My database schema is somewhat like
{
"_id" : ObjectId("1"),
"createdAt" : ISODate("2017-03-10T00:00:00.000+0000"),
"user_list" : [
{
"id" : "a",
"some_flag" : 1,
},
{
"id" : "b",
"some_flag" : 0,
}
]
}
What I want to do is get the document where id is b & some_flag for the user b is 0.
My query is
db.collection.find({
'createdAt': {
$gte: new Date()
},
'user_list.id': 'b',
'user_list.some_flag': 1
}).sort({
createdAt: 1
})
When I run the query in shell. It returns the doc with id 1(which it shouldn't as the value of some_flag for b is 0)
The thing happening here is,
the query 'user_list.id': user_id matches with the nested object where "id" : b
'user_list.some_flag': 1 is matched with some_flag of nested object where "id": a (as the value of some_flag is 1 here)
What modifications should I make to compare the id & some_flag for the same nested object.
P.S. the amount of data is quite large & using aggregate will be a performance bottleneck
You should be using $elemMatch otherwise mongoDB queries are applied independently on array items, so in your case 'user_list.some_flag': 1 will be matched to array item with id a and 'user_list.id': 'b' will match array item with id b. So essentially if you want to query on array field with and logic use $elemMatch as following:
db.collection.find({
'createdAt': {
$gte: new Date()
},
user_list: {$elemMatch: {id: 'b', some_flag: 1}} // will only be true if a unique item of the array fulfill both of the conditions.
}).sort({
createdAt: 1
})
you need to try something like :
db.collection.find({
'createdAt': {
$gte: new Date()
},
user_list: {
$elemMatch: {
id: 'b',
some_flag: 1
}
}
}).sort({
createdAt: 1
});
This will match only user_list entries where _id is b and someflag is 1
Related
Hello i would like to display n countries with the oldest date in field independence but without null value.
I have something like this but i dont know how to exclude all null values.
db.countries.find({}, {
"independence": 1,
"_id": 0
}).sort({
"independence": 1
}).limit(10);
this my collection
{
"name" : "Example",
"code" : "AXB",
"independence" : null,
}
You can use $exists to select documents where independence field is present:
db.countries.find({
independence: {
$exists: true
}
}, {
"independence": 1,
"_id": 0
}).sort({
"independence": 1
}).limit(10);
If independence field can exist with null value and you need to fetch documents where that field is a date then you can use $type:
db.countries.find({
independence: {
$type: "date"
}
})
Is there such a query that gets multiple fields, and returns which of these exists in the collection?
For example, if the collection has only:
{id : 1}
{id : 2}
And I want to know which of [{id : 1} , {id : 3}] exists in it, then the result will be something like [{id : 1}].
You are looking for the $in-operator.
db.collection.find({ id: { $in: [ 1, 3 ] } });
This will get you any documents where the id-field (different from the special _id field) is 1 or 3. When you only want the values of the id field and not the whole documents:
db.collection.find({ id: { $in: [ 1, 3 ] } }, { _id: false, id:true });
If you want to check provided key with value is present or not in collection, you can simply check by matching values and combining conditions using $or operator.
By considering id is different than _id in mongo.
You can use $or to get expected output and query will be as following.
db.collection.find({$or:[{"id":1},{"id":3}]},{"_id":0,"id":1})
If you want to match _id then use following query:
db.collection.find({$or:[{"_id":ObjectId("557fda78d077e6851e5bf0d3")},{"_id":ObjectId("557fda78d077e6851e5bf0d5")}]}
Say I have the following four documents in a collection called "Store":
{ item: 'chair', modelNum: 1154, votes: 75 }
{ item: 'chair', modelNum: 1152, votes: 16 }
{ item: 'table', modelNum: 1017, votes: 24 }
{ item: 'table', modelNum: 1097, votes: 52 }
I would like to find only the documents with the highest number of votes for each item type.
The result of this simple example would return modelNum: 1154 and modelNum: 1097. Showing me the most popular model of chair and table, based on the customer inputed vote score.
What is the best way write this query and sort them by vote in descending order? I'm developing using meteor, but I don't think that should have an impact.
Store.find({????}).sort({votes: -1});
You can use $first or $last aggregation operators to achieve what you want. These operators are only useful when $group follows $sort. An example using $first:
db.collection.aggregate([
// Sort by "item" ASC, "votes" DESC
{"$sort" : {item : 1, votes : -1}},
// Group by "item" and pick the first "modelNum" (which will have the highest votes)
{"$group" : {_id : "$item", modelNum : {"$first" : "$modelNum"}}}
])
Here's the output:
{
"result" : [
{
"_id" : "table",
"modelNum" : 1097
},
{
"_id" : "chair",
"modelNum" : 1154
}
],
"ok" : 1
}
If you are looking to do this in Meteor and on the client I would just use an each loop and basic find. Minimongo keeps the data in memory so I don't think additional find calls are expensive.
like this:
Template.itemsList.helpers({
items: function(){
var itemNames = Store.find({}, {fields: {item: 1}}).map(
function( item ) { return item.item; }
);
var itemsMostVotes = _.uniq( itemNames ).map(
function( item ) {
return Store.findOne({item: item}, {sort: {votes: -1}});
}
);
return itemsMostVotes;
}
});
I have switched to findOne so this returns an array of objects rather than a cursor as find would. If you really want the cursor then you could query minimongo with the _ids from itemMostVotes.
You could also use the underscore groupBy and sortBy functions to do this.
You would need to use the aggregation framework.
So
db.Store.aggregate(
{$group:{_id:"$item", "maxVotes": {$max:"$votes"}}}
);
An example of the schema i have;
{ "_id" : 1234,
“dealershipName”: “Eric’s Mongo Cars”,
“cars”: [
{“year”: 2013,
“make”: “10gen”,
“model”: “MongoCar”,
“vin”: 3928056,
“mechanicNotes”: “Runs great!”},
{“year”: 1985,
“make”: “DeLorean”,
“model”: “DMC-12”,
“vin”: 8056309,
“mechanicNotes”: “Great Scott!”}
]
}
I wish to query and return only the value "vin" in "_id : 1234". Any suggestion is much appreciated.
You can use the field selection parameter with dot notation to constrain the output to just the desired field:
db.test.find({_id: 1234}, {_id: 0, 'cars.vin': 1})
Output:
{
"cars" : [
{
"vin" : 3928056
},
{
"vin" : 8056309
}
]
}
Or if you just want an array of vin values you can use aggregate:
db.test.aggregate([
// Find the matching doc
{$match: {_id: 1234}},
// Duplicate it, once per cars element
{$unwind: '$cars'},
// Group it back together, adding each cars.vin value as an element of a vin array
{$group: {_id: '$_id', vin: {$push: '$cars.vin'}}},
// Only include the vin field in the output
{$project: {_id: 0, vin: 1}}
])
Output:
{
"result" : [
{
"vin" : [
3928056,
8056309
]
}
],
"ok" : 1
}
If by query, you mean get the values of the vins in javascript, you could read the json into a string called theString (or any other name) and do something like:
var f = [], obj = JSON.parse(theString);
obj.cars.forEach(function(item) { f.push(item.vin) });
If your json above is part of a larger collection, then you'd need an outer loop.
I have collection messages with the following documents
{
"_id" : ObjectId("5164218f359f109fd4000012"),
"receiver_id" : ObjectId("5164211e359f109fd4000004"),
"sender_id" : ObjectId("5162de8a359f10cbf700000c"),
"body" : "Hello Billy!!!",
"readed" : false,
"updated_at" : ISODate("2013-04-09T14:11:27.17Z"),
"created_at" : ISODate("2013-04-09T14:11:27.17Z")
}
I need to make query for receive last messages(don't matter recieved or sended) for a given user (grouped by reciever_id+sender_id fields) and sorted by created_at.
To better explain the question, an example of how I did it in SQL:
SELECT DISTINCT ON (sender_id+receiver_id) * FROM messages
ORDER by (sender_id+receiver_id), created_at DESC
WHERE sender_id = given_user or receiver_id = given_user
I don't understand how to solve this problem with mondodb.
The Aggregation Framework in MongoDB 2.2+ provides the most obvious translation of your query. The MongoDB manual includes an SQL to Aggregation Framework Mapping Chart as a general guide, although there are definite differences in the two approaches.
Here's a commented example you can try in the mongo shell:
var given_user = ObjectId("5162de8a359f10cbf700000c");
db.messages.aggregate(
// match: WHERE sender_id = given_user or receiver_id = given_user
// NB: do the match first, because it can take advantage of an available index
{ $match: {
$or:[
{ sender_id: given_user },
{ receiver_id: given_user },
]
}},
{ $group: {
// DISTINCT ON (sender_id+receiver_id)
_id: { sender_id: "$sender_id", receiver_id: "$receiver_id" }
}},
// ORDER by (sender_id+receiver_id), created_at DESC
{ $sort: {
sender_id: 1,
receiver_id: 1,
created_at: -1
}}
)
Sample result:
{
"result" : [
{
"_id" : {
"sender_id" : ObjectId("5162de8a359f10cbf700000c"),
"receiver_id" : ObjectId("5164211e359f109fd4000004")
}
}
],
"ok" : 1
}
You may want to add additional fields on the grouping, such as a count of messages received.
If you actually want to combine the sender_id+receiver_id into a single field, you can use the $concat operator in MongoDB 2.4+.
There is no explicit way to do so. Let's review workarounds:
Way 1:
do the distinct at code level (after find), then just use find:
db.message.find({$or:[{sender_id:?}, {receiver_id:?}]})
Way 2:Using aggregation framework :
db.message.aggregate( [
{$match: {$or:[{sender_id:?}, {receiver_id:?}]},
$group: { _id: {sender:"$sender_id", receiver:"$receiver_id"},
other: { ... } } },
$sort: {sender_id,receiver_id,...}
] )
This way problem appears at sort level since sender_id, receiver_id is not the same as sender_id+receiver_id
Way 3: Introduce the surrogate field sender_id+receiver_id then use find or even distinct per Stennie hint.