Question

我有下表： -

id order_number product_number order_status schedule_datetime
1  001          001.1          SUCCESS      20180103
2  001          001.2          SUCCESS      20180102
3  111          111.1          SUCCESS      20171225
4  111          111.2          SUCCESS      20171224
5  222          222.1          INPROGRESS   20171122
6  222          222.2          ON_HOLD      20171121
7  222          222.3          PARTLY_SUCCESS 20171121

我在这里尝试的是，如果任何product_number与查询中给出的状态匹配，它应该显示order_numbers的年份和列表。

输入1：order_status = SUCCESS，输出低于

{
"order_statuses" : [
    "SUCCESS"
],
"year" : "2018",
"order_number" : "001"
},

{
 "order_statuses" : [
    "SUCCESS"
],
"year" : "2017",
"order_number" : "111"
}

输入2：order_status = PARTLY_SUCCESS，输出下面应该

{
"order_statuses" : [
    "PARTLY_SUCCESS",
    "ON_HOLD",
    "INPROGRESS"
],
"year" : "2017",
"order_number" : "222"

} 在此链接convert mysql query contains sum and group_concat to mongodb query的帮助下，我试图获得所需的结果，但在查询下方给出错误“字段名称'$ addFields'不能是运算符名称”。因为我不想显示计数我尝试添加到单独的组，但它没有帮助。

db.order_summary.aggregate([

{"$project":{
    "schedule_datetime":1,
    "order_number ":1,
    "order_status":{"$ifNull":["$order_status",""]}
}},
{"$group":{
    "_id":{
    "order_number ":"$order_number ",
    "order_status":"$order_status"
    },
    "study_date":{"$first": "$study_date"}
}},
{"$sort":{"_id.order_status": 1}},
{"$group":{
    "_id":{
    "order_number ":"$_id.order_number "
    },
    "study_date":{"$first": "$study_date"},
    "order_status":{"$push": "$_id.order_status"}
}},
{"$group":{
    "_id":{
        "$substr":["$study_date",0,4]
    },
    "count":{
        "$sum":{
            "$cond": [
                {"$in": ["$order_status",[["SUCCESS"],["INPROGRESS","SUCCESS"]]]}, 

                1,0
             ]
        }
    },
    "order_numbers":{"$push":"$_id.order_number "},
    "$addFields":{"$size":"$order_numbers"},


}},
{"$match":{"order_numbers":{"$gt":0}}},
{"$sort":{"_id":-1}}

]）

请按照上面的输出帮助显示年份和order_numbers列表，谢谢！

Answer 1

您可以尝试以下聚合查询。

首先$group获取不同的订单号和状态组合，然后选择最新的时间戳。

第二个$group按年份和状态进行分组，然后选择所有订单号。

$match将状态组与输入状态进行比较。

db.order_summary.aggregate([
  {"$project":{
    "schedule_datetime":1,
    "order_number ":1,
    "order_status":{"$ifNull":["$order_status",""]
    }
  }},
  {"$sort":{"schedule_datetime":-1}},
  {"$group":{
    "_id":{
      "order_number":"$order_number",
      "order_status":"$order_status"
    },
    "schedule_datetime":{"$first":"$schedule_datetime"}
  }},
  {"$group":{
    "_id":{
       "year":{"$substr":["$schedule_datetime",0,4]},
       "order_number":"$_id.order_number"
      },
     "order_statuses":{"$push":"$_id.order_status"}
   }},
  {"$redact": {"$cond":[{"$in": ["SUCCESS","$order_statuses"]},"$$KEEP","$$PRUNE"]}},
  {"$sort":{"_id.year":-1}},
  {"$project":{"_id":0, "year":"$_id.year","order_number":"$_id.order_number", "order_statuses":1 }},
])

比较聚合内的多个值并在mongodb中显示数组中的结果

1 个答案: