I have the following data structure on my game collection:
{
name: game1
date: 2010-10-10
media: [{
id: 1,
created: 2010-10-10 00:00:59
}, {
id: 2,
created: 2010-10-10 00:00:30
}]
},
{
name: game2
date: 2010-10-09
media: [{
id: 1,
created: 2010-10-09 00:10:40
}, {
id: 2,
created: 2010-10-09 09:01:00
}]
}
I want to get the game with the highest date, then get the related media with the highest created to get it's id. In the example above, the result would be
{
name: game1
date: 2010-10-10
media: [{
id: 1,
created: 2010-10-10 00:00:59
}]
}
I tried to use the find and find_one, and also aggregation, but I could not figure a way to make this query.
Any suggestions?
You will need to $unwind the media array in order to get the subdocument in that array where created is the highest then you $sort your documents by date and created all in descending order. Use $limit to output n documents which is 1 in our case.
In [26]: import pymongo
In [27]: conn = pymongo.MongoClient()
In [28]: db = conn.test
In [29]: col = db.gamers
In [30]: list(col.aggregate([{"$unwind": "$media"}, {"$sort": {"date": -1, "media.created": -1}}, {"$limit": 1}]))
Out[30]:
[{'_id': ObjectId('553323ec0acf450bc6b7438c'),
'date': '2010-10-10',
'media': {'created': '2010-10-10 00:00:59', 'id': 1},
'name': 'game1'
}]
Related
I've Just started to work with MongoDB, so you might find my question really stupid.I tried to search a lot before posting my query here, Any Help would be Appreciated.
I also came across this link StackOverFlow Link, which advised to apply .sort() on every query, but that would increase the query time.
So I tried to index my collection using .createIndexes({_id:-1}), to sort data in descending order of creation time(newest to oldest), After that when I used the .find() method to get data in sorted format(newest to Oldest) I did'nt get the desired result , I still had to sort the data :( .
// connecting db
mongoose.connect(dbUrl, dbOptions);
const db = mongoose.connection;
// listener on db events
db.on('open', ()=>{console.log('DB SUCESSFULLY CONNECTED !!');});
db.on('error', console.error.bind(console, 'connection error:'));
// creating Schema for a person
const personSchma = new mongoose.Schema(
{ name: String,
age : Number}
)
// creating model from person Schema
const person = mongoose.model('person', personSchma);
// Chronological Order of Insertion Of Data
// {name: "kush", age:22}
// {name: "clutch", age:22}
// {name: "lauv", age:22}
person.createIndexes({_id:-1}, (err)=>{
if (err){
console.log(err);
}
})
person.find((err, persons)=>{
console.log(persons)
// Output
// [
// { _id: 6026eadd58a2b124d85b0f8d, name: 'kush', age: 22, __v: 0 },
// { _id: 6026facdf200f8261005f8e0, name: 'clutch', age: 22, __v: 0 },
// { _id: 6026facdf200f8261005f8e1, name: 'lauv', age: 22, __v: 0 }
// ]
})
person.find().sort({_id:-1}).lean().limit(100).then((persons)=>{
console.log(persons);
// Output
// [
// { _id: 6026facdf200f8261005f8e1, name: 'lauv', age: 22, __v: 0 },
// { _id: 6026facdf200f8261005f8e0, name: 'clutch', age: 22, __v: 0 },
// { _id: 6026eadd58a2b124d85b0f8d, name: 'kush', age: 22, __v: 0 }
// ]
})
Indexes are special data structure, which can be used to run the queries efficiently. While running the query, MongoDB tries to see which index should be used for running the query efficiently and then that index will be used.
Creating an index with {_id:-1} will create an auxiliary data structure(index) which will be sorted newest first. It doesn't affect the order of the data which we are storing.
To sort the data in descending order(newest first) we will have to explicitly add the sort operation in your query and make sure that an index for descending order _id is present.
I have a model like this:
[{item: {
_id: 123,
field1: someValue,
price: {
[_id: 456,
field1: anotherValue,
type: []],
[_id: 789,
field1: anotherValue,
type: ['super']]
}
}]
I need to find an item by 3 parameters: item _id, price _id, and check if price type array is empty. And check it in one price field.
Model.findOneAndUpdate({_id: 123, "price._id": 456, "price.type": {size:0})
This query always returns item, cause search in different prices.
Model.findOneAndUpdate({_id: 123, price: {id: 456, type: {size:0})
This query returns error (cast array value or something like this).
tried to build query with $in, $and, but still getting an error
Use $elemMatch:
The $elemMatch operator matches documents that contain an array field
with at least one element that matches all the specified query
criteria.
db.inventory.find({
price: {
"$elemMatch": {
_id: 456,
type: {
$size: 0
}
}
}
})
Using Mongo 3.2.
Let's say I have a collection with this schema:
{ _id: 1, type: a, source: x },
{ _id: 2, type: a, source: y },
{ _id: 3, type: b, source: x },
{ _id: 4, type: b, source: y }
Of course that my db is much larger and with many more types and sources.
I have created 4 indexes combinations of type and source (even though 1 should be enough):
{type: 1}
{source: 1},
{type: 1, source: 1},
{source: 1, type: 1}
Now, I am running this distinct query:
db.test.distinct("source", {type: "a"})
The problem is that this query takes much more time that it should take.
If I run it with runCommand:
db.runCommand({distinct: 'test', key: "source", query: {type: "a"}})
this is the result i get:
{
"waitedMS": 0,
"values": [
"x",
"y"
],
"stats": {
"n": 19400840,
"nscanned": 19400840,
"nscannedObjects": 19400840,
"timems": 14821,
"planSummary": "IXSCAN { type: 1 }"
},
"ok": 1
}
For some reason, mongo use only the type: 1 index for the query stage.
It should use the index also for the distinct stage.
Why is that? Using the {type: 1, source: 1} index would be much better, no? right now it is scanning all the type: a documents while it has an index for it.
Am I doing something wrong? Do I have a better option for this kind of distinct?
As Alex mentioned, apparently MongoDB doesn't support this right now.
There is an open issue for it:
https://jira.mongodb.org/browse/SERVER-19507
Just drop first 2 indexes. You don't need them. Mongo can use {type: 1, source: 1} in any query that may need {type: 1} index.
Suppose, I have the following database:
{
_id: 1,
name: 'Alice',
courses: [
{
_id: 'DB103',
credits: 6
},
{
_id: 'ML203',
credits: 4
}
]
},
{
_id: 2,
name: 'Bob',
courses: []
}
I now want to 'upsert' the document with the course id 'DB103' in both documents. Although the _id field should remain the same, the credits field value should change (i.e. to 4). In the first document, the respective field should be changed, in the second one, {_id: 'DB103', credits: 4} should be inserted into the courses array.
Is there any possibility in MongoDB to handle both cases?
Sure, I could search with $elemMatch in courses for 'DB103' and if I haven't found it, insert, otherwise update the value. But these are two steps and I would like to do both in just one.
I have a collection like this
{
_id: "qex6zPLgigq4KGpFP",
active: true,
name: "event name",
questions: [{
_id: "cff4cc7f1ae67282f1ea142b",
createdBy: "izRFAkhxgemihkRMF",
likes: 2
}, {
_id: "2cf23241ca456703f50a9ce4",
createdBy: "izRFAkhxgemihkRMF",
likes: 0
}]
}
and I initialize my collection in my Meteor application like this:
Events = new Mongo.Collection("events");
Now I want to remove a subdocument from the 'questions' array and it works in the shell with
db.events.update({_id:"qex6zPLgigq4KGpFP"}, {$pull: {'questions': {_id:"cff4cc7f1ae67282f1ea142b"}}})
but NOT in Meteor by calling
Events.update({_id:"qex6zPLgigq4KGpFP"}, {$pull: {'questions': {_id:"cff4cc7f1ae67282f1ea142b"}}})
The return value is always '1', but the document is still there.
It's confusing. Any ideas?