Question

我有一个包含数据的集合

match n where has(n.`longitude`) return n

我按“team.name”进行过滤，想知道状态是什么，并计算每个状态如下样本结果

{ "_id" : "...." , "team" : [ {"name" : "A", "state" : "Active"} , {"name" : "B", "state" : "Deactive", {"name" : "C", "state" : "Unknown"} }  ]},
{ "_id" : "...." , "team" : [ {"name" : "A", "state" : "Unknown"} , {"name" : "B", "state" : "Deactive", {"name" : "C", "state" : "Unknown"} }  ]},
{ "_id" : "...." , "team" : [ {"name" : "A", "state" : "Active"} , {"name" : "B", "state" : "Deactive", {"name" : "C", "state" : "Unknown"} }  ]}

是否可以仅使用MongoDB的聚合函数而不使用任何代码？

Answer 1

是否可以仅使用MongoDB的聚合函数码？

是

TL; DR 展开并分组

db.getCollection('foo').aggregate([
   {$unwind:"$team"},
   {
       $group: {
           _id: "$team.name",
           "Active": {$sum: {$cond:[{$eq:["$team.state","Active"]},1,0]}},
           "Deactive": {$sum: {$cond:[{$eq:["$team.state","Deactive"]},1,0]}},
           "Unknown": {$sum: {$cond:[{$eq:["$team.state","Unknown"]},1,0]}}
       }
   },
   {$project: {_id:0, name:"$_id", Active:1, Deactive:1, Unknown:1}},
   {$sort: {name:1}}
])

样品的输出：

 [
    { "name" : "A", "Active" : 2, "Deactive" : 0, "Unknown" : 1 },
    { "name" : "B", "Active" : 0, "Deactive" : 3, "Unknown" : 0 },
    { "name" : "C", "Active" : 0, "Deactive" : 0, "Unknown" : 3 }
 ]

我应该在MongoDB中做什么聚合？

1 个答案: