[
{ "$match": {
"created":{
"$gte": ISODate("2015-07-19T07:26:49.045Z")
},
"created":{
"$lte": ISODate("2015-07-20T07:37:56.045Z")
}
}},
{ "$group:{
"_id":{
"ln":"$l.ln",
"cid":"$cid"
},
"appCount":{ "$sum": 1 }
}},
{ "$group": {
"_id": { "ln":"$_id.ln" },
"cusappCount": { "$sum": 1 }
}},
{ "$sort":{ "_id.ln":1 } }
]
In above mongo db query I am not able to display the appcount in result.. I am able to display cusappCount. Could anyone please help me on this
The $match is wrong to start with and does not do what you think. It is only selecting the "second" statement:
"created":{
"$lte": ISODate("2015-07-20T07:37:56.045Z")
}
So your selections are incorrect to start with.
That and other corrections below:
[
{ "$match": {
"created": {
"$gte": ISODate("2015-07-19T07:26:49.045Z"),
"$lte": ISODate("2015-07-20T07:37:56.045Z")
}
}},
{ "$group":{
"_id": {
"ln":"$l.ln",
"cid":"$cid"
},
"appCount":{ "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.ln",
"cusappCount": { "$sum": "$appCount" },
"distinctCustCount": { "$sum": 1 }
}},
{ "$sort":{ "_id": 1 } }
]
Which seems to be what you are trying to do.
So your earier "count" is then passed to $sum when grouping at a "broader" level. The "second" count is just for the "distinct" items in the earlier key.
If you are trying to "retain" the values of "appCount", then the problem here is that your "grouping" is "taking away" the detail level that appears at. So for what it is woth, then this is where you use "arrays" in an output structure:
[
{ "$match": {
"created": {
"$gte": ISODate("2015-07-19T07:26:49.045Z"),
"$lte": ISODate("2015-07-20T07:37:56.045Z")
}
}},
{ "$group":{
"_id": {
"ln":"$l.ln",
"cid":"$cid"
},
"appCount":{ "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.ln",
"cusappCount": { "$sum": 1 },
"custs": { "$push": {
"cid": "$_id.cid", "appCount": "$appCount"
}}
}},
{ "$sort":{ "_id": 1 } }
]
Related
I'm new in mongoDB.
This is one example of record from collection:
{
supplier: 1,
type: "sale",
items: [
{
"_id": ObjectId("60ee82dd2131c5032342070f"),
"itemBuySum": 10
},
{
"_id": ObjectId("60ee82dd2131c50323420710"),
"itemBuySum": 10,
},
{
"_id": ObjectId("60ee82dd2131c50323420713"),
"itemBuySum": 10
},
{
"_id": ObjectId("60ee82dd2131c50323420714"),
"itemBuySum": 20
}
]
}
I need to group by TYPE field and get the SUM. This is output I need:
{
supplier: 1,
sales: 90,
returns: 170
}
please check Mongo playground for better understand. Thank you!
$match - Filter documents.
$group - Group by type and add item into data array which leads to the result like:
[
[/* data 1 */],
[/* data 2 */]
]
$project - Decorate output document.
3.1. First $reduce is used to flatten the nested array to a single array (from Result (2)) via $concatArrays.
3.2. Second $reduce is used to aggregate $sum the itemBuySum.
db.collection.aggregate({
$match: {
supplier: 1
},
},
{
"$group": {
"_id": "$type",
"supplier": {
$first: "$supplier"
},
"data": {
"$push": "$items"
}
}
},
{
"$project": {
_id: 0,
"supplier": "$supplier",
"type": "$_id",
"returns": {
"$reduce": {
"input": {
"$reduce": {
input: "$data",
initialValue: [],
in: {
"$concatArrays": [
"$$value",
"$$this"
]
}
}
},
"initialValue": 0,
"in": {
$sum: [
"$$value",
"$$this.itemBuySum"
]
}
}
}
}
})
Sample Mongo Playground
db.collection.aggregate([
{
$match: {
supplier: 1
},
},
{
"$group": {
"_id": "$ID",
"supplier": {
"$first": "$supplier"
},
"sale": {
"$sum": {
"$cond": {
"if": {
"$eq": [
"$type",
"sale"
]
},
"then": {
"$sum": "$items.itemBuySum"
},
"else": {
"$sum": 0
}
}
}
},
"returns": {
"$sum": {
"$sum": {
"$cond": {
"if": {
"$eq": [
"$type",
"return"
]
},
"then": {
"$sum": "$items.itemBuySum"
},
"else": {
"$sum": 0
}
}
}
}
}
}
},
{
"$project": {
_id: 0,
supplier: 1,
sale: 1,
returns: 1
}
}
])
My collection's data are something like this :
[
{
ANumberAreaCode: "+98",
BNumberAreaCode: "+1",
AccountingTime: 1629754886,
Length: 123
},
{
ANumberAreaCode: "+44",
BNumberAreaCode: "+98",
AccountingTime: 1629754786,
Length: 123
},
{
ANumberAreaCode: "+98",
BNumberAreaCode: "+96",
AccountingTime: 1629754886,
Length: 998
}
]
I'm going to group on countries codes and count result (summing country codes in ANumberAreaCode and BNumberAreaCode ) .
This is my group sample :
{ "$group": {
"_id": {
"ANumberAreaCode": "$ANumberAreaCode",
},
"count": { "$sum": 1 }
}},
{ "$group": {
"_id": {
"BNumberAreaCode": "$BNumberAreaCode",
},
"count": { "$sum": 1 }
}},
now , how can i summing count result of two above queries for common countries ?
I'm looking for a query that give me this result :
+98 : 3
+44 : 1
+1 :1
+96 :1
You can use this aggregation pipeline:
$facet to get both group, by A and B. This creates two objects: groupA and groupB.
Then using $concatArrays into $project stage it will concat two ouputs.
Deconstructs the array using $unwind
And $group again by values using $sum to get the total.
db.collection.aggregate([
{
"$facet": {
"groupA": [
{
"$group": {
"_id": "$ANumberAreaCode",
"total": {
"$sum": 1
}
}
}
],
"groupB": [
{
"$group": {
"_id": "$BNumberAreaCode",
"total": {
"$sum": 1
}
}
}
]
}
},
{
"$project": {
"result": {
"$concatArrays": [
"$groupA",
"$groupB"
]
}
}
},
{
"$unwind": "$result"
},
{
"$group": {
"_id": "$result._id",
"total": {
"$sum": "$result.total"
}
}
}
])
Example here
Now i get this table(it can't be more than two kinds of {A,B,C} to appear in the same data at the same time.):
{_id:1,A:a}
{_id:2,B:b}
{_id:3,C:a}
{_id:4,A:a}
{_id:5,A:b}
{_id:6,A:c}
{_id:7,C:a}
How to get this result?
a:4
b:2
c:1
you can get this result with mongodb aggregation framework,
first, you'll need to add all the value in a single field, and the perform a $group on that field:
db.collection.aggregate([{
"$project": {
"v": ["$A", "$B", "$C"]
}
}, {
"$unwind": "$v"
}, {
"$match": {
"v": {
"$ne": null
}
}
}, {
"$group": {
"_id": "$v",
"count": {
"$sum": 1
}
}
}])
result:
[
{
"_id": "c",
"count": 1
},
{
"_id": "b",
"count": 2
},
{
"_id": "a",
"count": 4
}
]
you can try it here: mongoplayground.net/p/rGHUPWsw2ee
I want to write a group by query to written active user and total count(both active and inactive) grouped by a date column in mongodb. I am able to run them as two separate scripts but how to retrieve the same information in one script
db.user.aggregate(
{
"$match": { 'phoneInfo.verifiedFlag': true}
},
{
"$project": {
yearMonthDayUTC: { $dateToString: { format: "%Y-%m-%d", date: "$createdOn" } }
}
},
{
"$group": {
"_id": {day: "$yearMonthDayUTC"},
count: {
"$sum": 1
}
}
},
{
$sort: {
"_id.day": 1,
}
})
You can use the $cond operator in your group to create a conditional count as follows (assuming the inactive/active values are in a field called status):
db.user.aggregate([
{ "$match": { 'phoneInfo.verifiedFlag': true} },
{
"$group": {
"_id": { "$dateToString": { "format": "%Y-%m-%d", "date": "$createdOn" } },
"total": { "$sum": 1 },
"active_count": {
"$sum": {
"$cond": [ { "$eq": [ "$status", "active" ] }, 1, 0 ]
}
},
"inactive_count": {
"$sum": {
"$cond": [ { "$eq": [ "$status", "inactive" ] }, 1, 0 ]
}
}
}
},
{ "$sort": { "_id": 1 } }
])
For different values you can adapt the following pipeline:
db.user.aggregate([
{ "$match": { 'phoneInfo.verifiedFlag': true} },
{
"$group": {
"_id": {
"day": {
"$dateToString": {
"format": "%Y-%m-%d",
"date": "$createdOn"
}
},
"status": { "$toLower": "$status" }
},
"count": { "$sum": 1 }
}
},
{
"$group": {
"_id": "$_id.day",
"counts": {
"$push": {
"status": "$_id.status",
"count": "$count"
}
}
}
},
{ "$sort": { "_id": 1 } }
])
I have documents like:
{
"platform":"android",
"install_date":20151029
}
platform - can have one value from [android|ios|kindle|facebook ] .
