Is there any way to remove all the documents except one from a collection based on condition.
I am using MongoDB version 2.4.9
You can do this in below way,
db.inventory.remove( { type : "food" } )
Above query will remove documents with type equals to "food"
To remove document that not matches condition you can do,
db.inventory.remove( { type : { $ne: "food" } } )
or
db.inventory.remove( { type : { $nin: ["Apple", "Mango"] } } )
Check here for more info.
To remove all documents except one, we can use the query operator $nin (not in) over a specified array containing the values related to the documents that we want to keep.
db.collections.remove({"field_name":{$nin:["valueX"]}})
The advantage of $nin array is that we can use it to delete all documents except one or two or even many other documents.
To delete all documents except two:
db.collections.remove({"field_name":{$nin:["valueX", "valueY"]}})
To delete all documents except three:
db.collections.remove({"field_name":{$nin:["valueX", "valueY", "valueZ"]}})
Query
db.collection.remove({ "fieldName" : { $ne : "value"}})
As stated above by Taha EL BOUFFI, the following worked for me.
db.collection.remove({"fieldName" : { $nin: ["value"]}});
Related
How can I return a set of documents, each not containing a specific item in an inner array?
My data scheme is:
Posts:
{
"_id" : ObjectId("57f91ec96241783dac1e16fe"),
"votedBy" : [
{
"userId" : "101",
"vote": 1
},
{
"userId" : "202",
"vote": 2
}
],
"__v" : NumberInt(0)
}
I want to return a set of posts, non of which contain a given userId in any of the votedBy array items.
The official documentation implies that this is possible:
MongoDB documentation: Field with no specific array index
Though it returns an empty set (for the more simple case of finding a document with a specific array item).
It seems like I have to know the index for a correct set of results, like:
votedBy.0.userId.
This Question is the closest I found, with this solution (Applied on my scheme):
db.collection.find({"votedBy": { $not: {$elemMatch: {userId: 101 } } } })
It works fine if the only inner document in the array matches the one I wish not to return, but in the example case I specified above, the document returns, because it finds the userId=202 inner document.
Just to clarify: I want to return all the documents, that NONE of their votedBy array items have the given userId.
I also tried a simpler array, containing only the userId's as an array of Strings, but still, each of them receives an Id and the search process is just the same.
Another solution I tried is using a different collection for uservotes, and applying a lookup to perform a SQL-similar join, but it seems like there is an easier way.
I am using mongoose (node.js).
User $ne on the embedded userId:
db.collection.find({'votedBy.userId': {$ne: '101'}})
It will filter all the documents with at least one element of userId = "101"
I have two schemas, a Profile and a LevelOfNeed.
Profile
{
"_id" : ObjectId("56d35960a695dfa140137fca"),
. . .
"levelOfNeedServiced" : ObjectId("56d35828a695dfa140137fc7")
}
Level of Need
{
"_id" : ObjectId("56d35828a695dfa140137fc7"),
"sortOrder" : 2,
"description" : "Moderate Needs",
"additionalCost" : 3,
"__v" : 0
}
I currently have 4 documents for LevelOfNeed. What I need to do is select all of the Profile documents where the levelOfNeedServiced.sortOrder is >= a value.
Example:
db.getCollection('profiles').find({
'levelOfNeedServiced.sortOrder': { $gte: 2 }
})
Given my data, I would expect to see the example Profile, but this returns no results. What am I doing wrong?
Update 1
Previously, I was running MongoDB 3.0.9. I've since upgraded to 3.2.3, however I'm still getting the same results. According to the docs, I should be able to query on an embedded document field value.
Update 2
The aggregate function solution works as expected, but since I already had an array of LevelOfNeed objects, I was able to use that to get to the related documents I needed using the $in operator.
Unfortunately mongodb does not support joins until version 3.2. In version 3.2 it provides the $lookup aggregation operator to lookup referenced documents across collections.
You could use it as below:
db.Profile.aggregate([
{
$lookup:{
"from":"LevelOfNeed",
"localField":"levelOfNeedServiced",
"foreignField":"_id",
"as":"joined"
}
},
{
$match:{
"joined.sortOrder":{$gte:2}
}
},
{
$project:{"levelOfNeedServiced":1,...} //include things you want to project.
}
])
Your code:
db.getCollection('profiles').find({
'levelOfNeedServiced.sortOrder': { $gte: 2 }
})
does not work as intended because, the field levelOfNeedServiced is identified as a field containing an ObjectId and not the resolved LevelOfNeed document.
I am new to MongoDB. I have a collection called person. I'm trying to get all the records without an _id field with this query:
db.person.find({}{_id:0})
but the error is
syntax error: unexpected {
but if i write
db.person.find()
it works perfectly.
Consider following documents inserted in person collection as
db.person.insert({"name":"abc"})
db.person.insert({"name":"xyz"}
If you want to find exact matching then use query as
db.person.find({"name":"abc"})
this return only matched name documents
If you want all names without _id then use projeciton id query as
db.person.find({},{"_id":0})
which return
{ "name" : "abc" }
{ "name" : "xyz" }
According to Mongodb manual you have little wrong syntax, you forgot to give comma after {}
Try this :
db.person.find({}, { _id: 0 } )
I have some data structured like this in a mongo collection:
{ "list" : [ { "update_time" : NumberLong(1426690563), "provider" : NumberLong(4) } ] }
What I would like to do is query for all of the entries where the list includes a provider value of 4.
If it helps, all of the list arrays contain only one element.
What I am trying right now is this:
db.collection.find(
{
list: {
provider: 4
}
}
)
This does not work though, and always returns an empty set
Use the dot notation i.e. concatenate the name of the field that contains the array, with a dot (.) and the name of the field in the embedded document and use that as your query:
db.collection.find({"list.provider": 4})
You can also use the $elemMatch operator to specify multiple criteria on an array of embedded documents such that at least one embedded document satisfies all the specified criteria:
db.collection.find({
list: {
$elemMatch: {
provider: 4
}
}
})
Consider the following document:
{
"_id" : "ID_01",
"code" : ["001", "002", "003"],
"Others" : "544554"
}
I went through this MongoDB doc for elemmatch-query & elemmatch-projection, but not able to figure it out how to use the same for the above document.
Could anyone tell me how can I use $elemMatch for the field code?
You'll want to use the $in operator rather than $elemMatch in this case as $in can be used to search for a value (or values) inside a specific field. $in requires a list of values to be passed as an array. Additionally, and for your case, it will find either a single value, or by searching in an array of values. The entire matching document is returned.
For example, you might use it like this:
db.mycodes.find( { code: { $in: ["001"] } } )
Which could be simplified to just be:
db.mycodes.find({ code: "001" })
As MongoDB will look in an array for a single match like above ("001").
Or if you want to search for "001" or "002":
db.mycodes.find( { code: { $in: ["001", "002"] } } )
$in documentation
If you're simply looking to match all documents with an array containing a given value, you can just specify the value on the reference to that array, e.g.
db.mycodes.find( { code: '001' } )
Which thus would return you all documents that contained '001' in their code array