如何在mongodb中获得特定月份的明智logging？

这是我的模式

empname: {type: String}, soldby: {type: String}, expenseprice:{type:Number}, expensedesc:{type:String}, expensetype:{type:String}, createdat: {type: Date, default: Date.now} 

我试过这个查询

 db.expenses.aggregate([ { $project: { _id: 1, year: { $year: "$createdat" }, month: { $month: "$createdat" }, day: { $dayOfMonth: "$createdat" }, expenseprice: 1 } }, { $group: { _id: { year: "$year", month: "$month", day: "$day" }, sum: { $sum: "$expenseprice" } } } ]) 

我得到的输出为

 { "_id" : { "year" : 2015, "month" : 8, "day" : 15 }, "sum" : 200 } { "_id" : { "year" : 2016, "month" : 5, "day" : 20 }, "sum" : 150 } { "_id" : { "year" : 2016, "month" : 6, "day" : 29 }, "sum" : 250 } 

我只想要特定年份和特定月份的logging，并且在这个月份，像这样的日子

 { "_id" : { "year" : 2016, "month" : 6, "day" : 28 }, "sum" : 150 } { "_id" : { "year" : 2016, "month" : 6, "day" : 29 }, "sum" : 200 } 

我也试过$match

 db.expenses.aggregate([ { $project: { _id: 1, year: { $year: "$createdat" }, month: { $month: "$createdat" }, day: { $dayOfMonth: "$createdat" }, expenseprice: 1 } }, { $match: { $eq: ["$month", 06] } }, { $group: { _id: { year: "$year", month: "$month", day: "$day" }, sum: { $sum: "$expenseprice" } } } ]) 

但是我得到这样的错误

 assert: command failed: { "ok" : 0, "errmsg" : "bad query: BadValue unknown top level operator: $eq", "code" : 16810 } : aggregate failed _getErrorWithCode@src/mongo/shell/utils.js:23:13 doassert@src/mongo/shell/assert.js:13:14 assert.commandWorked@src/mongo/shell/assert.js:266:5 DBCollection.prototype.aggregate@src/mongo/shell/collection.js:1215:5 @(shell):1:1 2016-06-29T16:20:47.754+0530 E QUERY [thread1] Error: command failed: { "ok" : 0, "errmsg" : "bad query: BadValue unknown top level operator: $eq", "code" : 16810 } : aggregate failed : _getErrorWithCode@src/mongo/shell/utils.js:23:13 doassert@src/mongo/shell/assert.js:13:14 assert.commandWorked@src/mongo/shell/assert.js:266:5 DBCollection.prototype.aggregate@src/mongo/shell/collection.js:1215:5 @(shell):1:1 

Answer 1

请参考下面的示例，了解$ group中的年份和月份是如何引用的。 根据我的理解，这是您需要的最后一点。

 db.expenses.aggregate([ {$project: {_id:1,year:{$year:"$createdat"},month:{$month:"$createdat"}, day:{$dayOfMonth:"$createdat"},expenseprice:1}}, {$group: {_id:{year:"$year",month:"$month",day:"$day"}, sum:{$sum:"$expenseprice"}}}, {$match : {"_id.year" : 2016, "_id.month" : 6}}]) 

我不确定你为什么在比赛结束后再增加一个小组。 从我最初的理解来看，可能并不需要。 如果上述解决scheme不符合预期，请更新评论或要求。 我会相应地调整答案。

我用一些示例数据testing了我的查询。 数据被过滤了一个月，我得到明智的分解。