Get position of selected document in collection [mongoDB] - mongodb

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.

Related

Issue with cosmos DB collection order

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.

How to get the particuler object fields using ReativeMongo without a case class [duplicate]

In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.

MongoDB get object id by finding on another column value

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;
})

Search full document in mongodb for a match

Is there a way to match a value with every array and sub document inside the document in mongodb collection and return the document
{
"_id" : "2000001956",
"trimline1" : "abc",
"trimline2" : "xyz",
"subtitle" : "www",
"image" : {
"large" : 0,
"small" : 0,
"tiled" : 0,
"cropped" : false
},
"Kytrr" : {
"count" : 0,
"assigned" : 0
}
}
for eg if in the above document I am searching for xyz or "ab" or "xy" or "z" or "0" this document should be returned.
I actually have to achieve this at the back end using C# driver but a mongo query would also help greatly.
Please advice.
Thanks
You could probably do this using '$where'
db.mycollection({$where:"JSON.stringify(this).indexOf('xyz')!=-1"})
I'm converting the whole record to a big string and then searching to see if your element is in the resulting string. Probably won't work if your xyz is in the fieldnames!
You can make it iterate through the fields to make a big string and then search it though.
This isn't the most elegant way and will involve a full tablescan. It will be faster if you look through the individual fields!
While Malcolm's answer above would work, when your collection gets large or you have high traffic, you'll see this fall over pretty quickly. This is because of 2 things. First, dropping down to javascript is a big deal and second, this will always be a full table scan because $where can't use an index.
MongoDB 2.6 introduced text indexing which is on by default (it was in beta in 2.4). With it, you can have a full text index on all the fields in the document. The documentation gives the following example where a text index is created for every field and names the index "TextIndex".
db.collection.ensureIndex(
{ "$**": "text" },
{ name: "TextIndex" }
)

using an Object (subdocument) with varying fields as _id

Our (edX) original Mongo persistence representation uses a bson dictionary (aka object or subdocument) as the _id value (see, mongo/base.py). This id is missing a field.
Can some documents' _id values have more subfields than others without totally screwing up indexing?
What's the best way to handle existing documents without the additional field? Remove and replace them? Try to query w/ new _id format and if fails fall over to query w/o the new field? Try to query with both new and old _id format in one query?
To be more specific, the current format is
{'_id': {
'tag': 'i4x', // yeah, it's always this fixed value
'org': your_school_x,
'course': a_catalog_number,
'category': the_xblock_type,
'name': uniquifier_within_course
}}
I need to add 'run': the_session_or_term_for_course_run or 'course_id': org/course/run.
Documents within a collection need not have values for _id that are of the same structure. Hence, it is perfectly acceptable to have the following documents within a collection:
> db.foo.find()
{ "_id" : { "a" : 1 } }
{ "_id" : { "a" : 1, "b" : 2 } }
{ "_id" : { "c" : 1, "b" : 2 } }
Note that because the index is on only _id, only queries that specify a value for _id will use the index:
db.foo.find({_id:1}) // will use the index on _id
db.foo.find({_id:{state:"Alaska"}) // will use the index on _id
db.foo.find({"_id.a":1}) // will NOT use the index on _id
Note also that only a complete match of the "value" of _id will return a document. So this returns no documents for the collection above:
db.foo.find({_id:{c:1}})
Hence, for your case, you are welcome to add fields to the object that is the value for the _id key. And it does not matter that all documents have a different structure. But if you are hoping to query the collection by_id and have it be efficient, you are going to need to add indexes for all relevant sub parts that might be used in isolation. That is not super efficient.
_id is no different than any other key in this regard.