复的条目是分开的并且必须分组来解决它的！在我的固定查询下面。

  SELECT u.id,
         COUNT(d.id) AS num_devices,
         SUM(messages) AS num_messages,
         MAX(most_recent) AS most_recent_message
    FROM users u
    JOIN devices d ON d.user_id = u.id
    JOIN (SELECT device_id,
                 COUNT() AS messages,
                 MAX(TIME) AS most_recent_message
            FROM messages
        GROUP BY device_id) m ON m.device_id = d.id
GROUP BY u.id

为什么我的汇总返回重复数据？

1 个答案: