Question

我下面列出了两个收藏夹：-

table1

{"_id" : ObjectId("5b9a.."), "item_id" :"1.1", "m_date" : "20130401","ref_id":"12R","sub_item_id":"1.1.1"}
{"_id" : ObjectId("5c37.."), "item_id" :"1.1", "m_date" :  "20140401","ref_id":"12R","sub_item_id":"1.1.2"}
{"_id" : ObjectId("123cb.."), "item_id" :"1.2", "m_date" : "20140401","ref_id":"12R","sub_item_id":"1.1.3"}

table2

{"_id" : ObjectId("7cb3.."), "item_id" :"1.1", "m_date" : "20130401","ref_id":"12R","sub_item_id":"1.1.1"}
{"_id" : ObjectId("8f34.."), "item_id" :"1.1", "m_date" : "20140401","ref_id":"13R","sub_item_id":"1.1.2"}
{"_id" : ObjectId("5ec8b.."), "item_id" :"1.2", "m_date" : "20150401","ref_id":"14R","sub_item_id":"1.1.3"}

我想显示table1 :item_id, m_date, sub_item_id和table2 : ref_id的字段，其中item_id：1.1必须在两个表中。因此，预期结果应显示为：-

{"item_id" :"1.1", "m_date" : "20130401","sub_item_id":"1.1.1","ref_id":"12R"}
{"item_id" :"1.1", "m_date" : "20140401","sub_item_id":"1.1.2","ref_id":"13R"}

我尝试使用$lookup在下面的查询中编写内容，但发现0个文档

db.table1.aggregate([

{$project:{
    item_id:1,
     m_date: 1, 
     sub_item_id : 1,
    ref_id :1
}},
{
    $lookup: {
        from: 'table2',
        localField: 'item_id',
        foreignField: 'item_id',
        as: 'table2_values'
     },
 },
{$unwind:'$table2_values'},

{ $group: { 
    _id: {ref_id: "$table2_values.ref_id", m_date:  "$m_date"
    ,sub_item_id:'$sub_item_id' },

}},    
{$project:{_id:0,m_date:'$_id.m_date',ref_id:'$_id.ref_id'
,sub_item_id:'$_id.sub_item_id',item_id:1}},
{
 $match: {"table2_values.item_id": "1.1"}
 }

])

请帮助我获得上述预期结果

Answer 1

您可以尝试使用mongodb 3.6

进行以下聚合

case

连接两个表与选定的字段和mongodb中的where子句

1 个答案: