I'm trying to order my collection using the following query:
db.getCollection('trip').find().sort({'itinerary.0.timestamp': 1})
The result is not being correctly sorted, however I exported the full collection to a local mongoDB database and the same query works like a charm. In order to perform that sort in cosmos DB I had to create the index 'itinerary.0.timestamp'.
data example:
{
"_id" : ObjectId("6087104ca68f171ce7715448"),
"tripId" : NumberLong(38533184),
"itinerary" : [
{
"transId" : NumberLong(39800097),
"timestamp" : NumberLong(1619372446291)
},
{
"transId" : NumberLong(39800576),
"timestamp" : NumberLong(1619372446321)
},
],
"results" : [],
"tripTimeSent" : ISODate("2021-04-29T14:44:53.253Z")
}
What am I missing?
Thanks!!
The solution was to create a new field, itiTimestamp, outside the array containing the value 'itinerary.0.timestamp'. Then just order by itiTimestamp
It's true that you need to create an index for the sort field. Here's the doc related:
To apply a sort to a query, you must create an index on the fields
used in the sort operation.
==========================================
I've tested in my side, after creating wildcard index on itinerary, sort query could be executed but has no luck. I also refer to this answer(new BasicDBObject("labels.0.value", 1)) and this one(db.testCollection.find().sort({"someArray.0": 1})), they all don't work for the date format Op provided.
But when I added a properity "score":[20,55,80] in each item in the collection, I found it can be sorted by the first item when sort by score directly.
I assume that this feature hasn't supported.
Related
Update command updates the key with provided Json. I want to update only the object that is not present in db and changed value. How can I do that?
"data" : [
{
"_id" : "5bb6253d861d057857ec3ff0",
"name" : "C"
},
{
"_id" : "5bb625fc861d057857ec3ff1",
"name" : "B"
},
{
"_id" : "5bb625fe861d057857ec3ff2",
"name" : "A"
}
]
my data is like this. So, if one more array object comes in json of only 2 new object comes then it should insert the two data along with the 3 data.
Update the object that is not present in DB:
Use upsert: Upsert creates a new document when no document matches the query criteria. Alternatively, you can add null checks in your query e.g { user_id:null }. This will allow to update the data where a record for the user is not present in DB.
Update changed value:
This can be implemented maintaining a key to store last_updated_at. If the last_updated_at value does not match to the previously_updatede_at that record can be treated at modified
You can implement Change Streams, introduced in MongoDB 3.6 from which you can receive real time changes on your data. You can receive only the data that was changed, filtering by the "update" operation. Furthermore you can also filter for data that is newly inserted filtering by the "insert" operation. Please see Change Streams.
My Mongodb dataset is like this
{
"_id" : ObjectId("5a27cc4783800a0b284c7f62"),
"action" : "1",
"silent" : "0",
"createdate" : ISODate("2017-12-06T10:53:59.664Z"),
"__v" : 0
}
Now I have to find the data whose Action value is 1 and silent value is 0. one more thing is that all the data returns is descending Order.
My Mongodb Query is
db.collection.find({'action': 1, 'silent': 0}).sort({createdate: -1}).exec(function(err, post) {
console.log(post.length);
});
Earlier It works Fine for me But Now I have 121000 entry on this Collection. Now it returns null.
I know there is some confusion on .sort()
If i remove the sort Query then everything is fine. Example
db.collection.find({'action': 1, 'silent': 0}).exec(function(err, post) {
console.log(post.length);// Now it returns data but not on Descending order
});
MongoDB limits the amount of data it will attempt to sort without an index .
This is because Mongo has to sort the data in memory or on disk, both of which can be expensive operations, particularly for queries run frequently.
In most cases, this can be alleviated by creating indexes on the fields you sort on.
you can create index with :-
db.myColl.createIndex( { createdate: 1 })
thanks !
I have created a collection and added just a name field and tried to apply the following index.
db.names.createIndex({"name":1})
Even after applying the index I see the below result.
db.names.find()
{ "_id" : ObjectId("57d14139eceab001a19f7e82"), "name" : "kkkk" } {
"_id" : ObjectId("57d1413feceab001a19f7e83"), "name" : "aaaa" } {
"_id" : ObjectId("57d14144eceab001a19f7e84"), "name" : "zzzz" } {
"_id" : ObjectId("57d14148eceab001a19f7e85"), "name" : "dddd" } {
"_id" : ObjectId("57d1414ceceab001a19f7e86"), "name" : "rrrrr" }
What am I missing here.
Khans...
the way you built your index is correct however building an ascending index on names wont return the results in ascending order.
if you need results to be ordered by name you have to use
{db.names.find().sort({names:1})}
what happens when you build an index is that when you search for data the Mongo process perform the search behind the scenes in an ordered fashion for faster outcomes.
Please note: if you just want to see output in sorted order. you dont even need an index.
You won't be able to see if an index has been successfully created (unless there is a considerable speed performance) by running a find() command.
