Question : mongodb group by having
Answered by : annoyed-antelope-87f5ykzp3ikm
#It's the equivalent of the following SQL instruction:
# SELECT COUNT(*) FROM Table
# GROUP BY your_field
# HAVING COUNT(*) > N
query = db.collection.aggregate([ { "$group": { "_id": "$your_field", #GROUP BY your_field "count": {"$sum":1} } #COUNT(*) }, { "$match": { "count": { "$gt": N } } } #HAVING COUNT(*) > N
])Source : | Last Update : Thu, 04 Mar 21
Question : mongodb aggregate group
Answered by : annoyed-antelope-87f5ykzp3ikm
#equivalent of the following SQL instruction:
# SELECT COUNT(*) FROM Table
# GROUP BY your_field
query = db.collection.aggregate([ { "$group": { "_id": "$your_field", #GROUP BY your_field "total": {"$sum":1} #COUNT(*) } }
])Source : | Last Update : Thu, 04 Mar 21
Question : aggregate method, mongodb, group
Answered by : xenophobic-xenomorph-tabr4ghwwuco
# Encontrar el valor máximo para el ancho del pétalo - Opción 3
agg_result = Collection.aggregate([ {"$group": { "_id":"_id", "std": { "$stdDevPop": "$petalWidth" } } } ])Source : | Last Update : Wed, 28 Dec 22
Question : $group in mongodb
Answered by : delightful-deer-zbof2n31yd7v
{
"_id":"26/04/2015 09:50",
"reservations":130,
"Event_types":[ { "type":"Party", "events":[ { "eventName":"After Party", "total_count":130, "by":[ { "siteName":"club8", "countArray":[ { "bucket":"default", "value":40 } ] }, { "siteName":"PostParty", "countArray":[ { "bucket":"1", "value":70 }, { "bucket":"2", "value":20 } ] } ] } ] } ]
}Source : https://stackoverflow.com/questions/29879825/mongodb-group-composite-key-with-nested-fields | Last Update : Tue, 27 Dec 22