Question

我是MongoDB集合，我按一个批次ID存储多个记录。我将批次ID保留为Date()对象的毫秒数。如何在单个MongoDB查询中检索具有最近/最大批次ID的所有记录。

现在，我正在两个顺序查询中检索记录：

查找最大/最近的批次ID-recentBatchID = db.users.find().sort( { 'batchID': -1 }).limit(1)
按最大/最近批次ID db.users.find({'batchID':recentBatchID})

我们如何在单个Mongo查询中进行优化？

样本数据：

[{
        "username": "user1",
        "batchID": "00001"
    },
    {
        "username": "user2",
        "batchID": "00001"
    },
    {
        "username": "user3",
        "batchID": "00001"
    },
    {
        "username": "user4",
        "batchID": "00002"
    },
    {
        "username": "user5",
        "batchID": "00002"
    },
    {
        "username": "user6",
        "batchID": "00002"
    },
]

我正在尝试仅获取最大/最近的user4, user5, user6的{{1}}。

Answer 1

根据新更新，您可以使用聚合，然后：

db.getCollection('tests').aggregate([
  {$group: {
     _id : "$batchID",
     doc : {$push : "$$ROOT"},
   }},
   {$sort : { '_id': -1 }},
   {$limit: 1},
   {$unwind: "$doc"}
])

输出：

/* 1 */
{
    "_id" : "00002",
    "doc" : {
        "_id" : ObjectId("5bd2dcced79cc5d8b1c6296e"),
        "username" : "user4",
        "batchID" : "00002"
    }
}

/* 2 */
{
    "_id" : "00002",
    "doc" : {
        "_id" : ObjectId("5bd2dcd3d79cc5d8b1c6296f"),
        "username" : "user5",
        "batchID" : "00002"
    }
}

/* 3 */
{
    "_id" : "00002",
    "doc" : {
        "_id" : ObjectId("5bd2dcdbd79cc5d8b1c62970"),
        "username" : "user6",
        "batchID" : "00002"
    }
}

如何在MongoDB中将两个顺序查询变成单个查询

1 个答案: