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Find last record of each day
get last record less than or equal in meteor? This means that I want to download all records for every item, but only the last record of each itemHow to aggregate this query correctly in MongoDb?Group result by 15 minutes time interval in MongoDbRandom record from MongoDBFind objects between two dates MongoDBHow to get the last N records in mongodb?How can I use mongodump to dump out records matching a specific date range?“Large data” work flows using pandasFind MongoDB records where array field is not emptyLatest record by date for each item mongodb groupSort the data in mongo collection based on TimestampHow to Get last 5 Days Data with all FieldsHow do I get the earliest record by date for a unique constraint?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I use mongodb (I'm new in mongodb) to store data about my power consumption , each minute there is a new record, here is an example :
"id":"5309d4cae4b0fbd904cc00e1","adco":"O","hchc":7267599,"hchp":10805900,"hhphc":"g","ptec":"c","iinst":13,"papp":3010,"imax":58,"optarif":"s","isousc":60,"motdetat":"Á","date":1393156826114
so I have around 1440 records a day.
I want to compute the cost by day, but the problem for this is that I need the last record of the day, because this record can give me the amount of absolute kWh (kiloWatt-Hour). So if I remove the kwh amount of the last record of the yesterday, I have the amount of kWh for the day.
The field hchp
gives me this absolute kWh. The field date
corresponds to the time of the measure in millisecond.
consumption of the day = absolute consumption of the end of the day - absolute consumption of the end of yesterday.
How is it possible to get the last record of each day in mongodb ?
Note : I use mongodb in spring java, so I need a query like this :
Example to get all measures :
@Query(" 'date' : $gt : ?0 ")
public List<Mesure> findByDateGreaterThan(Date date, Sort sort);
mongodb mongodb-query aggregation-framework
add a comment |
I use mongodb (I'm new in mongodb) to store data about my power consumption , each minute there is a new record, here is an example :
"id":"5309d4cae4b0fbd904cc00e1","adco":"O","hchc":7267599,"hchp":10805900,"hhphc":"g","ptec":"c","iinst":13,"papp":3010,"imax":58,"optarif":"s","isousc":60,"motdetat":"Á","date":1393156826114
so I have around 1440 records a day.
I want to compute the cost by day, but the problem for this is that I need the last record of the day, because this record can give me the amount of absolute kWh (kiloWatt-Hour). So if I remove the kwh amount of the last record of the yesterday, I have the amount of kWh for the day.
The field hchp
gives me this absolute kWh. The field date
corresponds to the time of the measure in millisecond.
consumption of the day = absolute consumption of the end of the day - absolute consumption of the end of yesterday.
How is it possible to get the last record of each day in mongodb ?
Note : I use mongodb in spring java, so I need a query like this :
Example to get all measures :
@Query(" 'date' : $gt : ?0 ")
public List<Mesure> findByDateGreaterThan(Date date, Sort sort);
mongodb mongodb-query aggregation-framework
Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00
add a comment |
I use mongodb (I'm new in mongodb) to store data about my power consumption , each minute there is a new record, here is an example :
"id":"5309d4cae4b0fbd904cc00e1","adco":"O","hchc":7267599,"hchp":10805900,"hhphc":"g","ptec":"c","iinst":13,"papp":3010,"imax":58,"optarif":"s","isousc":60,"motdetat":"Á","date":1393156826114
so I have around 1440 records a day.
I want to compute the cost by day, but the problem for this is that I need the last record of the day, because this record can give me the amount of absolute kWh (kiloWatt-Hour). So if I remove the kwh amount of the last record of the yesterday, I have the amount of kWh for the day.
The field hchp
gives me this absolute kWh. The field date
corresponds to the time of the measure in millisecond.
consumption of the day = absolute consumption of the end of the day - absolute consumption of the end of yesterday.
How is it possible to get the last record of each day in mongodb ?
Note : I use mongodb in spring java, so I need a query like this :
Example to get all measures :
@Query(" 'date' : $gt : ?0 ")
public List<Mesure> findByDateGreaterThan(Date date, Sort sort);
mongodb mongodb-query aggregation-framework
I use mongodb (I'm new in mongodb) to store data about my power consumption , each minute there is a new record, here is an example :
"id":"5309d4cae4b0fbd904cc00e1","adco":"O","hchc":7267599,"hchp":10805900,"hhphc":"g","ptec":"c","iinst":13,"papp":3010,"imax":58,"optarif":"s","isousc":60,"motdetat":"Á","date":1393156826114
so I have around 1440 records a day.
I want to compute the cost by day, but the problem for this is that I need the last record of the day, because this record can give me the amount of absolute kWh (kiloWatt-Hour). So if I remove the kwh amount of the last record of the yesterday, I have the amount of kWh for the day.
The field hchp
gives me this absolute kWh. The field date
corresponds to the time of the measure in millisecond.
consumption of the day = absolute consumption of the end of the day - absolute consumption of the end of yesterday.
How is it possible to get the last record of each day in mongodb ?
Note : I use mongodb in spring java, so I need a query like this :
Example to get all measures :
@Query(" 'date' : $gt : ?0 ")
public List<Mesure> findByDateGreaterThan(Date date, Sort sort);
mongodb mongodb-query aggregation-framework
mongodb mongodb-query aggregation-framework
edited Jun 26 '17 at 10:31
Neil Lunn
105k24 gold badges194 silver badges198 bronze badges
105k24 gold badges194 silver badges198 bronze badges
asked Mar 5 '14 at 16:06
elolozoneelolozone
163 bronze badges
163 bronze badges
Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00
add a comment |
Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00
Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00
Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
A bit more modern than the original answer:
db.collection.aggregate([
"$sort": "date": 1 ,
"$group":
"_id":
"$subtract": ["$date","$mod": ["$date",86400000]]
,
"doc": "$last": "$$ROOT"
,
"$replaceRoot": "newDocument": "$doc"
])
The same principle applies that you essentially $sort
the collection and then $group
on the required grouping key picking up the $last
data from the grouping boundary.
Making things a bit clearer since the original writing is that you can use $$ROOT
instead of specifying every document property, and of course the $replaceRoot
stage allows you to restore that data fully as the original document form.
But the general solution is still $sort
first, then $group
on the common key that is required and keep the $last
or $first
depending on sort order occurrences from the grouping boundary for the properties that are required.
Also for BSON Dates as opposed to a timestamp value as in the question, see Group result by 15 minutes time interval in MongoDb for different approaches on how to accumulate for different time intervals actually using and returning BSON Date values.
Not quite sure what you are going for here but you could do this in aggregate if my understanding is right. So to get the last record for each day:
db.collection.aggregate([
// Sort in date order as ascending
"$sort": "date": 1 ,
// Date math converts to whole day
"$project":
"adco": 1,
"hchc": 1,
"hchp": 1,
"hhphc": 1,
"ptec": 1,
"iinst": 1,
"papp": 1,
"imax": 1,
"optarif": 1,
"isousc": 1,
"motdetat": 1,
"date": 1,
"wholeDay": "$subtract": ["$date","$mod": ["$date",86400000]]
,
// Group on wholeDay ( _id insertion is monotonic )
"$group":
"_id": "$wholeDay",
"docId": "$last": "$_id" ,
"adco": "$last": "$adco" ,
"hchc": "$last": "$hchc" ,
"hchp": "$last": "$hchp" ,
"hhphc": "$last": "$hhphc" ,
"ptec": "$last": "$ptec" ,
"iinst": "$last": "$iinst" ,
"papp": "$last": "$papp" ,
"imax": "$last": "$imax" ,
"optarif": "$last": "$optarif",
"isousc": "$last": "$isouc" ,
"motdetat": "$last": "$motdetat" ,
"date": "$last": "$date" ,
])
So the principle here is that given the timestamp value, do the date math to project that as the midnight time at the beginning of each day. Then as the _id
key on the document is already monotonic (always increasing), then simply group on the wholeDay
value while pulling the $last
document from the grouping boundary.
If you don't need all the fields then only project and group on the ones you want.
