Question

如何匹配属于连接表的列的条件？

For eg:
Consider two collections A and B:
Collection A: 
id|name|status
++++++++++++++
1 |Rock|1
2 |Sam |1
3 |Jack|1
Collection B: 
id|userId|      loc       | status
+++++++++++++++++++++++++++++++++++
1| 1     |[11.111,22.321] | -1
2| 1     |[16.22,33.213]  |  1
3| 2     |[334.11,242.321]|  1
4| 3     |[1.111,224.321] |  1

我想根据集合B中的状态字段获取所有用户的当前位置，即用户的当前位置在B中的状态为1，之前的位置的状态为-1。

So, the result I want is like:
+++++++++++++++++
id:1, name: Rock, status: 1, userLocTable: [{id:2, userId:1, loc: [16.22,33.213], status: 1}]

我在做的是：

db.collectionA.aggregate([
    {
        $lookup: {
            from: collectionB,
            localField: id,
            foreignField: userId,
            as: userLocTable
        }
    },
    {
        $match:{
           userLocTable.status: 1
        }
    }
])

但是我从查找表（B）获取所有行（状态1和-1），而不是只有状态为1的记录。可能是什么问题。任何帮助都值得赞赏。谢谢！

Answer 1

试试这个，

db.collectionA.aggregate([
    { 
        "$lookup": { 
            "from": collectionB,
            "localField": id,
            "foreignField": userId,
            "as": userLocTable
        } 
    },
    {
        "$addFields": {
            "userLocTable": {
                        "$filter": {
                            "input": "$userLocTable",
                            "as": "userLoc",
                            "cond": {
                                "$eq": [ "$$userLoc.status", "1" ]
                            }
                        }
            }
        }
    }
])

Mongodb匹配查找表不起作用

1 个答案: