我的收藏品包含
等记录{ "type" : "me", "tid" : "1" }
{ "type" : "me", "tid" : "1" }
{ "type" : "me", "tid" : "1" }
{ "type" : "you", "tid" : "1" }
{ "type" : "you", "tid" : "1" }
{ "type" : "me", "tid" : "2" }
{ "type" : "me", "tid" : "2"}
{ "type" : "you", "tid" : "2"}
{ "type" : "you", "tid" : "2" }
{ "type" : "you", "tid" : "2"}
我想要下面的结果
[
{"tid" : "1","me" : 3,"you": 2},
{"tid" : "2","me" : 2,"you": 3}
]
我试过小组和;聚合查询无法获得所需的结果格式。
以下是群组查询。
db.coll.group({
key: {tid : 1,type:1},
cond: { tid : { "$in" : [ "1","2"]} },
reduce: function (curr,result) {
result.total = result.total + 1
},
initial: { total : 0}
})
结果就像
[
{"tid" : "1", "type" : "me" ,"total": 3 },
{"tid" : "1","type" : "you" ,"total": 2 },
{"tid" : "2", "type" : "me" ,"total": 2 },
{"tid" : "2","type" : "you" ,"total": 3 }
]
以下是汇总查询
db.coll.aggregate([
{$match : { "tid" : {"$in" : ["1","2"]}}},
{$group : { _id : {tid : "$tid",type : "$type"},total : {"$sum" : 1}}}
])
给出以下结果
{
"result" :
[
{"_id" : {"tid" : "1","type" : "me"},"total" : 3},
{"_id" : {"tid" : "2","type" : "me" },"total" : 2},
{"_id" : {"tid" : "2","type" : "you"},"total" : 3}
]
"ok" : 1
}
可以获得我指定的结果,或者我必须在我的代码中进行一些操作。
由于
答案 0 :(得分:1)
如果您将聚合更改为:
db.so.aggregate([
{ $match : { "tid" : { "$in" : ["1", "2"] } } },
{ $group : {
_id : { tid : "$tid", type : "$type" },
total : { "$sum" : 1 }
} },
{ $group : {
_id : "$_id.tid",
values: { $push: { type: "$_id.type", total: '$total' } }
} }
])
然后你的输出是:
{
"result" : [
{
"_id" : "1",
"values" : [
{ "type" : "you", "total" : 2 },
{ "type" : "me", "total" : 3 }
]
},
{
"_id" : "2",
"values" : [
{ "type" : "me", "total" : 2 },
{ "type" : "you", "total" : 3 }
]
}
],
"ok" : 1
}
虽然这与你想要的不一样,但它将是你能得到的最接近的。在您的应用程序中,您可以轻松地提取与您喜欢的相同的值,以便摆脱它。
请注意,一般情况下,您无法将值(you
,me
)提升为键 - 除非您的密钥设置有限(最多3-4项。