And yes you can do this in the spring data framework. I'm sure there is a wrapped command in there. But otherwise, the incantation to get to the native command goes something like this:
mongoOps.getCollection("yourCollection").aggregate( ... )
For the record, if you actually had BSON date types rather than a timestamp as a number, then you can skip the date math:
db.collection.aggregate([
"$group":
"_id":
"year": "$year": "$date" ,
"month": "$month": "$date" ,
"day": "$dayOfMonth": "$date"
,
"hchp": "$last": "$hchp"
])
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A bit more modern than the original answer:
db.collection.aggregate([
"$sort": "date": 1 ,
"$group":
"_id":
"$subtract": ["$date","$mod": ["$date",86400000]]
,
"doc": "$last": "$$ROOT"
,
"$replaceRoot": "newDocument": "$doc"
])
The same principle applies that you essentially $sort
the collection and then $group
on the required grouping key picking up the $last
data from the grouping boundary.
Making things a bit clearer since the original writing is that you can use $$ROOT
instead of specifying every document property, and of course the $replaceRoot
stage allows you to restore that data fully as the original document form.
But the general solution is still $sort
first, then $group
on the common key that is required and keep the $last
or $first
depending on sort order occurrences from the grouping boundary for the properties that are required.
Also for BSON Dates as opposed to a timestamp value as in the question, see Group result by 15 minutes time interval in MongoDb for different approaches on how to accumulate for different time intervals actually using and returning BSON Date values.
Not quite sure what you are going for here but you could do this in aggregate if my understanding is right. So to get the last record for each day:
db.collection.aggregate([
// Sort in date order as ascending
"$sort": "date": 1 ,
// Date math converts to whole day
"$project":
"adco": 1,
"hchc": 1,
"hchp": 1,
"hhphc": 1,
"ptec": 1,
"iinst": 1,
"papp": 1,
"imax": 1,
"optarif": 1,
"isousc": 1,
"motdetat": 1,
"date": 1,
"wholeDay": "$subtract": ["$date","$mod": ["$date",86400000]]
,
// Group on wholeDay ( _id insertion is monotonic )
"$group":
"_id": "$wholeDay",
"docId": "$last": "$_id" ,
"adco": "$last": "$adco" ,
"hchc": "$last": "$hchc" ,
"hchp": "$last": "$hchp" ,
"hhphc": "$last": "$hhphc" ,
"ptec": "$last": "$ptec" ,
"iinst": "$last": "$iinst" ,
"papp": "$last": "$papp" ,
"imax": "$last": "$imax" ,
"optarif": "$last": "$optarif",
"isousc": "$last": "$isouc" ,
"motdetat": "$last": "$motdetat" ,
"date": "$last": "$date" ,
])
So the principle here is that given the timestamp value, do the date math to project that as the midnight time at the beginning of each day. Then as the _id
key on the document is already monotonic (always increasing), then simply group on the wholeDay
value while pulling the $last
document from the grouping boundary.
If you don't need all the fields then only project and group on the ones you want.
And yes you can do this in the spring data framework. I'm sure there is a wrapped command in there. But otherwise, the incantation to get to the native command goes something like this:
mongoOps.getCollection("yourCollection").aggregate( ... )
For the record, if you actually had BSON date types rather than a timestamp as a number, then you can skip the date math:
db.collection.aggregate([
"$group":
"_id":
"year": "$year": "$date" ,
"month": "$month": "$date" ,
"day": "$dayOfMonth": "$date"
,
"hchp": "$last": "$hchp"
])
add a comment |
A bit more modern than the original answer:
db.collection.aggregate([
"$sort": "date": 1 ,
"$group":
"_id":
"$subtract": ["$date","$mod": ["$date",86400000]]
,
"doc": "$last": "$$ROOT"
,
"$replaceRoot": "newDocument": "$doc"
])
The same principle applies that you essentially $sort
the collection and then $group
on the required grouping key picking up the $last
data from the grouping boundary.
Making things a bit clearer since the original writing is that you can use $$ROOT
instead of specifying every document property, and of course the $replaceRoot
stage allows you to restore that data fully as the original document form.
But the general solution is still $sort
first, then $group
on the common key that is required and keep the $last
or $first
depending on sort order occurrences from the grouping boundary for the properties that are required.
Also for BSON Dates as opposed to a timestamp value as in the question, see Group result by 15 minutes time interval in MongoDb for different approaches on how to accumulate for different time intervals actually using and returning BSON Date values.
Not quite sure what you are going for here but you could do this in aggregate if my understanding is right. So to get the last record for each day:
db.collection.aggregate([
// Sort in date order as ascending
"$sort": "date": 1 ,
// Date math converts to whole day
"$project":
"adco": 1,
"hchc": 1,
"hchp": 1,
"hhphc": 1,
"ptec": 1,
"iinst": 1,
"papp": 1,
"imax": 1,
"optarif": 1,
"isousc": 1,
"motdetat": 1,
"date": 1,
"wholeDay": "$subtract": ["$date","$mod": ["$date",86400000]]
,
// Group on wholeDay ( _id insertion is monotonic )
"$group":
"_id": "$wholeDay",
"docId": "$last": "$_id" ,
"adco": "$last": "$adco" ,
"hchc": "$last": "$hchc" ,
"hchp": "$last": "$hchp" ,
"hhphc": "$last": "$hhphc" ,
"ptec": "$last": "$ptec" ,
"iinst": "$last": "$iinst" ,
"papp": "$last": "$papp" ,
"imax": "$last": "$imax" ,
"optarif": "$last": "$optarif",
"isousc": "$last": "$isouc" ,
"motdetat": "$last": "$motdetat" ,
"date": "$last": "$date" ,
])
So the principle here is that given the timestamp value, do the date math to project that as the midnight time at the beginning of each day. Then as the _id
key on the document is already monotonic (always increasing), then simply group on the wholeDay
value while pulling the $last
document from the grouping boundary.
If you don't need all the fields then only project and group on the ones you want.
And yes you can do this in the spring data framework. I'm sure there is a wrapped command in there. But otherwise, the incantation to get to the native command goes something like this:
mongoOps.getCollection("yourCollection").aggregate( ... )
For the record, if you actually had BSON date types rather than a timestamp as a number, then you can skip the date math:
db.collection.aggregate([
"$group":
"_id":
"year": "$year": "$date" ,
"month": "$month": "$date" ,
"day": "$dayOfMonth": "$date"
,
"hchp": "$last": "$hchp"
])
add a comment |
A bit more modern than the original answer:
db.collection.aggregate([
"$sort": "date": 1 ,
"$group":
"_id":
"$subtract": ["$date","$mod": ["$date",86400000]]
,
"doc": "$last": "$$ROOT"
,
"$replaceRoot": "newDocument": "$doc"
])
The same principle applies that you essentially $sort
the collection and then $group
on the required grouping key picking up the $last
data from the grouping boundary.
Making things a bit clearer since the original writing is that you can use $$ROOT
instead of specifying every document property, and of course the $replaceRoot
stage allows you to restore that data fully as the original document form.
But the general solution is still $sort
first, then $group
on the common key that is required and keep the $last
or $first
depending on sort order occurrences from the grouping boundary for the properties that are required.
Also for BSON Dates as opposed to a timestamp value as in the question, see Group result by 15 minutes time interval in MongoDb for different approaches on how to accumulate for different time intervals actually using and returning BSON Date values.
Not quite sure what you are going for here but you could do this in aggregate if my understanding is right. So to get the last record for each day:
db.collection.aggregate([
// Sort in date order as ascending
"$sort": "date": 1 ,
// Date math converts to whole day
"$project":
"adco": 1,
"hchc": 1,
"hchp": 1,
"hhphc": 1,
"ptec": 1,
"iinst": 1,
"papp": 1,
"imax": 1,
"optarif": 1,
"isousc": 1,
"motdetat": 1,
"date": 1,
"wholeDay": "$subtract": ["$date","$mod": ["$date",86400000]]
,
// Group on wholeDay ( _id insertion is monotonic )
"$group":
"_id": "$wholeDay",
"docId": "$last": "$_id" ,
"adco": "$last": "$adco" ,
"hchc": "$last": "$hchc" ,
"hchp": "$last": "$hchp" ,
"hhphc": "$last": "$hhphc" ,
"ptec": "$last": "$ptec" ,
"iinst": "$last": "$iinst" ,
"papp": "$last": "$papp" ,
"imax": "$last": "$imax" ,
"optarif": "$last": "$optarif",
"isousc": "$last": "$isouc" ,
"motdetat": "$last": "$motdetat" ,
"date": "$last": "$date" ,
])
So the principle here is that given the timestamp value, do the date math to project that as the midnight time at the beginning of each day. Then as the _id
key on the document is already monotonic (always increasing), then simply group on the wholeDay
value while pulling the $last
document from the grouping boundary.
