Question

Mongodb has a record table with a field DateTime in the table. The type is Date. I want to find out which dates have records
Because the time in the database is UTC time, which is 8 hours different from China time, I will add 8 hours before grouping. SQL is as follows

db.getCollection('discerns').aggregate([
 {
 $project: {
 'date': {
 $add:['$DateTime',28800000]
 }
 }
 },{
 $project: {
 'year': {
 $year: '$date'
 },
 'month': {
 $month: '$date'
 },
 'day': {
 $dayOfMonth: '$date'
 }
 }
 },{
 $group: {
 _id: {
 year:'$year',
 month: '$month',
 day:'$day'
 }
 }
 },{
 $sort: {'_id': -1}
 }
 ]);

I wonder if it is possible to optimize two $projects into one $ project. is this possible?
That is, whether it can be done, first add 8 hours to the time, and then get the year, month and day of the calculated time, all of which are written in a $project

Answer 1

db.getCollection('discerns').aggregate([{
 $project: {
 'year': {
 $year: {
 $add: ['$DateTime', 28800000]
 }
 },
 'month': {
 $month: {
 $add: ['$DateTime', 28800000]
 }
 },
 'day': {
 $dayOfMonth: {
 $add: ['$DateTime', 28800000]
 }
 }
 }
 }])

Simply write it together. Of course, this will count three times.$add: ['$DateTime', 28800000], for obsessive-compulsive disorder patients is difficult to accept, so we can use variable definition method to play again:

db.getCollection('discerns').aggregate([{
 $project: {
 dateParts: {
 $let: {
 vars: {
 date: {
 $add: ['$DateTime', 28800000]
 }
 },
 in: {
 year: {
 $year: "$$date"
 },
 month: {
 $month: "$$date"
 },
 day: {
 $dayOfMonth: "$$date"
 }
 }
 }
 }
 }
 }])

About$letPlease refer to the documentation for the usage of:$let

The results thus obtained are actually nested one layer, which is not the same as the initial results, but for the latter ones$groupNo big deal, I believe you can handle it.