Question

我遇到MongoDB聚合问题。

在我的工作＆＃34; document，它有creatorParent（单值）和Children（mongodb对象id的数组）。在＆＃34;用户＆＃34;用户可以在子文档中显示子项详细信息。

当用户请求检索此文档时，我想要聚合子详细信息，如果数组包含子ID。

我写了一个有一些帮助的聚合，它适用于creatorParent，但无论我尝试过什么，它都没有为孩子们工作。

db.getCollection('Jobs').aggregate([
    {
        $unwind: {
            path : "$children"
        }
    },
    {
        $lookup: {
            "from" : "Users",
            "localField" : "creatorparent",
            "foreignField" : "_id",
            "as" : "creatorparent"
        }
    },
    {
        $lookup: {
            "from" : "Users",
            "localField" : "children",
            "foreignField" : "children",
            "as" : "children"
        }
    }
])

用户文档：

{
    "_id" : ObjectId("58daf84877733645eaa9b44f"),
    "email" : "meto93@gmail.com",
    "password" : "vpGl+Fjnef616cRgNbCkwaFDpSI=",
    "passwordsalt" : "99397F4A9D3A499D96694547667E74595CE994D2E83345D6953EF866303E8B65",
    "children" : [ 
        {
            "_id" : ObjectId("58daf84977733645eaa9b450"),
            "name" : "Mert",
            "age" : 5,
            "additionalinformation" : "ilk cocuk",
            "creationtime" : ISODate("2017-03-28T23:56:56.952Z"),
            "userid" : ObjectId("58daf84877733645eaa9b44f"),
            "gender" : null
        }, 
        {
            "_id" : ObjectId("58daf84977733645eaa9b451"),
            "name" : "Sencer",
            "age" : 7,
            "additionalinformation" : "ikinci cocuk",
            "creationtime" : ISODate("2017-03-28T23:56:56.952Z"),
            "userid" : ObjectId("58daf84877733645eaa9b44f"),
            "gender" : null
        }
    ]
}

工作

{
    "_id" : ObjectId("58db0a2d77733645eaa9b453"),
    "creationtime" : ISODate("2017-03-29T01:13:17.509Z"),
    "startingtime" : ISODate("2017-04-03T13:00:00.000Z"),
    "endingtime" : ISODate("2017-04-03T17:00:00.000Z"),
    "children" : [ 
        ObjectId("58daf84977733645eaa9b450"), 
        ObjectId("58daf84977733645eaa9b451")
    ],
    "creatorparent" : ObjectId("58daf84877733645eaa9b44f"),
    "applicants" : []
}

Answer 1

试试这个：

db.jobs.aggregate(
    [
        {
            $unwind: {
                path : "$children",
            }
        },
        {
            $lookup: {
                "from" : "users",
                "localField" : "creatorparent",
                "foreignField" : "_id",
                "as" : "creatorparent"
            }
        },
        {
            $lookup: {
                "from" : "users",
                "localField" : "children",
                "foreignField" : "children._id",
                "as" : "children"
            }
        },
        {
            $addFields: {
                children : {$arrayElemAt : ["$children",0]}
            }
        },
        {
            $addFields: {
              "children":"$children.children"
            }
        },
        {
            $unwind: {
                path : "$children",
            }
        },
        {
            $group: {
                "_id": "$_id",
                "name": { "$first": "$name" },
                "jobstatus" : { "$first": "$jobstatus" },
                "hourlyrate" : { "$first": "$hourlyrate" },
                "creatorparent" : { "$first" : "$creatorparent" },
                "children": { "$addToSet": "$children"  }
            }
        },

    ]
);

Mongodb聚合从文档内部检索文档

1 个答案: