db.Rooms.find({"name" : "room3"}).pretty()
{
"_id" : ObjectId("57f50608ace5ceb9af033528"),
"name" : "room3",
"userData" : {
"user" : ObjectId("57f4d142ace5ceb9af033521"),
"date" : "Wed Oct 05 2016 15:54:16 GMT+0200"
},
"active" : true,
"users" : [
{
"uid" : ObjectId("57f383a6ace5ceb9af033511")
},
{
"uid" : ObjectId("57f4d142ace5ceb9af033521")
}
],
"messages" : [
{
"msg" : "first test since statement ",
"time" : "Wed Oct 05 2016 15:55:26 GMT+0200",
"user" : ObjectId("57f383a6ace5ceb9af033511")
},
{
"msg" : "second test since statement ",
"time" : "Wed Oct 05 2016 15:57:35 GMT+0200",
"user" : ObjectId("57f4d142ace5ceb9af033521")
},
{
"msg" : "third test since statement ",
"time" : "Wed Oct 05 2016 15:58:11 GMT+0200",
"user" : ObjectId("57f383a6ace5ceb9af033511")
}
]
}
我对Mongo很新,我在解决这个问题上遇到了麻烦。实际上我已经半天尝试了自己:(
我想要的是找到“仅”某个用户插入的消息..
这是我的收藏。
(如果它看起来很乱,我会链接到图像) image as link
我想要的是显示: 来自“user”的所有“msg”_id:ObjectId(“57f383a6ace5ceb9af033511”)
我希望有人可以指导我,甚至告诉我这个收藏品是不是很糟糕......
thx;)
答案 0 :(得分:1)
之前提出的类似问题
How to filter array in subdocument with MongoDB
对于使用聚合函数的问题
db.Room.aggregate(
{ $unwind: '$messages'},
{ $match: {'messages.user': {$eq: ObjectId("57f383a6ace5ceb9af033511")}}},
{ $group: {_id: '$_id', list: {$push: '$messages.msg'}}}
结果:
{
"_id" : ObjectId("57f50608ace5ceb9af033528"),
"list" : [
"first test since statement ",
"third test since statement "
]
}
答案 1 :(得分:0)
关于这个主题还有另一个问题:
How to select a single field in MongoDB?
在find方法中,您可以传递另一个对象(查询除外),该对象指示要返回的字段。
对你而言,它看起来像这样:
db.student.find({"_id" : ObjectId("57f50608ace5ceb9af033528")}, {messages:1})
1
中的{messages:1}
=选择"消息"字段。