我在MongoDB中有一些数据包含10分钟的期间最新总和:
db.test.insert({perEnd: ISODate('2013-06-05T18:00:00'), val: 7.3})
db.test.insert({perEnd: ISODate('2013-06-05T18:10:00'), val: 6.23})
db.test.insert({perEnd: ISODate('2013-06-05T18:20:00'), val: 4.1})
db.test.insert({perEnd: ISODate('2013-06-05T18:30:00'), val: 0.21})
db.test.insert({perEnd: ISODate('2013-06-05T18:40:00'), val: 12.1})
db.test.insert({perEnd: ISODate('2013-06-05T18:50:00'), val: 6.0})
db.test.insert({perEnd: ISODate('2013-06-05T19:00:00'), val: 8.9})
db.test.insert({perEnd: ISODate('2013-06-05T19:10:00'), val: .98})
db.test.insert({perEnd: ISODate('2013-06-05T19:20:00'), val: 14.7})
我想汇总以查找每个小时结束时段的总和,因此我应该得到以下值:
ending 2013-06-05 18:00:00 - 7.3
ending 2013-06-05 19:00:00 - 37.54
ending 2013-06-05 20:00:00 - 15.68
使用内置日期运算符不起作用,因为它们将所有日期向下舍入(截断)到最近的边界,我需要围绕向上:
> db.test.aggregate({$group: {_id: {Year: {$year: "$perEnd"},
Day: {$dayOfYear: "$perEnd"},
Hour: {$hour: "$perEnd"}},
sum: {$sum: "$val"}}})
{
"result" : [
{ "_id" : { "Year" : 2013,
"Day" : 156,
"Hour" : 19 },
"sum" : 24.58 },
{ "_id" : { "Year" : 2013,
"Day" : 156,
"Hour" : 18 },
"sum" : 35.940000000000005 }
],
"ok" : 1
}
任何人都能找到一种方法来实现这一目标吗?
答案 0 :(得分:0)
您可以使用mongodb map-reduce执行此操作:
var map = function(){
var date = new Date(this.perEnd.getTime());
if(date.getMinutes() > 0){
date.setHours(date.getHours() + 1, 0, 0, 0);
} else {
date.setHours(date.getHours(), 0, 0, 0);
}
emit(date, this.val);
};
var reduce = function(key, values){
return Array.sum(values)
};
db.collection.mapReduce(map, reduce, {out : {inline : 1}}, callback);
对于您的数据,我得到以下结果:
[ { _id: Wed Jun 05 2013 21:00:00 GMT+0300 (EEST), value: 7.3 },
{ _id: Wed Jun 05 2013 22:00:00 GMT+0300 (EEST), value: 37.54 },
{ _id: Wed Jun 05 2013 23:00:00 GMT+0300 (EEST), value: 15.68 } ]