install_date - there are many install_dates
There are also many fields.
Aim : I am calculating installs per platform on particular date.
So I am using group by in aggregation framework and make counts by platform. Document should look like like:
{
"install_date":20151029,
"platform" : {
"android":1000,
"ios": 2000,
"facebook":1500
}
}
I have done like:
db.collection.aggregate([
{
$group: {
_id: { platform: "$platform",install_date:"$install_date"},
count: { "$sum": 1 }
}
},
{
$group: {
_id: { install_date:"$_id.install_date"},
platform: { $push : {platform :"$_id.platform", count:"$count" } }
}
},
{
$project : { _id: 0, install_date: "$_id.install_date", platform: 1 }
}
])
which Gives document like:
{
"platform": [
{
"platform": "facebook",
"count": 1500
},
{
"platform": "ios",
"count": 2000
},
{
"platform": "android",
"count": 1000
}
],
"install_date": 20151027
}
Problem:
Projecting array to single object as "platform"
With MongoDb 3.4 and newer, you can leverage the use of $arrayToObject operator to get the desired result. You would need to run the following aggregate pipeline:
db.collection.aggregate([
{ "$group": {
"_id": {
"date": "$install_date",
"platform": { "$toLower": "$platform" }
},
"count": { "$sum": 1 }
} },
{ "$group": {
"_id": "$_id.date",
"counts": {
"$push": {
"k": "$_id.platform",
"v": "$count"
}
}
} },
{ "$addFields": {
"install_date": "$_id",
"platform": { "$arrayToObject": "$counts" }
} },
{ "$project": { "counts": 0, "_id": 0 } }
])
For older versions, take advantage of the $cond operator in the $group pipeline step to evaluate the counts based on the platform field value, something like the following:
db.collection.aggregate([
{ "$group": {
"_id": "$install_date",
"android_count": {
"$sum": {
"$cond": [ { "$eq": [ "$platform", "android" ] }, 1, 0 ]
}
},
"ios_count": {
"$sum": {
"$cond": [ { "$eq": [ "$platform", "ios" ] }, 1, 0 ]
}
},
"facebook_count": {
"$sum": {
"$cond": [ { "$eq": [ "$platform", "facebook" ] }, 1, 0 ]
}
},
"kindle_count": {
"$sum": {
"$cond": [ { "$eq": [ "$platform", "kindle" ] }, 1, 0 ]
}
}
} },
{ "$project": {
"_id": 0, "install_date": "$_id",
"platform": {
"android": "$android_count",
"ios": "$ios_count",
"facebook": "$facebook_count",
"kindle": "$kindle_count"
}
} }
])
In the above, $cond takes a logical condition as it's first argument (if) and then returns the second argument where the evaluation is true (then) or the third argument where false (else). This makes true/false returns into 1 and 0 to feed to $sum respectively.
So for example, if { "$eq": [ "$platform", "facebook" ] }, is true then the expression will evaluate to { $sum: 1 } else it will be { $sum: 0 }