【发布时间】:2020-04-09 14:23:11
【问题描述】:
我有两个收藏events & members:
事件架构:
{
name : String,
members: [{status : Number, memberId : {type: Schema.Types.ObjectId, ref: 'members'}]
}
事件示例文档:
"_id" : ObjectId("5e8b0bac041a913bc608d69d")
"members" : [
{
"status" : 4,
"_id" : ObjectId("5e8b0bac041a913bc608d69e"),
"memberId" : ObjectId("5e7dbf5b257e6b18a62f2da9"),
"date" : ISODate("2020-04-06T10:59:56.997Z")
},
{
"status" : 1,
"_id" : ObjectId("5e8b0bf2041a913bc608d6a3"),
"memberId" : ObjectId("5e7e2f048f80b46d786bfd67"),
"date" : ISODate("2020-04-06T11:01:06.463Z")
}
],
成员架构:
{
firstname : String
photo : String
}
成员示例文档:
[{
"_id" : ObjectId("5e7dbf5b257e6b18a62f2da9"),
"firstname" : "raed",
"photo" : "/users/5e7dbf5b257e6b18a62f2da9/profile/profile-02b13aef6e.png"
},
{
"_id" : ObjectId("5e7e2f048f80b46d786bfd67"),
"firstname" : "sarra",
"photo" : "/5e7e2f048f80b46d786bfd67/profile/profile-c79f91aa2e.png"
}]
我使用聚合进行查询,并查找以获取成员的填充数据,并且我想通过字符串连接成员的照片字段,但出现错误, 我该怎么做?
查询:
db.getCollection('events').aggregate([
{ $match: { _id: ObjectId("5e8b0bac041a913bc608d69d")}},
{
"$lookup": {
"from": "members",
"localField": "members.memberId",
"foreignField": "_id",
"as": "Members"
}
},
{
$project: {
"Members.firstname" : 1,
"Members.photo": 1,
//"Members.photo": {$concat:["http://myurl", "$Members.photo"]},
"Members._id" : 1,
},
}
])
没有 concat 的结果:
{
"_id" : ObjectId("5e8b0bac041a913bc608d69d"),
"Members" : [
{
"_id" : ObjectId("5e7dbf5b257e6b18a62f2da9"),
"firstname" : "raed",
"photo" : "/users/5e7dbf5b257e6b18a62f2da9/profile/profile-02b13aef6e.png"
},
{
"_id" : ObjectId("5e7e2f048f80b46d786bfd67"),
"firstname" : "sarra",
"photo" : "/5e7e2f048f80b46d786bfd67/profile/profile-c79f91aa2e.png"
}
]
}
错误:
$concat only supports strings, not array
【问题讨论】:
标签: mongodb mongodb-query aggregation-framework