I have users' collection whose schema is like:
{
_id: unique number,
name: 'asdf',
age: '12',
gender: 'm',
address: [
{area: 'sdf',
city: 'sdq',
state: 'wfw'},
{area: 'asdf',
city: 'sdfs',
state: 'vfdwd'}
]
}
I want to find out the users for whom all the values of state
in address
should be the value I pass. If even one of the state
value doesn't match with the value I pass the user shouldn't be returned. I tried simple find, aggregation framework with $unwind
, $match
but nothing seemed to get solution. Can you please help me out...
Thanks
PS please bear with multiple addresses for the sake of question. :)
To find out if all array entries match the state "wfw", do an aggregation like the following:
db.users.aggregate([
{ "$project" : {
"test" : {
"$allElementsTrue" : [{
"$map" : {
"input" : "$address",
"as" : "a",
"in" : { "$eq" : ["wfw", "$$a.state"] }
}
}]
}
} },
{ "$match" : { "test" : true } }
])
This aggregation takes each document, maps "state equals 'wfw'" over the address array to get a boolean array, and tests if the entire array is true, storing the result in `test, and then filtering the results based on test. You will need MongoDB 2.6 for support of some of the operators.
I don't know if I understand.
I replicated your document. When you want to retrieve an user by state
you can do in many ways
If you search with single value you can do
db.g.find({ "address.state": "wfw" })
and retrieve an user
You can use $all
db.g.find( { "address.state": { $all: ["wfw","vfdwd"] } } ) // retrieve User
db.g.find( { "address.state": { $all: ["wfw","vfdwd","foo"] } } ) // don't retrieve User
or you can use $and
db.g.find( { $and: [ { "address.state":"wfw" },{ "address.state":"vfdwd" }] } )
But I don't know if I understand your question
Update and the correct answer
db.g.find( { "address.state": { $nin: ["wfw"] } } )
Let me Know
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.