Question

我试图将3个集合合并为一个，但不能使其工作（与全部合并相同）。例如，我有3个收藏集：

1：活跃用户

[{_id:'', client_id:'',created:''}]

2：非活跃用户

[{_id:'', visitor_id:'',created:''}]

3：被阻止的用户

[{_id:'', blocked_id:'',created:''}]

我要做的是将所有3个集合合并为一个，按日期对它们进行排序，并确保获得前10个结果，诸如此类：

[{_id:'', client_id:'',created:''},
 {_id:'', blocked_id:'',created:''},
 {_id:'', visitor_id:'',created:''}]

我想到了以下代码：

collection('active).aggregate([        
       {  
          "$facet":{  
             "active":[  
                {"$match":{  "client_id": 5cfe83820c19ee3c50c8f323 }}
             ],
             "not_active":[  
                {"$match":{"cust_id":"5cfe83820c19ee3c50c8f323" }}
             ],
             "blocked":[  
                {"$match":{"blocked_id":"5cfe83820c19ee3c50c8f323" }}
             ]
          }
       },
       {  
          "$skip":0
       },
       {  
          "$limit":10
       },
       {  
          "$sort":{  
             "created":-1
          }
       }
])

但是此代码返回空数组：

[ { active: [], not_active: [], blocked : []} ]

我做错了什么？谢谢

Answer 1

最后，我想出了完成这项工作的这段代码：

mongo.clients.aggregate([
         {$limit:1},{$project:{_id:1}},{$project:{_id:0}},
         {$lookup : {from:'active',as:'active',pipeline:[{$match:{}}]}},
         {$lookup : {from:'not_active',as:'not_active',pipeline:[{$match:{}}]}},
         {$lookup : {from:'blocked',as:'blocked',pipeline:[{$match:{}}]}},
         {$project:{Union:{$concatArrays:['$active','$not_active','$blocked']}}},
         {$unwind:"$Union"},
         {$replaceRoot:{newRoot:"$Union"}},
         {$skip:0},
         {$limit:10},
         {$sort:{'created': -1}}
]).then(x);

这里有更详细的答案： enter link description here

MongoDB Union All？

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