I'm using mongodb with the following collection sample
{
"_id" : ObjectId("5703750ca9c436386c4814c9"),
"user_id" : NumberLong(17),
"activitytype_id" : NumberLong(1),
"created_date" : ISODate("2015-10-03T03:52:03.000Z")
},
{
"_id" : ObjectId("5703750ca9c436386c4814ca"),
"s_id" : NumberLong(132919),
"user_id" : NumberLong(17),
"activitytype_id" : NumberLong(4),
"created_date" : ISODate("2016-03-18T17:13:43.000Z")
},
{
"_id" : ObjectId("5703750ca9c436386c4814cb"),
"s_id" : NumberLong(215283),
"user_id" : NumberLong(17),
"activitytype_id" : NumberLong(4),
"created_date" : ISODate("2015-10-03T04:12:33.000Z")
}
,
{
"_id" : ObjectId("5703750ca9c436386c4814cc"),
"s_id" : NumberLong(360888),
"user_id" : NumberLong(17),
"activitytype_id" : NumberLong(4),
"created_date" : ISODate("2015-10-03T04:12:41.000Z")
}
This is my aggregation pipeline
db.activitylogs.aggregate([
{ $group: {
_id: {
user_id: "$user_id",
activitytype_id: "$activitytype_id"
},
activity_log_docs: {
$addToSet: {
s_id: "$s_id",
friend_id: "$friend_id",
playlist_id: "$playlist_id",
created_date:"$created_date"
}
}
}},
])
I need to get distinct s_id in activity_log_docs.
here is a screenshot for the result,
screen shot for the result
i need to avoid duplicated s_id in activity_log_docs array, so i will get distinct s_id
I think something like this should do :
db.activitylogs.aggregate([
{ $group: {
_id: {
user_id: "$user_id",
activitytype_id: "$activitytype_id" ,
s_id:"$s_id"
},
friend_id: {$first:"$friend_id"}}},
playlist_id: {$first:"$playlist_id"}}},
created_date: {$first:"$created_date"}}},
{ $group: {
_id: {
user_id: "$_id.user_id",
activitytype_id: "$_id.activitytype_id"
},
activity_log_docs: {
$addToSet: {
s_id: "$_id.s_id",
friend_id: "$friend_id",
playlist_id: "$playlist_id",
created_date:"$created_date"
}
}
}},
])
But please double check your own field's name.
Related
I have collection like this.
[{
"_id" : ObjectId("62bae0858e4132ca723f00d4"),
"appliedDate" : ISODate("2022-06-28T00:00:00Z"),
"status" : "Approved"
},
{
"_id" : ObjectId("62bae0858e4132ca723f00d4"),
"appliedDate" : ISODate("2022-06-24T00:00:00Z"),
"status" : "Applied"
},
{
"_id" : ObjectId("62bae0858e4132ca723f00d4"),
"appliedDate" : ISODate("2022-06-25T00:00:00Z"),
"status" : "Applied"
},
{
"_id" : ObjectId("62bae0858e4132ca723f00d4"),
"appliedDate" : ISODate("2022-06-25T00:00:00Z"),
"status" : "Absent"
}]
I need to sort the status by Applied 1st and then by appliedDate
db.leaverequest.aggregate([
{ $match: { $text: { $search: "Applied" } } },
{ $sort: { score: { $meta: "textScore" } } }
]).pretty()
I tried the above aggregate query but its not working as expected
You can do like this
db.collection.aggregate([
{
"$sort": {
"status": 1, //sort by status
"appliedDate": 1 //if same, use appliedDate for collision resolution
}
}
])
Here my records for mongdb
{ "_id" : 1,"userId" : "x", "name" : "Central", "borough": "Manhattan"},
{ "_id" : 2,"userId" : "x", "name" : "Rock", "borough" : "Queens"},
{ "_id" : 3,"userId" : "y", "name" : "Empire", "borough" : "Brooklyn"},
{ "_id" : 4,"userId" : "y", "name" : "Stana", "borough" : "Manhattan"},
{ "_id" : 5,"userId" : "y", "name" : "Jane", "borough" :"Brooklyn"}
how can we take result with aggregate by userId field like this
[
{
x : [{"_id":1,"name":"Central"},{"_id":2,"name":"Rock"}],
y:[{ "_id" : 3,"name":"Empire"},{ "_id" : 4,"name":"Stana"},{ "_id" : 5,"name":"Jane"}]
}
]