Instead, use db.names.getIndexes() to see if the index has been created (it may take some time if you're running the index in the background for it to appear in the index list)
I am new to querying dbs and especially mongodb.If I run :
db.<customers>.find({"contact_name: Anny Hatte"})
I get:
{
"_id" : ObjectId("55f7076079cebe83d0b3cffd"),
"company_name" : "Gap",
"contact_name" : "Anny Hatte",
"email" : "ahatte#gmail.com"
}
I wish to get the value of the "_id" attribute from this query result. How do I achieve that?
Similarly, if I have another collection, named items, with the following data:
{
"_id" : ObjectId("55f7076079cebe83d0b3d009"),
"_customer" : ObjectId("55f7076079cebe83d0b3cfda"),
"school" : "St. Patrick's"
}
Here, the "_customer" field is the "_id" of the customer collection (the previous collection). I wish to get the "_id", the "_customer" and the "school" field values for the record where "_customer" of items-collection equals "_id" of customers-collection.
How do I go about this?
I wish to get the value of the "_id" attribute from this query result.
How do I achieve that?
The find() method returns a cursor to the results, which you can iterate and retrieve the documents in the result set. You can do this using forEach().
var cursor = db.customers.find({"contact_name: Anny Hatte"});
cursor.forEach(function(customer){
//access all the attributes of the document here
var id = customer._id;
})
You could make use of the aggregation pipeline's $lookup stage that has been introduced as part of 3.2, to look up and fetch the matching rows in some other related collection.
db.customers.aggregate([
{$match:{"contact_name":"Anny Hatte"}},
{$lookup:{
"from":"items",
"localField":"_id",
"foreignField":"_customer",
"as":"items"
}}
])
In case you are using a previous version of mongodb where the stage is not supported, then, you would need to fire an extra query to lookup the items collection, for each customer.
db.customers.find(
{"contact_name":"Anny Hatte"}).map(function(customer){
customer["items"] = [];
db.items.find({"_customer":customer._id}).forEach(function(item){
customer.items.push(item);
})
return customer;
})
How to get position (index) of selected document in mongo collection?
E.g.
this document: db.myCollection.find({"id":12345})
has index 3 in myCollection
myCollection:
id: 12340, name: 'G'
id: 12343, name: 'V'
id: 12345, name: 'A'
id: 12348, name: 'N'
If your requirement is to find the position of the document irrespective of any order, that is not
possible as MongoDb does not store the documents in specific order.
However,if you want to know the index based on some field, say _id , you can use this method.
If you are strictly following auto increments in your _id field. You can count all the documents
that have value less than that _id, say n , then n + 1 would be index of the document based on _id.
n = db.myCollection.find({"id": { "$lt" : 12345}}).count() ;
This would also be valid if documents are deleted from the collection.
As far as I know, there is no single command to do this, and this is impossible in general case (see Derick's answer). However, using count() for a query done on an ordered id value field seems to work. Warning: this assumes that there is a reliably ordered field, which is difficult to achieve in a concurrent writer case. In this example _id is used, however this will only work with a single writer case.:
MongoDB shell version: 2.0.1
connecting to: test
> use so_test
switched to db so_test
> db.example.insert({name: 'A'})
> db.example.insert({name: 'B'})
> db.example.insert({name: 'C'})
> db.example.insert({name: 'D'})
> db.example.insert({name: 'E'})
> db.example.insert({name: 'F'})
> db.example.find()
{ "_id" : ObjectId("4fc5f040fb359c680edf1a7b"), "name" : "A" }
{ "_id" : ObjectId("4fc5f046fb359c680edf1a7c"), "name" : "B" }
{ "_id" : ObjectId("4fc5f04afb359c680edf1a7d"), "name" : "C" }
{ "_id" : ObjectId("4fc5f04dfb359c680edf1a7e"), "name" : "D" }
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
{ "_id" : ObjectId("4fc5f053fb359c680edf1a80"), "name" : "F" }
> db.example.find({_id: ObjectId("4fc5f050fb359c680edf1a7f")})
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
> db.example.find({_id: {$lte: ObjectId("4fc5f050fb359c680edf1a7f")}}).count()
5
>
This should also be fairly fast if the queried field is indexed. The example is in mongo shell, but count() should be available in all driver libs as well.
This might be very slow but straightforward method. Here you can pass as usual query. Just I am looping all the documents and checking if condition to match the record. Here I am checking with _id field. You can use any other single field or multiple fields to check it.
var docIndex = 0;
db.url_list.find({},{"_id":1}).forEach(function(doc){
docIndex++;
if("5801ed58a8242ba30e8b46fa"==doc["_id"]){
print('document position is...' + docIndex);
return false;
}
});
There is no way that MongoDB can return this as it does not keep documents in order in the database, just like MySQL f.e. doesn't name row numbers.
The ObjectID trick from jhonkola will only work if only one client creates new elements, as the ObjectIDs are generated on the client side, with the first part being a timestamp. There is no guaranteed order if different clients talk to the same server. Still, I would not rely on this.
I also don't quite understand what you are trying to do though, so perhaps mention that in your question? I can then update the answer.
Restructure your collection to include the position of any entry i.e {'id': 12340, 'name': 'G', 'position': 1} then when searching the database collection(myCollection) using the desired position as a query
The queries I use that return the entire collection all use sort to get a reproducible order, find.sort.forEach works with the script above to get the correct index.