Is it possible to rename the name of fields returned in a find query? I would like to use something like $rename, however I wouldn't like to change the documents I'm accessing. I want just to retrieve them differently, something that works like SELECT COORINATES AS COORDS in SQL.
What I do now:
db.tweets.findOne({}, {'level1.level2.coordinates': 1, _id:0})
{'level1': {'level2': {'coordinates': [10, 20]}}}
What I would like to be returned is:
{'coords': [10, 20]}
So basically using .aggregate() instead of .find():
db.tweets.aggregate([
{ "$project": {
"_id": 0,
"coords": "$level1.level2.coordinates"
}}
])
And that gives you the result that you want.
MongoDB 2.6 and above versions return a "cursor" just like find does.
See $project and other aggregation framework operators for more details.
For most cases you should simply rename the fields as returned from .find() when processing the cursor. For JavaScript as an example, you can use .map() to do this.
From the shell:
db.tweets.find({},{'level1.level2.coordinates': 1, _id:0}).map( doc => {
doc.coords = doc['level1']['level2'].coordinates;
delete doc['level1'];
return doc;
})
Or more inline:
db.tweets.find({},{'level1.level2.coordinates': 1, _id:0}).map( doc =>
({ coords: doc['level1']['level2'].coordinates })
)
This avoids any additional overhead on the server and should be used in such cases where the additional processing overhead would outweigh the gain of actual reduction in size of the data retrieved. In this case ( and most ) it would be minimal and therefore better to re-process the cursor result to restructure.
As mentioned by #Neil Lunn this can be achieved with an aggregation pipeline:
And starting Mongo 4.2, the $replaceWith aggregation operator can be used to replace a document by a sub-document:
// { level1: { level2: { coordinates: [10, 20] }, b: 4 }, a: 3 }
db.collection.aggregate(
{ $replaceWith: { coords: "$level1.level2.coordinates" } }
)
// { "coords" : [ 10, 20 ] }
Since you mention findOne, you can also limit the number of resulting documents to 1 as such:
db.collection.aggregate([
{ $replaceWith: { coords: "$level1.level2.coordinates" } },
{ $limit: 1 }
])
Prior to Mongo 4.2 and starting Mongo 3.4, $replaceRoot can be used in place of $replaceWith:
db.collection.aggregate(
{ $replaceRoot: { newRoot: { coords: "$level1.level2.coordinates" } } }
)
As we know, in general, $project stage takes the field names and specifies 1 or 0/true or false to include the fields in the output or not, we also can specify the value against a field instead of true or false to rename the field. Below is the syntax
db.test_collection.aggregate([
{$group: {
_id: '$field_to_group',
totalCount: {$sum: 1}
}},
{$project: {
_id: false,
renamed_field: '$_id', // here assigning a value instead of 0 or 1 / true or false effectively renames the field.
totalCount: true
}}
])
Stages (>= 4.2)
$addFields : {"New": "$Old"}
$unset : {"$Old": 1}
Related
I have a collection T, with 2 fields: Grade1 and Grade2, and I want to select those with condition Grade1 > Grade2, how can I get a query like in MySQL?
Select * from T Where Grade1 > Grade2
You can use a $where. Just be aware it will be fairly slow (has to execute Javascript code on every record) so combine with indexed queries if you can.
db.T.find( { $where: function() { return this.Grade1 > this.Grade2 } } );
or more compact:
db.T.find( { $where : "this.Grade1 > this.Grade2" } );
UPD for mongodb v.3.6+
you can use $expr as described in recent answer
You can use $expr ( 3.6 mongo version operator ) to use aggregation functions in regular query.
Compare query operators vs aggregation comparison operators.
Regular Query:
db.T.find({$expr:{$gt:["$Grade1", "$Grade2"]}})
Aggregation Query:
db.T.aggregate({$match:{$expr:{$gt:["$Grade1", "$Grade2"]}}})
If your query consists only of the $where operator, you can pass in just the JavaScript expression:
db.T.find("this.Grade1 > this.Grade2");
For greater performance, run an aggregate operation that has a $redact pipeline to filter the documents which satisfy the given condition.
The $redact pipeline incorporates the functionality of $project and $match to implement field level redaction where it will return all documents matching the condition using $$KEEP and removes from the pipeline results those that don't match using the $$PRUNE variable.
Running the following aggregate operation filter the documents more efficiently than using $where for large collections as this uses a single pipeline and native MongoDB operators, rather than JavaScript evaluations with $where, which can slow down the query:
db.T.aggregate([
{
"$redact": {
"$cond": [
{ "$gt": [ "$Grade1", "$Grade2" ] },
"$$KEEP",
"$$PRUNE"
]
}
}
])
which is a more simplified version of incorporating the two pipelines $project and $match:
db.T.aggregate([
{
"$project": {
"isGrade1Greater": { "$cmp": [ "$Grade1", "$Grade2" ] },
"Grade1": 1,
"Grade2": 1,
"OtherFields": 1,
...
}
},
{ "$match": { "isGrade1Greater": 1 } }
])
With MongoDB 3.4 and newer:
db.T.aggregate([
{
"$addFields": {
"isGrade1Greater": { "$cmp": [ "$Grade1", "$Grade2" ] }
}
},
{ "$match": { "isGrade1Greater": 1 } }
])
In case performance is more important than readability and as long as your condition consists of simple arithmetic operations, you can use aggregation pipeline. First, use $project to calculate the left hand side of the condition (take all fields to left hand side). Then use $match to compare with a constant and filter. This way you avoid javascript execution. Below is my test in python:
import pymongo
from random import randrange
docs = [{'Grade1': randrange(10), 'Grade2': randrange(10)} for __ in range(100000)]
coll = pymongo.MongoClient().test_db.grades
coll.insert_many(docs)
Using aggregate:
%timeit -n1 -r1 list(coll.aggregate([
{
'$project': {
'diff': {'$subtract': ['$Grade1', '$Grade2']},
'Grade1': 1,
'Grade2': 1
}
},
{
'$match': {'diff': {'$gt': 0}}
}
]))
1 loop, best of 1: 192 ms per loop
Using find and $where:
%timeit -n1 -r1 list(coll.find({'$where': 'this.Grade1 > this.Grade2'}))
1 loop, best of 1: 4.54 s per loop
How can I get documents from mongo with an array containing some elements but IN THE SAME ORDER?
I know that $all do the job but ignoring the order of elements. The order in my case is important and I can't sort my arrays since it's describing a path that I want to keep the order.
111,222,333 is not the same as 222,111,333
Is there a way to do it using $all or maybe another operator in mongo aggregation framework?
You can avoid the first "intersect" field, is just to give you back as debug what MongoDB make with this command. You should create the $and operator dynamically.
db.Test6.aggregate([
{
$project: {
_id:1,
pages:1,
intersect: {$setIntersection: [[111,666], "$pages"]},
theCondition: {$let: {
vars: {
intersect: {$setIntersection: [[111,666], "$pages"]}
},
in: {
$cond:[ {$and:[
{$eq:[{$arrayElemAt:["$$intersect", 0]}, 111]},
{$eq:[{$arrayElemAt:["$$intersect", 1]}, 666]}
]} , true, false]
}
}
}
}
}
]);
I should first point out I am not attacking Mongo here. I like Mongo. Mongo is cool.
If I have data like this:
{
a: 2,
b: 2
},
{
a:3,
b:4
}
I can use aggregation with $project to format the results:
db.collection.aggregate({
$project: {
c: {
$add: ["$a", "$b"]
}
}
})
which will yield
{
c: 4
},
{
c:7
}
I notice the find function also supports projection as a second argument. However if I do this:
db.collection.find({}, {
c: {
$add: ["$a", "$b"]
}
});
The query fails.
Can I enable expression based projection in find, and if not, is this by accident or design, or my own dimwittery?
The projection inside find and $project work differently. As you may find in the documentation, the $add is defined for and under the aggregation framework.
By the way, another difference is that for example, in find projection you can exclude fields using the {field: 0} notation, but that is not possible in the aggregation pipeline $project.
I have the following data in MongoDB (simplified for what is necessary to my question).
{
_id: 0,
actions: [
{
type: "insert",
data: "abc, quite possibly very very large"
}
]
}
{
_id: 1,
actions: [
{
type: "update",
data: "def"
},{
type: "delete",
data: "ghi"
}
]
}
What I would like is to find the first action type for each document, e.g.
{_id:0, first_action_type:"insert"}
{_id:1, first_action_type:"update"}
(It's fine if the data structured differently, but I need those values present, somehow.)
EDIT: I've tried db.collection.find({}, {'actions.action_type':1}), but obviously that returns all elements of the actions array.
NoSQL is quite new to me. Before, I would have stored all this in two tables in a relational database and done something like SELECT id, (SELECT type FROM action WHERE document_id = d.id ORDER BY seq LIMIT 1) action_type FROM document d.
You can use $slice operator in projection. (but for what you do i am not sure that the order of the array remain the same when you update it. Just to keep in mind))
db.collection.find({},{'actions':{$slice:1},'actions.type':1})
You can also use the Aggregation Pipeline introduced in version 2.2:
db.collection.aggregate([
{ $unwind: '$actions' },
{ $group: { _id: "$_id", first_action_type: { $first: "$actions.type" } } }
])
Using the $arrayElemAt operator is actually the most elegant way, although the syntax may be unintuitive:
db.collection.aggregate([
{ $project: {first_action_type: {$arrayElemAt: ["$actions.type", 0]}
])
Mongo DB: I'm looking to make one query to return both the first and last element of an array. I realize that I can do this multiple queries, but I would really like to do it with one.
Assume a collection "test" where each objects has an array "arr" of numbers:
db.test.find({},{arr:{$slice: -1},arr:{$slice: 1}});
This will result in the following:
{ "_id" : ObjectId("xxx"), "arr" : [ 1 ] } <-- 1 is the first element
Is there a way to maybe alias the results? Similar to what the mysql AS keyword would allow in a query?
This is not possible at the moment but will be with the Aggregation Framework that's in development now if I understand your functional requirement correctly.
You have to wonder about your schema if you have this requirement in the first place though. Are you sure there isn't a more elegant way to get this to work by changing your schema accordingly?
This can be done with the aggregation framework using the operators $first and $last as follows:
db.test.aggregate([
{ '$addFields': {
'firstElem': { '$first': '$arr' },
'lastElem': { '$last': '$arr' }
} }
])
or using $slice as
db.test.aggregate([
{ '$addFields': {
'firstElem': { '$slice': [ '$arr', 1 ] },
'lastElem': { '$slice': [ '$arr', -1 ] }
} }
])