Question

我是Mongo Db的新手，希望对此查询有所帮助。我的mongodb数据横断面相同id的数目基于create_date如何显示每个横断面id的第一记录

 db.collection.aggregate([{
     {"$project": {
     "RESOURCE_ID": 1,
     "TRANSACTION_ID":1,
     "CREATE_DATE":1
   }}

  ])

输出：

      RESOURCE_ID         TRANSACTION_ID              CREATE_DATE   
1      "100-101"      "0:ffff0a0a0983UY:-142"   "Fri Sep 07 16:51:30IST2018"
2      "100-101"      "0:ffff0a0a0983UY:-142"   "Fri Sep 07 16:51:29IST2018"
3      "100-101"      "0:ffff0a0a0983UY:-142"   "Fri Sep 07 16:51:29IST2018"
4      "100-102"      "0:ffff0a0a0983UY:-111"   "Fri Sep 06 16:51:29IST2018"
5      "100-102"      "0:ffff0a0a0983UY:-111"   "Fri Sep 06 16:51:28IST2018"

预期输出：

       RESOURCE_ID         TRANSACTION_ID              CREATE_DATE   
1      "100-101"      "0:ffff0a0a0983UY:-142"   "Fri Sep 07 16:51:30IST2018"
2      "100-102"      "0:ffff0a0a0983UY:-111"   "Fri Sep 06 16:51:29IST2018"

Answer 1

这是您想要的： onSingleTapConfirmed

它的作用-按日期升序对所有文档进行排序，因为我们首先需要最早的文档（请注意db.collection.aggregate([{"$sort": {CREATE_DATE: 1}},{$group:{"_id": "$RESOURCE_ID", TRANSACTION_ID: {$first: "$TRANSACTION_ID"}, "CREATE_DATE": {$first: "$CREATE_DATE"}}}])子句）。
然后，它通过字段“ RESOURCE_ID”（标记为新的sort）对文档进行分组，并使用_id TRANSACTION_ID和$first CREATE_DATE。

请注意，您必须将时间戳转换为实际时间戳，才能使mongo真正了解时间戳的顺序。否则mongo会将它们排序为字符串，这不是您想要的

一个例子：

$first

此外，值得一提的是，您应该为> db.collection.find().pretty() { "_id" : ObjectId("5bd2bf353ca22147747ec212"), "RESOURCE_ID" : "100-101", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-142", "CREATE_DATE" : ISODate("2017-10-13T10:53:53Z") } { "_id" : ObjectId("5bd2bf3c3ca22147747ec213"), "RESOURCE_ID" : "100-101", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-142", "CREATE_DATE" : ISODate("2017-10-14T10:53:53Z") } { "_id" : ObjectId("5bd2bf3c3ca22147747ec214"), "RESOURCE_ID" : "100-102", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-111", "CREATE_DATE" : ISODate("2017-10-13T10:53:53Z") } { "_id" : ObjectId("5bd2bf3c3ca22147747ec215"), "RESOURCE_ID" : "100-102", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-111", "CREATE_DATE" : ISODate("2017-10-14T10:53:53Z") } > db.collection.aggregate([{"$sort": {CREATE_DATE: 1}},{$group:{"_id": "$RESOURCE_ID", TRANSACTION_ID: {$first: "$TRANSACTION_ID"}, "CREATE_DATE": {$first: "$CREATE_DATE"}}}]) { "_id" : "100-102", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-111", "CREATE_DATE" : ISODate("2017-10-13T10:53:53Z") { "_id" : "100-101", "TRANSACTION_ID" : "0:ffff0a0a0983UY:-142", "CREATE_DATE" : ISODate("2017-10-13T10:53:53Z")添加索引，因为您要按此字段进行排序；而应该为CREATE_DATE添加索引，因为mongo必须对其进行排序才能对其进行分组

如何避免重复值？

1 个答案: