Generally we can select a column in mongodb like follows,
db.TestData.find({"_id" : ObjectId("53eba30740adad086ce3599e")},{_id:1})
In sql we are selecting a same column twice like follows,
Select column1,column1 from table1
I need a mongodb query to select like above.
How can I do that?
thanks in advance. . . .
You could either use a javascript function to manipulate it, or do a basic aggregate query using the project operator:
db.accounts.aggregate({ $project: { _id: 1, _id2: '$_id' }});
// Or using a match to filter the results first
db.accounts.aggregate(
{ $match: { _id: ObjectId("53eba30740adad086ce3599e") } },
{ $project: { _id: 1, _id2: '$_id' } }
);
Related
I have a schema which has one field named ownerId and a field which is an array named participantIds. In the frontend users can select participants. I'm using these ids to filter documents by querying the participantIds with the $all operator and the list of participantsIds from the frontend. This is perfect except that the participantsIds in the document don't include the ownerId. I thought about using aggregate to add a new field which consists of a list like this one: [participantIds, ownerId] and then querying against this new field with $all and after that delete the field again since it isn't need in the frontend.
How would such a query look like or is there any better way to achieve this behavior? I'm really lost right now since I'm trying to implement this with mongo_dart for the last 3 hours.
This is how the schema looks like:
{
_id: ObjectId(),
title: 'Title of the Event',
startDate: '2020-09-09T00:00:00.000',
endDate: '2020-09-09T00:00:00.000',
startHour: 1,
durationHours: 1,
ownerId: '5f57ff55202b0e00065fbd10',
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13'],
classesIds: [],
categoriesIds: [],
roomsIds: [],
creationTime: '2020-09-10T16:42:14.966',
description: 'Some Desc'
}
Tl;dr I want to query documents with the $all operator on the participantsIds field but the ownerId should be included in this query.
What I want is instead of querying against:
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13']
I want to query against:
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13', '5f57ff55202b0e00065fbd10']
Having fun here, by the way, it's better to use Joe answer if you are doing the query frequently, or even better a "All" field on insertion.
Additional Notes: Use projection at the start/end, to get what you need
https://mongoplayground.net/p/UP_-IUGenGp
db.collection.aggregate([
{
"$addFields": {
"all": {
$setUnion: [
"$participantsIds",
[
"$ownerId"
]
]
}
}
},
{
$match: {
all: {
$all: [
"5f57ff55202b0e00065fbd14",
"5f57ff55202b0e00065fbd15",
"5f57ff55202b0e00065fbd13",
"5f57ff55202b0e00065fbd10"
]
}
}
}
])
Didn't fully understand what you want to do but maybe this helps:
db.collection.find({
ownerId: "5f57ff55202b0e00065fbd10",
participantsIds: {
$all: ['5f57ff55202b0e00065fbd14',
'5f57ff55202b0e00065fbd15',
'5f57ff55202b0e00065fbd13']
})
You could use the pipeline form of update to either add the owner to the participant list or add a new consolidated field:
db.collection.update({},[{$set:{
allParticipantsIds: {$setUnion: [
"$participantsIds",
["$ownerId"]
]}
}}])
I tried to show particular columns in mongodb colletion.but its not working.how to show particular columnns.
user_collection
[{
"user_name":"hari",
"user_password":"123456"
}]
find_query
db.use_collection.find({},{projection:{user_name:1}})
I got output
[{
"user_name":"hari",
"user_password":"123456"
}]
Excepted output
[{
"user_name":"hari",
}]
Try:
db.use_collection.find({}, {user_name:1, _id: 0 })
In that way you get the field user_name and exclude the _id.
Extra info:
project fields and project fields excluding the id
With aggregate:
db.use_collection.aggregate( [ { $project : { _id: 0, user_name : 1 } } ] )
You can try this
Mongo query:
db.users.aggregate([
{
"$project":
{
"_id": 0,
"first_name": 1,
}
}
])
Or in ruby (Mongoid)
User.collection.aggregate(
[
"$project":
{
"_id": 0,
"first_name": 1,
}
]
)
If you try to inspect the record, you can convert it into an array first (e.g. User.collection.aggregate(...).to_a)
You can use the official mongodb reference when writing in Mongoid, usually you just need to use double quote on the property name on the left hand side, to make it work on Mongoid.
Try:
db.use_collection.find({}, {user_password:0, _id: 0 ,user_name:1 })
In my MongoDB backend I want to create an endpoint that returns all the unique values for a property called department. Now, if I were doing this in IntelliShell I would just do something like:
db.staffmembers.distinct( "department" )
This will return an array of all the values for department.
But how do you return all unique values within a Mongoose find() query like this one?
Staffmember.find({ name: 'john', age: { $gte: 18 }});
In other words, what would the syntax look like if I want to use a find() like above, to return all unique values for department within the "staffmembers" collection?
You can use .aggregate() and pass your condition into $match stage and then use $addToSet within $group to get unique values.
let result = await Staffmember.aggregate([
{ $match: { name: 'john', age: { $gte: 18 }} },
{ $group: { _id: null, departments: { $addToSet: "$department" } } }
]);
We can use find and distinct like this for the above scenario. Aggregate might be a little overkill. I have tried both the solutions below.
Staffmember.find({name: 'john', age: {$gte: 18}}).distinct('department',
function(error, departments) {
// departments is an array of all unique department names
}
);
Staffmember.distinct('department', {name:'john', age:{ $gte: 18 }},
function(error, departments) {
// departments is an array of all unique department names
}
);
This link just nails it with all different possibilities:
How do I query for distinct values in Mongoose?
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]}
])
I am trying to put a limit on the number of distinct results returned from a mongoid query.
Place.where({:tags.in => ["food"]}).distinct(:name).limit(2)
But this throws up the following error:
NoMethodError: undefined method 'limit' for ["p1", "p2", "p3", "p4"]:Array
How do I put a limit in the mongoid query instead of fetching the entire result set and then selecting the limited number of items from the array.
Thanks
distinct in MongoDB is a command and commands don't return cursors. Only cursors support limit() whereas command results don't, just like sort() isn't supported here, and I think sort is important enough to run first, otherwise you never know which "first two" distinct items you get
There is a way around this, but using the aggregation framework. In plain MongoDB query speak (as you use on the MongoDB shell), you'd use:
db.places.aggregate( [
{ $match: { 'tags' : { $in: [ 'food' ] } } },
{ $group: { '_id': '$name' } },
{ $sort: { 'name': 1 } },
{ $limit: 2 },
] );
In Mongoid I suspect you can change the first line to:
Place.collection.aggregate( [