mongo: query array field size is greather than [duplicate] - mongodb

Question

I have a MongoDB collection with documents in the following format:
{
"_id" : ObjectId("4e8ae86d08101908e1000001"),
"name" : ["Name"],
"zipcode" : ["2223"]
}
{
"_id" : ObjectId("4e8ae86d08101908e1000002"),
"name" : ["Another ", "Name"],
"zipcode" : ["2224"]
}
I can currently get documents that match a specific array size:
db.accommodations.find({ name : { $size : 2 }})
This correctly returns the documents with 2 elements in the name array. However, I can't do a $gt command to return all documents where the name field has an array size of greater than 2:
db.accommodations.find({ name : { $size: { $gt : 1 } }})
How can I select all documents with a name array of a size greater than one (preferably without having to modify the current data structure)?

Answer 1

There's a more efficient way to do this in MongoDB 2.2+ now that you can use numeric array indexes (0 based) in query object keys.
// Find all docs that have at least two name array elements.
db.accommodations.find({'name.1': {$exists: true}})
You can support this query with an index that uses a partial filter expression (requires 3.2+):
// index for at least two name array elements
db.accommodations.createIndex(
{'name.1': 1},
{partialFilterExpression: {'name.1': {$exists: true}}}
);

Answer 2

Update:
For mongodb versions 2.2+ more efficient way to do this described by #JohnnyHK in another answer.
Using $where
db.accommodations.find( { $where: "this.name.length > 1" } );
But...
Javascript executes more slowly than the native operators listed on
this page, but is very flexible. See the server-side processing page
for more information.
Create extra field NamesArrayLength, update it with names array length and then use in queries:
db.accommodations.find({"NamesArrayLength": {$gt: 1} });
It will be better solution, and will work much faster (you can create index on it).

Answer 3

I believe this is the fastest query that answers your question, because it doesn't use an interpreted $where clause:
{$nor: [
{name: {$exists: false}},
{name: {$size: 0}},
{name: {$size: 1}}
]}
It means "all documents except those without a name (either non existant or empty array) or with just one name."
Test:
> db.test.save({})
> db.test.save({name: []})
> db.test.save({name: ['George']})
> db.test.save({name: ['George', 'Raymond']})
> db.test.save({name: ['George', 'Raymond', 'Richard']})
> db.test.save({name: ['George', 'Raymond', 'Richard', 'Martin']})
> db.test.find({$nor: [{name: {$exists: false}}, {name: {$size: 0}}, {name: {$size: 1}}]})
{ "_id" : ObjectId("511907e3fb13145a3d2e225b"), "name" : [ "George", "Raymond" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225c"), "name" : [ "George", "Raymond", "Richard" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225d"), "name" : [ "George", "Raymond", "Richard", "Martin" ] }
>

Answer 4

You can use aggregate, too:
db.accommodations.aggregate(
[
{$project: {_id:1, name:1, zipcode:1,
size_of_name: {$size: "$name"}
}
},
{$match: {"size_of_name": {$gt: 1}}}
])
// you add "size_of_name" to transit document and use it to filter the size of the name

Answer 5

You can use $expr ( 3.6 mongo version operator ) to use aggregation functions in regular query.
Compare query operators vs aggregation comparison operators.
db.accommodations.find({$expr:{$gt:[{$size:"$name"}, 1]}})

Answer 6

Try to do something like this:
db.getCollection('collectionName').find({'ArrayName.1': {$exists: true}})
1 is number, if you want to fetch record greater than 50 then do ArrayName.50
Thanks.

Answer 7

MongoDB 3.6 include $expr
https://docs.mongodb.com/manual/reference/operator/query/expr/
You can use $expr in order to evaluate an expression inside a $match, or find.
{ $match: {
$expr: {$gt: [{$size: "$yourArrayField"}, 0]}
}
}
or find
collection.find({$expr: {$gte: [{$size: "$yourArrayField"}, 0]}});

Answer 8

None of the above worked for me. This one did so I'm sharing it:
db.collection.find( {arrayName : {$exists:true}, $where:'this.arrayName.length>1'} )

Answer 9

db.accommodations.find({"name":{"$exists":true, "$ne":[], "$not":{"$size":1}}})

Answer 10

Although the above answers all work, What you originally tried to do was the correct way, however you just have the syntax backwards (switch "$size" and "$gt")..
Correct:
db.collection.find({items: {$gt: {$size: 1}}})

Answer 11

I found this solution, to find items with an array field greater than certain length
db.allusers.aggregate([
{$match:{username:{$exists:true}}},
{$project: { count: { $size:"$locations.lat" }}},
{$match:{count:{$gt:20}}}
])
The first $match aggregate uses an argument thats true for all the documents. If blank, i would get
"errmsg" : "exception: The argument to $size must be an Array, but was of type: EOO"

Answer 12

You can MongoDB aggregation to do the task:
db.collection.aggregate([
{
$addFields: {
arrayLength: {$size: '$array'}
},
},
{
$match: {
arrayLength: {$gt: 1}
},
},
])

Answer 13

This will work in Compass also. This is the fastest of all i have tried without indexing.
{$expr: {
$gt: [
{
$size: { "$ifNull": [ "$name", [] ] }
},
1
]
}}

Answer 14

you can use $expr to cover this
// $expr: Allows the use of aggregation expressions within the query language.
// syntax: {$expr: {<expression>}}
db.getCollection("person_service").find(
{
"$expr" : {
// services is Array, find services length gt 3
"$gt" : [
{
"$size" : "$services"
},
3.0
]
}
}
)