$group by userId and construct docs array with required fields
$group by null and construct array of docs in key-value format
$arrayToObject convert docs array to object
$replaceRoot to replace above converted object to root
db.collection.aggregate([
{
$group: {
_id: "$userId",
docs: {
$push: {
_id: "$_id",
name: "$name"
}
}
}
},
{
$group: {
_id: null,
docs: {
$push: {
k: "$_id",
v: "$docs"
}
}
}
},
{ $replaceRoot: { newRoot: { $arrayToObject: "$docs" } } }
])
Playground
We have nested document and trying to group by array element. Our document structure looks like
/* 1 */
{
"_id" : ObjectId("5a690a4287e0e50010af1432"),
"slug" : [
"true-crime-the-10-most-infamous-american-murder-mysteries",
"10-most-infamous-american-murder-mysteries"
],
"tags" : [
{
"id" : "59244aa6b1be5055278e9b5b",
"name" : "true crime",
"_id" : "59244aa6b1be5055278e9b5b"
},
{
"id" : "5924524db1be5055278ebd6e",
"name" : "Occult Museum",
"_id" : "5924524db1be5055278ebd6e"
},
{
"id" : "5a690f0fc1a72100110c2656",
"_id" : "5a690f0fc1a72100110c2656",
"name" : "murder mysteries"
},
{
"id" : "59244d71b1be5055278ea654",
"name" : "unsolved murders",
"_id" : "59244d71b1be5055278ea654"
}
]
}
We want to find list of all slugs group by tag name. I am trying with following and it gets result but it isn't accurate. We have hundreds of records with each tag but i only get few with my query. I am not sure what i am doing wrong here.
Thanks in advance.
// Requires official MongoShell 3.6+
db.getCollection("test").aggregate(
[
{
"$match" : {
"item_type" : "Post",
"site_id" : NumberLong(2),
"status" : NumberLong(1)
}
},
{$unwind: "$tags" },
{
"$group" : {
"_id" : {
"tags᎐name" : "$tags.name",
"slug" : "$slug"
}
}
},
{
"$project" : {
"tags.name" : "$_id.tags᎐name",
"slug" : "$_id.slug",
"_id" : NumberInt(0)
}
}
],
{
"allowDiskUse" : true
}
);
Expected output is
TagName Slug
----------
true crime "true-crime-the-10-most-infamous-american-murder-mysteries",
"10-most-infamous-american-murder-mysteries"
"All records where tags true crime"
Instead of using slug as a part of _id you should use $push or $addToSet to accumulate them, try:
db.test.aggregate([
{
$unwind: "$tags"
},
{
$unwind: "$slug"
},
{
$group: {
_id: "$tags.name",
slugs: { $addToSet: "$slug" }
}
},
{
$project: {
_id: 1,
slugs: {
$reduce: {
input: "$slugs",
initialValue: "",
in: {
$concat: [ "$$value", ",", "$$this" ]
}
}
}
}
}
])
EDIT: to get comma separated string for slugs you can use $reduce with $concat
Output:
{ "_id" : "murder mysteries", "slugs" : ",10-most-infamous-american-murder-mysteries,true-crime-the-10-most-infamous-american-murder-mysteries" }
{ "_id" : "Occult Museum", "slugs" : ",10-most-infamous-american-murder-mysteries,true-crime-the-10-most-infamous-american-murder-mysteries" }
{ "_id" : "unsolved murders", "slugs" : ",10-most-infamous-american-murder-mysteries,true-crime-the-10-most-infamous-american-murder-mysteries" }
{ "_id" : "true crime", "slugs" : ",10-most-infamous-american-murder- mysteries,true-crime-the-10-most-infamous-american-murder-mysteries" }
I can use this query to get the average sqmPrice for a myArea
db.getCollection('sold').aggregate([
{$match:{}},
{$group: {_id: "$myArea", "sqmPrice": {$avg: "$sqmPrice"} }}
])
Output:
[
{
"_id" : "Yttre Aspudden",
"sqmPrice" : 48845.7777777778
},
{
"_id" : "Hägerstensåsen",
"sqmPrice" : 120
}
]
I would like to group this by year, ideally an object that looks like this:
{
"Yttre Aspudden": {
2008: 1232,
2009: 1244
...
