Mongodb每两小时聚合一次
我有一个如下形式的聚合查询Mongodb每两小时聚合一次,mongodb,mongodb-query,aggregation-framework,Mongodb,Mongodb Query,Aggregation Framework,我有一个如下形式的聚合查询 db.mycollection.aggregate([ { "$match": { "Time": { $gte: ISODate("2016-01-30T00:00:00.000+0000") } } }, { "$group": { "_id": { "day": { "$dayOfYear": "$Time" }, "h
db.mycollection.aggregate([
{
"$match":
{
"Time": { $gte: ISODate("2016-01-30T00:00:00.000+0000") }
}
},
{
"$group":
{
"_id":
{
"day": { "$dayOfYear": "$Time" },
"hour": { "$hour": "$Time" }
},
"Dishes": { "$addToSet": "$Dish" }
}
},
{
"$group":
{
"_id": "$_id.hour",
"Food":
{
"$push":
{
"Day": "$_id.day",
"NumberOfDishes": { "$size":"$Dishes" }
}
}
}
},
{
"$project":
{
"Hour": "$_id",
"Food": "$Food",
"_id" : 0
}
},
{
"$sort": { "Hour": 1 }
}
]);
我希望能够在两小时内完成这项工作,而不是像上面那样在一小时内完成,例如0-1,1-2,2-3,3-4,4-5,…,23-24。e、 g.0-2,2-4,4-6,…,22-24。有办法吗?提示:在 比如说,
H=floor(hour/2)
,其中hour
是从文档日期算起的实际小时数。然后,您可以通过在此日期应用and运算符来获得H
"H": { $floor: { $divide: [ { "$hour": "$Time" }, 2 ] } }
这里,H
对应于一对小时(hours=[0,2)=>H=0
,hours=[2,4)=>H=1
,hours=[22,24)=>H=11
,等等),您可以使用
$group: { "_id": { "day": { $dayOfYear: "$Time" }, "H": "$H" } }
然后,您可以使用输出特定H
的小时数对
"Hours": [ { $multiply: [ "$H", 2 ] }, { $sum: [ { $multiply: [ "$H", 2 ] }, 2 ] } ]
给定的文件集
{ "Time" : ISODate("2016-01-30T01:00:00Z"), "Dish" : "dish1" }
{ "Time" : ISODate("2016-01-30T02:00:00Z"), "Dish" : "dish2" }
{ "Time" : ISODate("2016-01-30T03:00:00Z"), "Dish" : "dish3" }
{ "Time" : ISODate("2016-01-30T04:00:00Z"), "Dish" : "dish4" }
{ "Time" : ISODate("2016-01-30T05:00:00Z"), "Dish" : "dish5" }
{ "Time" : ISODate("2016-01-30T06:00:00Z"), "Dish" : "dish6" }
{ "Time" : ISODate("2016-01-30T07:00:00Z"), "Dish" : "dish7" }
{ "Time" : ISODate("2016-01-30T08:00:00Z"), "Dish" : "dish8" }
{ "Time" : ISODate("2016-01-30T09:00:00Z"), "Dish" : "dish9" }
并在其上使用下一个聚合
db.mycollection.aggregate([
{
"$match":
{
"Time": { $gte: ISODate("2016-01-30T00:00:00.000+0000") }
}
},
{
"$project":
{
"Dish": 1,
"Time": 1,
"H": { $floor: { $divide: [ { $hour: "$Time" }, 2 ] } }
}
},
{
"$group":
{
"_id":
{
"day": { $dayOfYear: "$Time" },
"H": "$H"
},
"Dishes": { $addToSet: "$Dish" }
}
},
{
"$group":
{
"_id": "$_id.H",
"Food":
{
"$push":
{
"Day": "$_id.day",
"NumberOfDishes": { $size: "$Dishes" }
}
}
}
},
{
"$sort": { "_id": 1 }
},
{
"$project":
{
"Hours": [ { $multiply: [ "$_id", 2 ] }, { $sum: [ { $multiply: [ "$_id", 2 ] }, 2 ] } ],
"Food": "$Food",
"_id": 0
}
}
]);
提供结果
{ "Food" : [ { "Day" : 30, "NumberOfDishes" : 1 } ], "Hours" : [ 0, 2 ] }
{ "Food" : [ { "Day" : 30, "NumberOfDishes" : 2 } ], "Hours" : [ 2, 4 ] }
{ "Food" : [ { "Day" : 30, "NumberOfDishes" : 2 } ], "Hours" : [ 4, 6 ] }
{ "Food" : [ { "Day" : 30, "NumberOfDishes" : 2 } ], "Hours" : [ 6, 8 ] }
{ "Food" : [ { "Day" : 30, "NumberOfDishes" : 2 } ], "Hours" : [ 8, 10 ] }
我使用mongo 3.0。地板不是我的选择。还有什么我可以使用的吗?可能会有帮助