Question

我有这样的数据收集：

> db.LogBuff.find()
{ "_id" : ObjectId("578899d5d2b76f77d083f16c"), "SUBJECT" : "DD", "SYS" : "A" }
{ "_id" : ObjectId("578899d5d2b76f77d083f16d"), "SUBJECT" : "AA", "SYS" : "B" }
{ "_id" : ObjectId("578899d5d2b76f77d083f16e"), "SUBJECT" : "BB", "SYS" : "A" }
{ "_id" : ObjectId("578899d5d2b76f77d083f16f"), "SUBJECT" : "AA", "SYS" : "C" }
{ "_id" : ObjectId("578899d5d2b76f77d083f170"), "SUBJECT" : "BB", "SYS" : "A" }
{ "_id" : ObjectId("578899d5d2b76f77d083f171"), "SUBJECT" : "BB", "SYS" : "A" }
{ "_id" : ObjectId("578899d5d2b76f77d083f172"), "SUBJECT" : "CC", "SYS" : "B" }

我想提取如下输出（使用不同的＆＃34; SYS＆＃34;值）

{"SUBJECT" : "AA", "SYS" : ["A","B","C","D"]}
{"SUBJECT" : "BB", "SYS" : ["A","B","C","D"]}
{"SUBJECT" : "CC", "SYS" : ["A","B","C"]}

这是我的代码，我被困在中间，请帮我排序

db.LogBuff.aggregate([{
     "$unwind": "$SYS"
}, {
     "$group": {
             _id: {
                     "_id": "$SUBJECT"
             },
             SYST: {
                     $addToSet: "$SYS"
             }
     }
}, {
     "$unwind": "$SYST"
}, {
     "$group": {
             _id: {
             "SUBJECT": "$_id",
                     "SYST":"$SYST"
             }
     }
}])

Answer 1

只需按_id和addToSet SYS值分组：

db.LogBuff.aggregate([
    {
     "$group": {
             _id: {
                     "_id": "$SUBJECT"
             },
             SYST: {
                     $addToSet: "$SYS"
             }
        }
    }
])

无需放松，一个group可以获得所需的结果。

示例数据的组聚合结果：

{ "_id" : { "_id" : "CC" }, "SYST" : [ "B" ] }
{ "_id" : { "_id" : "BB" }, "SYST" : [ "A" ] }
{ "_id" : { "_id" : "AA" }, "SYST" : [ "C", "B" ] }
{ "_id" : { "_id" : "DD" }, "SYST" : [ "A" ] }

MongoDB数据分组与聚合

1 个答案: