Related
Kind of a complex one here and i'm pretty new to Mongo, so hopefully someone can help. I have a db of users. Each user has a state/province listed. I'm trying to create another collection of the total users in each state/province. Because users sign up pretty regularly, this will be an growing total i'm trying to generate and display on a map.
I'm able to query the database to find total number of users in a specific state, but i want to do this for all users and come out with a list of totals in all states/provinces and have a separate collection in the DB with all states/provinces listed and the field TOTAL to be dynamically populated with the count query of the other collection. But i'm not sure how to have a query be the result of a field in another collection.
used this to get users totals:
db.users.aggregate([
{"$group" : {_id:"$state", count:{$sum:1}}}
])
My main question is how to make the results of a query the value of a field in each corresponding record in another collection. Or if that's even possible.
Thanks for any help or guidance.
Looks like that On-Demand Materialized Views (just added on version 4.2 of MongoDB) should solve your problem!
You can create an On-Demand Materialized View using the $merge operator.
A possible definition of the Materialized View could be:
updateUsersLocationTotal = function() {
db.users.aggregate( [
{ $match: { <if you need to perform a match, like $state, otherwise remove it> } },
{ $group: { _id:"$state", users_quantity: { $sum: 1} } },
{ $merge: { into: "users_total", whenMatched: "replace" } }
] );
};
And then you perform updates just by calling updateUsersLocationTotal()
After that you can query the view just like a normal collection, using db.users_total.find() or db.users_total.aggregate().
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.
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.
I need help with filtering reactively minimongo collection with 3+ parameters. Firs, I have loaded minimongo from server's mongo collection with publish/subscribe. Now, I want to filter that collection and let user see only objects from collection that are same like filters. First of all I have search query filter that checks if input string is same as some filed in collection:
Job.find({ $or: [ { Name: regex }, { Description: regex }, ]});
This part is doing his job well. And now second filter: I have field in object that is true if that particular job is remote friendly and false if it is not, I wanted to if user enables that filter he only see just remote friendly job positions, and if he disables that, he can see all jobs available ( + search query of course):
if (remoteQuery === true ) {
return Job.find({ $or: [ { Name: regex }, { Description: regex }, ] , Remote: true});
}
This is working, but it is bad way for doing this. And now biggest problem comes with last filter: I have one more field that is storing "job" (collection object) type. Type can be 1, 2 , 3 or 4. So how could I say to minimongo e.g. "Show only jobs that have "Front-end" inside (search query), that are remote friendly, and have third filter 2 and 3 inside"
Job.find({ $or: [ { Name: "Front-end"}, { Description: "Front-end"}, ] , Remote: true, positionType: [2,3],});
Something like this? Thanks!
Sounds like you are looking for the MongoDB query $in operator:
The $in operator selects the documents where the value of a field equals any value in the specified array.
Therefore your 3rd query could look like:
Job.find({
positionType: {
$in: [2, 3] // Matches documents where `positionType` is either 2 or 3
},
// Other conditions…
});
This is my MongoDB query:
db.events.update({date:{$gte: ISODate("2014-09-01T00:00:00Z")}},{$set:{"artists.$.soundcloud_toggle":false}},{multi:true,upsert:false})
Apparently I cannot use "artists.$.soundcloud_toggle" to update all artist documents within the artists array:
"The $ operator can update the first array element that matches
multiple query criteria specified with the $elemMatch() operator.
http://docs.mongodb.org/manual/reference/operator/update/positional/"
I'm happy to run the query a number of times changing the index of the array in order to set the soundcloud_toggle property of every artist in every event that matches the query e.g
artists.0.soundcloud_toggle
artists.1.soundcloud_toggle
artists.2.soundcloud_toggle
artists.3.soundcloud_toggle
The problem is: when there is say, only one artist document in the artists array and I run the query with "artists.1.soundcloud_toggle" It will insert an artist document into the artist array with a single property:
{
"soundcloud_toggle" : true
},
(I have declared "upsert:false", which should be false by default anyways)
How do I stop the query from inserting a document and setting soundcloud_toggle:false when there is no existing document there? I only want it to update the property if an artist exists at the given artists array index.
If, like you said, you don't mind completing the operation with multiple queries, you can add an $exists condition to your filter.
E.g. in the 5th iteration, when updating index=4, add: "artists.4": {$exists: true}, like:
db.events.update(
{ date: {$gte: ISODate("2014-09-01T00:00:00Z")},
"artists.4": {$exists: true} },
{ $set:{ "artists.4.soundcloud_toggle" :false } },
{ multi: true, upsert: false }
)