排除$ lookup聚合中的字段

Question

我正在查询3个我想排除_id在输出中所有位置的集合之间

我的输出是：

{
    "_id" : ObjectId("5b6aed5f9bcdb5d4ae64aef5"),
    "userID" : "1",
    "skills" : [ 
        {
            "_id" : ObjectId("5b766b5f1365a4940bb6050f"),
            "skillID" : "javaid",
            "skillname" : "????",
            "languageID" : "hindiid"
        }, 
        {
            "_id" : ObjectId("5b766b8c1365a4940bb60535"),
            "skillID" : "pythonid",
            "skillname" : "?????",
            "languageID" : "hindiid"
        }
    ],

    "gender" : {
        "_id" : ObjectId("5b7687cd2a2329043e2383d5"),
        "genderID" : "femaleid",
        "gendername" : "?????",
        "languageID" : "hindiid"
    }
}

查询：

db.User.aggregate([
  { "$match": { "userID":"1" }},
  { "$lookup":{
    "from": "Skill",
    "pipeline": [
      { "$match": { "languageID": "hindiid", "skillID": { "$in": [ "javaid","pythonid" ] }}},
    ],
    "as": "skills"
  }},
  { "$lookup": {
    "from": "Gender",
    "pipeline": [
      { "$match": { "languageID": "hindiid", "genderID" : "femaleid" }},
    ],
    "as": "gender"
  }},
  { "$unwind": { "path": "$gender", "preserveNullAndEmptyArrays": true }},
  { "$project": { "userID": 1, "skills": 1, "gender": 1 }}
]) 

在输出中，每个对象都有_id。示例skill列表中的每个对象都有_id我要_id在任何地方排除字段。如何排除？

Answer 1

在mongodb 3.6中，您可以$project在$lookup管道内使用投影（）。

db.User.aggregate([
  { "$match": { "userID":"1" }},
  { "$lookup":{
    "from": "Skill",
    "pipeline": [
      { "$match": { "languageID": "hindiid", "skillID": { "$in": [ "javaid","pythonid" ] }}},
      { "$project": { "_id": 0 }}
    ],
    "as": "skills"
  }}
])