I have a mongodb collection with many fields. One field is 'date_time', which is in an ISO datetime format, Ex: ISODate("2014-06-11T19:16:46Z"), and another field is 'name'.
Given a name, how do I find out the oldest/youngest post in the collection?
Ex: If there are two posts in the collection 'data' :
[{'name' : 'John', 'date_time' : ISODate("2014-06-11T19:16:46Z")},
{'name' : 'John', 'date_time' : ISODate("2015-06-11T19:16:46Z")}]
Given the name 'John' how do I find out the oldest post in the collection i.e., the one with ISODate("2014-06-11T19:16:46Z")? Similarly for the youngest post.
Oldest:
db.posts.find({ "name" : "John" }).sort({ "date_time" : 1 }).limit(1)
Newest:
db.posts.find({ "name" : "John" }).sort({ "date_time" : -1 }).limit(1)
Index on { "name" : 1, "date_time" : 1 } to make the queries efficient.
You could aggregate it as below:
Create an index on the name and date_time fields, so that the
$match and $sort stage operations may use it.
db.t.ensureIndex({"name":1,"date_time":1})
$match all the records for the desired name(s).
$sort by date_time in ascending order.
$group by the name field. Use the $first operator to get the first
record of the group, which will also be the oldest. Use the $last
operator to get the last record in the group, which will also be the
newest.
To get the entire record use the $$ROOT system variable.
Code:
db.t.aggregate([
{$match:{"name":"John"}},
{$sort:{"date_time":1}},
{$group:{"_id":"$name","oldest":{$first:"$$ROOT"},
"youngest":{$last:"$$ROOT"}}}
])
o/p:
{
"_id" : "John",
"oldest" : {
"_id" : ObjectId("54da62dc7f9ac597d99c182d"),
"name" : "John",
"date_time" : ISODate("2014-06-11T19:16:46Z")
},
"youngest" : {
"_id" : ObjectId("54da62dc7f9ac597d99c182e"),
"name" : "John",
"date_time" : ISODate("2015-06-11T19:16:46Z")
}
}
db.t.find().sort({ "date_time" : 1 }).limit(1).pretty()
Related
I have a simple order table in mongoDB
{
"_id" : NumberInt(2),
"bar" : "Maggie Choos Bar"
},
{
"_id" : NumberInt(3),
"bar" : "Corona Bar"
{
I want to find the BIGGEST "_id" number in the table
db.getCollection("order").find({}).sort({"$_id":-1}).limit(1);
But no matter if I sort 1 or -1 I keep getting the result with _id : 2
Any ideas?
The field name prefixed with a dollar sign, like $_id, is for references in an aggregation pipeline. For a sort document, use just the field name:
db.getCollection("order").find({}).sort({"_id":-1}).limit(1);
I have a collection of Projects in where projects are like this:
{
"_id" : ObjectId("57e3e55c638cb8b971"),
"allocInvestor" : "Example Investor",
"fieldFoo" : "foo bar",
"name" : "GTP 3 (Roof Lease)"
}
I want to receive a list of projects grouped by allocInvestor field and only show fields: name and id
If I use aggregate and $group like this:
db.getCollection('projects').aggregate([
{"$group" : {
_id:"$allocInvestor", count:{$sum:1}
}
}
])
I receive a count of project per allocInvestor but I need is to receive the list of allocInvestor with subarray of projects per allocInvestor.
I'm using meteor by the way in case that helps. But I first want to get the query right on mongodb then try for meteor.
You can use $push or $addToSet to create a list of name and _id per every group.
$push allows duplicates and $addToSet does not add an element to the list again, if it is already there.
db.getCollection('projects').aggregate([
{ "$group" : { _id : "$allocInvestor",
count : {$sum : 1},
"idList" : {"$addToSet" : "$_id"},
"nameList" : {"$addToSet":"$name"}
}
}
]);
To get the name and _id data in a single list:
db.getCollection('projects').aggregate([
{ "$group" : { _id : "$allocInvestor", "projects" : {"$addToSet" : {id : "$_id", name: "$name"}}}},
{"$project" : {"_id" : 0, allocInvestor : "$_id", "projects" : 1 }}
]);
Use the $$ROOT operator to reference the entire document and then use project to eliminate the fields that you do not require.
db.projects.aggregate([
{"$group" : {
"_id":"$allocInvestor",
"projects" : {"$addToSet" : "$$ROOT"}
}
},
{"$project" : {
"_id":0,
"allocInvestor":"$_id",
"projects._id":1
"projects.name":1
}
}
])
I have documents with my internal id field inside of each document and date when this document was added. There could be number of documents with the same id (differents versions of the same document), but dates will always be different for those documents. I want in some query, to bring only one document from all versions of the same document (with same id field) that was relevant to specified date, and I want to display them with paging (50 rows in the page). So, is there any chance to do this in MongoDB (operations - query documents by some field, group them by id field, sort by date field and take only first, and all this should be with paging.) ?
Please see example :Those are documents, some of them different documents,like documents A,B and C, and some are versions of the same documents,
like _id: 1, 2 and 3 are all version of the same document A
Document A {
_id : 1,
"id" : "A",
"author" : "value",
"date" : "2015-11-05"
}
Document A {
_id : 2,
"id" : "A",
"author" : "value",
"date" : "2015-11-06"
}
Document A {
_id : 3,
"id" : "A",
"author" : "value",
"date" : "2015-11-07"
}
Document B {
_id : 4,
"id" : "B",
"author" : "value",
"date" : "2015-11-06"
}
Document B {
_id : 5,
"id" : "B",
"author" : "value",
"date" : "2015-11-07"
}
Document C {
_id : 6,
"id" : "C",
"author" : "value",
"date" : "2015-11-07"
}
And I want to query all documents that has "value" in the "author" field.
And from those documents to bring only one document of each with latest date for
the specified date, for example 2015-11-08. So, I expect the result to be :
_id : 3, _id : 5, _id : 6
And also paging , for example 10 documents in each page.
Thanks !!!!!
Two documents can't have the same _id. There is a unique index on _id by default.
As per 1. you need to have a compound _id field which includes the date:
{
"_id":{
docId: yourFormerIdValue,
date: new ISODate()
}
// other fields
}
To get the version valid at a specified date, the query becomes rather easy:
db.yourColl.find({
"_id":{
"docId": idToFind,
// get only the version valid up to a specific date...
"date":{ "$lte": someISODate }
}
})
// ...sort the results descending...
.sort("_id.date":-1)
// ...and get only the first and therefor newest entry
.limit(1)
I have a mongo document like
{ "_id" : 12, "location" : [ "Kannur","Hyderabad","Chennai","Bengaluru"] }
{ "_id" : 13, "location" : [ "Hyderabad","Chennai","Mysore","Ballary"] }
From this how can I get the location aggregation (distinct area count).
some thing like
Hyderabad 2,
Kannur 1,
Chennai 2,
Bengaluru 1,
Mysore 1,
Ballary 1
Using aggregation you cannot get the exact output that you want. One of the limitations of aggregation pipeline is its inability to transform values to keys in the output document.
For example, Kannur is one of the values of the location field, in the input document. In your desired output structure it needs to be the key("kannur":1). This is not possible using aggregation. While, this can be used achieving map-reduce, you can however get a very closely related and useful structure using aggregation.
Unwind the location array.
Group by the location fields, get the count of individual locations
using the $sum operator.
Group again all the documents once again to get a consolidated array
of results.
Code:
db.collection.aggregate([
{$unwind:"$location"},
{$group:{"_id":"$location","count":{$sum:1}}},
{$group:{"_id":null,"location_details":{$push:{"location":"$_id",
"count":"$count"}}}},
{$project:{"_id":0,"location_details":1}}
])
Sample o/p:
{
"location_details" : [
{
"location" : "Ballary",
"count" : 1
},
{
"location" : "Mysore",
"count" : 1
},
{
"location" : "Bengaluru",
"count" : 1
},
{
"location" : "Chennai",
"count" : 2
},
{
"location" : "Hyderabad",
"count" : 2
},
{
"location" : "Kannur",
"count" : 1
}
]
}
I have a mongo document which has structure like
{
"_id" : "THIS_IS_A_DHP_USER_ID+2014-11-26",
"_class" : "weight",
"items" : [
{
"dateTime" : ISODate("2014-11-26T08:08:38.716Z"),
"value" : 98.5
},
{
"dateTime" : ISODate("2014-11-26T08:18:38.716Z"),
"value" : 95.5
},
{
"dateTime" : ISODate("2014-11-26T08:28:38.663Z"),
"value" : 90.5
}
],
"source" : "MANUAL",
"to" : ISODate("2014-11-26T08:08:38.716Z"),
"from" : ISODate("2014-11-26T08:08:38.716Z"),
"userId" : "THIS_IS_A_DHP_USER_ID",
"createdDate" : ISODate("2014-11-26T08:38:38.776Z")
}
{
"_id" : "THIS_IS_A_DHP_USER_ID+2014-11-25",
"_class" : "weight",
"items" : [
{
"dateTime" : ISODate("2014-11-25T08:08:38.716Z"),
"value" : 198.5
},
{
"dateTime" : ISODate("2014-11-25T08:18:38.716Z"),
"value" : 195.5
},
{
"dateTime" : ISODate("2014-11-25T08:28:38.716Z"),
"value" : 190.5
}
],
"source" : "MANUAL",
"to" : ISODate("2014-11-25T08:08:38.716Z"),
"from" : ISODate("2014-11-25T08:08:38.716Z"),
"userId" : "THIS_IS_A_DHP_USER_ID",
"createdDate" : ISODate("2014-11-26T08:38:38.893Z")
}
The query that want to fire on this document structure,
finding documents for a particular user id
unwinding the embedded array
Grouping the documents based over _id with -
summing the items.value of the embedded array
getting the minimum of the items.dateTime of the embedded array
Note. The sum and min, I want to get as a object i.e. { value : sum , dateTime : min of the items.dateTime} inside an array of items
Can this be achieved in an single aggregation call using push or some other technique.
When you group over a particular _id, and apply aggregation operators such as $min and $sum, there exists only one record per group(_id), that holds the sum and the minimum date for that group. So there is no way to obtain a different sum and a different minimum date for the same _id, which also logically makes no sense.
What you would want to do is:
db.collection.aggregate([
{$match:{"userId":"THIS_IS_A_DHP_USER_ID"}},
{$unwind:"$items"},
{$group:{"_id":"$_id",
"values":{$sum:"$items.value"},
"dateTime":{$min:"$items.dateTime"}}}
])
But in case when you do not query for a particular userId, then you would have multiple groups, each group having its own sum and min date. Then it makes sense to accumulate all these results together in an array using the $push operator.
db.collection.aggregate([
{$unwind:"$items"},
{$group:{"_id":"$_id",
"result":{$sum:"$items.value"},
"dateTime":{$min:"$items.dateTime"}}},
{$group:{"_id":null,"result":{$push:{"value":"$result",
"dateTime":"$dateTime",
"id":"$_id"}}}},
{$project:{"_id":0,"result":1}}
])
you should use following aggregation may it works
db.collectionName.aggregate(
{"$unwind":"$items"},
{"$match":{"userId":"THIS_IS_A_DHP_USER_ID"}},
{"$group":{"_id":"$_id","sum":{"$sum":"$items.value"},
"minDate":{"$min":"$items.dateTime"}}}
)