If you don't need all the fields then only project and group on the ones you want.
And yes you can do this in the spring data framework. I'm sure there is a wrapped command in there. But otherwise, the incantation to get to the native command goes something like this:
mongoOps.getCollection("yourCollection").aggregate( ... )
For the record, if you actually had BSON date types rather than a timestamp as a number, then you can skip the date math:
db.collection.aggregate([
"$group":
"_id":
"year": "$year": "$date" ,
"month": "$month": "$date" ,
"day": "$dayOfMonth": "$date"
,
"hchp": "$last": "$hchp"
])
A bit more modern than the original answer:
db.collection.aggregate([
"$sort": "date": 1 ,
"$group":
"_id":
"$subtract": ["$date","$mod": ["$date",86400000]]
,
"doc": "$last": "$$ROOT"
,
"$replaceRoot": "newDocument": "$doc"
])
The same principle applies that you essentially $sort
the collection and then $group
on the required grouping key picking up the $last
data from the grouping boundary.
Making things a bit clearer since the original writing is that you can use $$ROOT
instead of specifying every document property, and of course the $replaceRoot
stage allows you to restore that data fully as the original document form.
But the general solution is still $sort
first, then $group
on the common key that is required and keep the $last
or $first
depending on sort order occurrences from the grouping boundary for the properties that are required.
Also for BSON Dates as opposed to a timestamp value as in the question, see Group result by 15 minutes time interval in MongoDb for different approaches on how to accumulate for different time intervals actually using and returning BSON Date values.
Not quite sure what you are going for here but you could do this in aggregate if my understanding is right. So to get the last record for each day:
db.collection.aggregate([
// Sort in date order as ascending
"$sort": "date": 1 ,
// Date math converts to whole day
"$project":
"adco": 1,
"hchc": 1,
"hchp": 1,
"hhphc": 1,
"ptec": 1,
"iinst": 1,
"papp": 1,
"imax": 1,
"optarif": 1,
"isousc": 1,
"motdetat": 1,
"date": 1,
"wholeDay": "$subtract": ["$date","$mod": ["$date",86400000]]
,
// Group on wholeDay ( _id insertion is monotonic )
"$group":
"_id": "$wholeDay",
"docId": "$last": "$_id" ,
"adco": "$last": "$adco" ,
"hchc": "$last": "$hchc" ,
"hchp": "$last": "$hchp" ,
"hhphc": "$last": "$hhphc" ,
"ptec": "$last": "$ptec" ,
"iinst": "$last": "$iinst" ,
"papp": "$last": "$papp" ,
"imax": "$last": "$imax" ,
"optarif": "$last": "$optarif",
"isousc": "$last": "$isouc" ,
"motdetat": "$last": "$motdetat" ,
"date": "$last": "$date" ,
])
So the principle here is that given the timestamp value, do the date math to project that as the midnight time at the beginning of each day. Then as the _id
key on the document is already monotonic (always increasing), then simply group on the wholeDay
value while pulling the $last
document from the grouping boundary.
If you don't need all the fields then only project and group on the ones you want.
And yes you can do this in the spring data framework. I'm sure there is a wrapped command in there. But otherwise, the incantation to get to the native command goes something like this:
mongoOps.getCollection("yourCollection").aggregate( ... )
For the record, if you actually had BSON date types rather than a timestamp as a number, then you can skip the date math:
db.collection.aggregate([
"$group":
"_id":
"year": "$year": "$date" ,
"month": "$month": "$date" ,
"day": "$dayOfMonth": "$date"
,
"hchp": "$last": "$hchp"
])
edited May 5 '18 at 3:48
answered Mar 6 '14 at 5:51
Neil LunnNeil Lunn
105k24 gold badges194 silver badges198 bronze badges
105k24 gold badges194 silver badges198 bronze badges
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Have you looked at getting a range representing the date, and then, sorting and limiting to the highest date value?
– WiredPrairie
Mar 5 '14 at 18:00