我正在查询3个集合之间的查询,我想在输出中的任何地方排除_id
我的输出是:
{
"_id" : ObjectId("5b6aed5f9bcdb5d4ae64aef5"),
"userID" : "1",
"skills" : [
{
"_id" : ObjectId("5b766b5f1365a4940bb6050f"),
"skillID" : "javaid",
"skillname" : "जावा",
"languageID" : "hindiid"
},
{
"_id" : ObjectId("5b766b8c1365a4940bb60535"),
"skillID" : "pythonid",
"skillname" : "पायथन",
"languageID" : "hindiid"
}
],
"gender" : {
"_id" : ObjectId("5b7687cd2a2329043e2383d5"),
"genderID" : "femaleid",
"gendername" : "महिला",
"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
。对于skill
列表的示例,每个对象都有_id
,我想在任何地方排除_id
字段。如何排除?
答案 0 :(得分:4)