日期范围内的Mongodb Sum字段和其他过滤器

Question

假设我的mongodb中具有以下文档结构：

{
  _id: "tMSfNq9JR85XDaQe5"
  customerid: "QDGvBQhS6vYgZtnRr"
  employeeid: "QDGvBQhS6vYgZtnRr" 
  date: Sun Dec 07 2014 19:50:21 GMT+0800 (HKT) 
  projectid: "S83NEGHnrefvfASrf" 
  hours: 2
}

查询特定日期范围内的数据时，是否可以获取按雇员和日期分组的特定项目和客户的总工时，因此输出将例如：

{ "projectid":"XXX", "date":"Dec 1, 2014", "employeeid":"AAA", "totalHrs":"20" }
{ "projectid":"XXX", "date":"Dec 1, 2014", "employeeid":"BBB", "totalHrs":"11" }
{ "projectid":"XXX", "date":"Dec 2, 2014", "employeeid":"AAA", "totalHrs":"3" }
{ "projectid":"XXX", "date":"Dec 2, 2014", "employeeid":"BBB", "totalHrs":"5" }
{ "projectid":"XXX", "date":"Dec 2, 2014", "employeeid":"CCC", "totalHrs":"18" }

谢谢

Answer 1

嗨，据我所知，我认为您希望对projectid和employeeid进行分组，以查找所需的输出，因此在这种情况下，下面的汇总可能有效

db.collectionName.aggregate(
                       {"$group":{"_id":{"projectId":"$projectid","employeeId":"$employeeid","date":"$date"},
                       "totalHrs":{"$sum":"$hours"}}},
                      {"$project":{"projectid":"$_id.projectId",
                         "date":"$_id.date","employeeid":"$_id.employeeId",
                        "totalHrs":"$totalHrs",
                          "_id":0}}
                        )

Answer 2

您需要按projectid，date，employeeid分组

db.collectionname.aggregate({"$project":{"projectid":1,"date":1,"employeeid":1,"totalHrs":1}},
{"$group":{"_id":{"$projectId":"$projectid","date":"$date","employeeId":"$employeeid"}, 
"totalHrs":{"$sum":"$hours"}}})