}
...
}
but the formatting is not the most important.
Here is a sample object, I would like to use soldDate:
{
"_id" : ObjectId("5beca41c78f21248ab47f4a6"),
"location" : {
"address" : {
"streetAddress" : "Ljusstöparbacken 26C"
},
"position" : {
"latitude" : 59.31427884,
"longitude" : 18.00892421
},
"namedAreas" : [
"Hägersten-Liljeholmen"
],
"region" : {
"municipalityName" : "Stockholm",
"countyName" : "Stockholms län"
},
"distance" : {
"ocean" : 3777
}
},
"listPrice" : 1895000,
"rent" : 1959,
"floor" : 1,
"livingArea" : 38.5,
"source" : {
"name" : "Fastighetsbyrån",
"id" : 1573,
"type" : "Broker",
"url" : "http://www.fastighetsbyran.se/"
},
"rooms" : 1.5,
"published" : ISODate("2018-11-02T20:55:19.000Z"),
"constructionYear" : 1959,
"objectType" : "Lägenhet",
"booliId" : 3278478,
"soldDate" : ISODate("2018-11-14T00:00:00.000Z"),
"soldPrice" : 2620000,
"soldPriceSource" : "bid",
"url" : "https://www.booli.se/annons/3278478",
"publishedDays" : 1735,
"soldDays" : 1747,
"daysUp" : 160,
"street" : "Ljusstöparbacken",
"streetYear" : "Ljusstöparbacken Hägersten-Liljeholmen 1959",
"yearDay" : 318,
"yearWeek" : 46,
"roughSize" : 40,
"sqmPrice" : 49221,
"myArea" : "Gröndal",
"hotlist" : true
}
You need to generate your keys dynamically so you have to use $arrayToObject. To build an object which aggregates the data you need three $group stages and to create new root of your document you can use $replaceRoot, try:
db.sold.aggregate([
{ $group: {_id: { area: "$myArea", year: { $year: "$soldDate" } }, "sqmPrice": {$avg: "$sqmPrice"} }},
{ $group: { _id: "$_id.area", avgs: { $push: { k: { $toString: "$_id.year" }, v: "$sqmPrice" } } } },
{ $group: { _id: null, areas: { $push: { k: "$_id", v: { $arrayToObject: "$avgs" } } } } },
{ $replaceRoot: { newRoot: { $arrayToObject: "$areas" } } }
])
I have a database of teachers details as given
{ "_id" : ObjectId("5bcc0a44f2752576a8545d99"), "Teacher_id" : "Pic002", "Teacher_Name" : "Ravi Kumar", "Dept_Name" : "IT", "Salary" : 40000, "Status" : "A" }
{ "_id" : ObjectId("5bcc0a5af2752576a8545d9a"), "Teacher_id" : "Pic003", "Teacher_Name" : "Akshay", "Dept_Name" : "Comp", "Salary" : 25500, "Status" : "N" }
{ "_id" : ObjectId("5bcc0a85f2752576a8545d9b"), "Teacher_id" : "Pic003", "Teacher_Name" : "Akshay", "Dept_Name" : "Comp", "Salary" : 25500, "Status" : "N" }
{ "_id" : ObjectId("5bcc0a9af2752576a8545d9c"), "Teacher_id" : "Pic004", "Teacher_Name" : "Sumit", "Dept_Name" : "Mech", "Salary" : 35000, "Status" : "N" }
How would I list down complete details of a teacher whose Department Name is distinct?
Basically, I want to display the details of the first first and last document in this collection.
You can achieve this via this aggregation:
db.collection.aggregate([{
$group: {
_id: "$Dept_Name",
docs: {
$addToSet: "$$CURRENT"
},
count: {
$sum: 1
}
}
},
{
$match: {
"count": {
"$eq": 1
}
}
},
{
$unwind: "$docs"
},
{
$replaceRoot: {
newRoot: "$docs"
}
}
])
The idea is first to $group and at the same time keep the objects via $addToSet.
Then filter (via $match) on those which count is 1 and then $unwind & $replaceRoot.
See it working here