Mongodb 按关联文档的编号排序

我有两个收集，身份和干预。身份有很多干预。我必须按三种类型分类：首先是积极干预的身份，然后是消极干预的身份，最后是没有干预的身份。每种类型都必须按字母顺序排序

以下是我的数据：

> db.identities.find().pretty()

{
        "_id" : ObjectId("56c2ba0377656280a1000008"),
        "screenname" : "User without intervention 1"
}
{
        "_id" : ObjectId("56c2ba0677656280a100000c"),
        "screenname" : "User with inactive intervention 2"
}
{       
        "_id" : ObjectId("56c2ba49776562aca000076b"),
        "screenname" : "User with active intervention 1"
}
{       
        "_id" : ObjectId("56c432fa776562356200011a"),
        "screenname" : "User with active intervention 2"
}
{       
        "_id" : ObjectId("56c2ba49776562aca000076b"),
        "screenname" : "User without intervention 2"
}
{       
        "_id" : ObjectId("56c42cc7776562209500006a"),
        "screenname" : "User with inactive intervention 1"
}

> db.interventions.find().pretty()

{
        "_id" : ObjectId("56c4448a7765622817000045"),
        "identity_id" : ObjectId("56c2ba0377656280a1000008"),
        "active" : true
}
{
        "_id" : ObjectId("56c44a1b77656282f0000451"),
        "identity_id" : ObjectId("56c42f54776562b78a00032f"),
        "active" : false
}
{
        "_id" : ObjectId("56c450c8776562336a000776"),
        "identity_id" : ObjectId("56c432fa776562356200011a"),
        "active" : true
}
{
        "_id" : ObjectId("56c450c8776562336a000776"),
        "identity_id" : ObjectId("56c42cc7776562209500006a"),
        "active" : false
}

我希望得到以下结果：

{
        "_id" : ObjectId("56c2ba49776562aca000076b"),
        "screenname" : "User with active intervention 1"
}
{
        "_id" : ObjectId("56c432fa776562356200011a"),
        "screenname" : "User with active intervention 2"
}
{       
        "_id" : ObjectId("56c2ba49776562aca000076b"),
        "screenname" : "User with inactive intervention 1"
}
{       
        "_id" : ObjectId("56c2ba0677656280a100000c"),
        "screenname" : "User with inactive intervention 2"
}
{       
        "_id" : ObjectId("56c2ba0377656280a1000008"),
        "screenname" : "User without intervention 1"
}
{       
        "_id" : ObjectId("56c2ba49776562aca000076b"),
        "screenname" : "User without intervention 2"
}

我对MongoDb不是很在行。目前，我刚刚：

db['domain_test#identities'].aggregate({ $project: { screenname: { $toLower: "$screenname" } } }, {$sort: { screenname: 1 }})

它只按屏幕名称排序，但不检查其他表。我想我必须向

$project

添加其他值，但我不知道如何添加

---

您有解决方案吗？

您可以尝试下面3.4中的聚合

用于获取匹配的干预数组（一对一），并在

$addFields

中输出第一个元素

用于定位搜索的布尔值在值列表中的位置，该值仅为0或1，缺少字段为-1，后跟将输出索引保留在文档中的额外字段中，并在排序字段上降序以对文档进行排序

通过排除删除排序字段并进行干预以获得预期输出

db.identities.aggregate([
  {"$lookup":{
    "from":"interventions",
    "localField":"_id",
    "foreignField":"identity_id",
    "as":"interventions"
  }},
  {"$addFields":{
    "sortorder":{
      "$let":{
        "vars":{"intervention":{"$arrayElemAt":["$interventions",0]}},
        "in":{"$indexOfArray":[[false,true],"$$intervention.active"]}
      }
    }
  }},
  {"$sort":{"sortorder":-1}},
  {"$project":{"sortorder":0,"interventions":0}}
])

---

您的_id有重复项，您只需按

db.identies.find（）.sort（{screenname:1}）

您的mongo版本是什么？@Saravana很抱歉重复。这不仅仅是复制/粘贴，我编辑了数据。如果只是按屏幕名称排序在这种情况下起作用，那只是巧合。我希望先有积极干预的用户，然后是不积极干预的用户，然后是没有干预的用户。@Veeram我有mongo 3.4.9。