Question

我有2个收藏品，如下所示。

DATA1：

{ "_id" : , "timestamp" : ISODate("2016-01-05T07:42:37.312Z"), "Prof_Name" : "Jack ", "SUBJECT" : "Maths, Chemistry, Machinery1, Ele1" }
{ "_id" : , "timestamp" : ISODate("2016-01-05T07:42:37.312Z"), "Prof_Name" : "Mac", "SUBJECT" : "Chemistry, CS, German" }

DATA2：

{ "_id" : ObjectId(""), timestamp" : ISODate("2016-08-05T07:42:37.312Z", "SUBJECT_ID" : "Maths", "ID" : "OI-12", "Rating" : 6, "UUID" : 8123 }
{ "_id" : ObjectId(""), timestamp" : ISODate("2017-09-05T07:42:37.312Z", "SUBJECT_ID" : "Maths, Machinery1, German", "ID" : "OI-134", "Rating" : 6, "UUID" : 8123 }
{ "_id" : ObjectId(""), timestamp" : ISODate("2016-01-05T07:42:37.312Z", "SUBJECT_ID" : "Machinery1, Maths, French, German", "ID" : "OI-32", "Rating" : 3, "UUID" : 8123 }
{ "_id" : ObjectId(""), timestamp" : ISODate("2016-01-05T07:42:37.312Z", "SUBJECT_ID" : "CS, Chemistry", "ID" : "OI-36", "Rating" : , "UUID" : 8124 }

我希望在2016年1月至11月210日的时间戳之间获得3个集合，其中对于来自＆＃34; data1＆＃34;的SUBJECT中的每个Prof_Name和主题，检查它是否存在于＆＃34; data2＆＃34;并将UUID和UUID计数为1，如果在下一条记录中找到相同的主题，则使UUID计数= 2，依此类推。 Herre是我的收藏品的样子..

DATA3：

{ "_id" : ,
"Prof_Name" : "Jack", 
"Subjects_list" : [ "Maths", "Chemistry", "Machinery1"], 
"UUID_list" : [8123, 8124 ], 
"UUID_count" : 3,   // Because UUID 8123 has present in 2 records which comes under 2016 timestamp
"subject_count" : 3 } // Ele1 is not mentioned because it has not been seen in any of the data2 record
{ "_id" : , 
"Prof_Name" : "Mac", 
"Subjects_list" : [ "CS"], 
"UUID_list" : [8124 ],  
"UUID_count" : 1,   // Because UUID 8123 has present in 2 records which comes under 2016 timestamp
"subject_count" : 1 }

我的聚合查询是：

db.data1.aggregate([
  {
    "$addFields": {
      "SUBJECT": {
        "$split": [
          "$SUBJECT",
          ", "
        ]
      }
    }
  },
  {
    "$unwind": "$SUBJECT"
  },
  {
    "$lookup": {
      "from": "data2",
      "let": {
        "subject": "$SUBJECT"
      },
      "pipeline": [
        {
          "$addFields": {
            "SUBJECT_ID": {
              "$split": [
                "$SUBJECT_ID",
                ", "
              ]
            }
          }
        },
        {
          "$match": {
            "$expr": {
              "$in": [
                "$$subject",
                "$SUBJECT_ID"
              ]
            }
          }
        },
        {
          "$project": {
            "UUID": 1,
            "_id": 0
          }
        }
      ],
      "as": "ref_data"
    }
  },
  {
    "$unwind": {
      "path": "$ref_data",
      "preserveNullAndEmptyArrays": true
    }
  },
  {
    "$group": {
      "_id": "$Prof_Name",
      "subjects_list": {
        "$addToSet": "$SUBJECT"
      },
      "UUID_list": {
        "$addToSet": "$ref_data.UUID"
      }
    }
  },
  {
    "$addFields": {
      "Prof_Name": "$_id",
      "UUID_count": {
        "$size": "$UUID_list"
      },
      "subject_count": {
        "$size": "$subjects_list"
      }
    }
  },
  {
    "$project": {
      "_id": 0
    }
  },
  {
    "$out": "data3"
  }
])

此查询需要进行哪些修改才能获得上述集合data3，mainy UUID_list和UUID-count以及Subject_list。

还想知道如何在下面的查询聚合查询中匹配给定月份和年份但不是iso的记录的时间戳。

试过这个：

    { "$project": {"year":{"$year":"$timestamp"},"month":{"$month":"$timestamp"}}},{ "$match":{"year" :"2016","month": "01"}}

但是工作。

Answer 1

您可以通过将主题从逗号分隔的值更改为数据库中的数组来简化聚合。

对于前"SUBJECT" : ["Maths", "", "Chemistry", "Machinery1", "Ele1"]

您可以使用以下聚合。

db.data1.aggregate([
{"$lookup":{
  "from":"data2",
  "localField":"SUBJECT",
  "foreignField":"SUBJECT_ID",
  "as":"ref_data"
}}, // outputs all the input documents where there is any match between two subjects array.
{"$unwind":{"path":"$ref_data","preserveNullAndEmptyArrays":true}},
{"$match":{"ref_data.timestamp":{"$gte":ISODate("2016-01-01T00:00:00.000Z"), "$lte":ISODate("2016-11-31T11:59:59.999Z")}}},
{"$addFields":{"SUBJECT":{"$setIntersection":["$SUBJECT","$ref_data.SUBJECT_ID"]}}}, // outputs the common subjects (matching) between two subjects array
{"$unwind":"$SUBJECT"},
{"$group":{
  "_id":{
    "Prof_Name":"$Prof_Name",
    "UUID":"$ref_data.UUID",
    "SUBJECT":"$SUBJECT"
  }
}},// outputs all the distinct combination of UUID and Subject
{"$group":{
  "_id":"$_id.Prof_Name",
  "UUID_count":{"$sum":1},
  "subjects_list":{"$push":"$_id.SUBJECT"},
  "UUID_distinct_list":{"$addToSet":"$_id.UUID"}
}}, // outputs the distinct uuid list, count the uuids & subjects list 
{"$addFields": {
  "Prof_Name": "$_id",
  "UUID_distinct_count": {
    "$size": "$UUID_distinct_list"
  },
  "subject_count": {
    "$size": "$subjects_list"
  }
}}, // Adds the subject list size
{"$project": {"_id": 0}},// excludes the id from final output
{"$out":"data3"}])

如果没有架构修改，您可以使用以下聚合查询。

db.data1.aggregate([
  {"$lookup":{
    "from":"data2",
    "let":{"subject":{"$split":["$SUBJECT",", "]}},
    "pipeline":[
      {"$match": {"expr":{"$and":[{"$eq":[{"$year":"$timestamp"}, 2016]}, {"$eq":[{"$month":"$timestamp"}, 1]}]}}},
      {"$addFields":{"SUBJECT_ID":{"$split":["$SUBJECT_ID",", "]},"SUBJECT":"$$subject"}},
      {"$unwind":"$SUBJECT"},
      {"$match":{"$expr":{"$in":["$SUBJECT","$SUBJECT_ID"]}}},
      {"$facet":{
        "UUID":[{"$group":{"_id":{"id":"$_id","UUID":"$UUID"}}},{"$count":"UUID_Count"}],
        "REST":[
          {"$group":{"_id":null,"subjects_list":{"$addToSet":"$SUBJECT"},"UUID_distinct_list":{"$addToSet":"$UUID"}}},
          {"$addFields":{"subject_count":{"$size":"$subjects_list"},"UUID_distinct_count":{"$size":"$UUID_distinct_list"}}},
          {"$project":{"_id":0}}
         ]
      }},
      {"$replaceRoot":{"newRoot":{"$mergeObjects":[{"$arrayElemAt":["$UUID",0]},{"$arrayElemAt":["$REST",0]}]}}}
    ],
    "as":"ref_data"
  }},
  {"$unwind":{"path":"$ref_data","preserveNullAndEmptyArrays":true}},
  {"$addFields":{"ref_data.Prof_Name":"$Prof_Name"}},
  {"$replaceRoot":{"newRoot":"$ref_data"}},
  {"$out":"data3"}
])

MongoDB聚合查询以获取记录中每个实例的唯一元素列表和计数

1 个答案: