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
}
}
})
Related
I'm new to mongodb.
I have this object in my collection:
{
"_id" : ObjectId("5b549be38d9f1c00160117d3"),
"name" : "Name of object",
"a" : {
"b" : {
"c" : 100
}
}
}
What interests me is the 100 value, I want to fetch it from the object.
When I query the collection like this:
db.getCollection('myCollection').find({}, {'name':1, 'a.b.c':1})
I only get the same object with the inner objects.
Is there a way to query it so that I will get a result like this:
{"Name": "Name of object", "c":100}
By using Mongo aggregate query you can get the result. In $project stage of Mongo aggregate query you can add the conditions as per requirement.
Please try this query, might you will get the result:
db.myCollection.aggregate({
$project: {
"name": "$name",
"c": "$a.b.c",
_id: 0
}
})
As suggested by #Mayuri, You can achieve this by using .aggregate() but if you wanted to look at difference between $project in aggregation (Vs) projection in .find(), check out :
So one first thing with $project is it is much more powerful & can accept a lot more features that are helpful to transform the fields. But projection in .find() is pretty straight forward that can accept only few things : projection i.e; value to a field in projection can be one of these :
1 or true to include the field in the return documents.
0 or false to exclude the field.
Projection Operators : $, $elemMatch, $slice, $meta
Actual Issue :
In your query when you're doing 'a.b.c':1 in projection it means you're returning c field in the output, As field c is nested in b & a the output structure doesn't change & you'll be getting the c value under the same structure, but if you use aggregation { $project: { "c": "$a.b.c" } } it means you're assigning value of "$a.b.c" to a field named c. How ? : So when $ is used against a field in aggregation it will access field's value, which helps for assignment.
I have multiple documents(3 documents in this example) in one collection that looks like this:
{
_id:123,
bizs:[{_id:'',name:'a'},{_id:'',name:'b'}]
},
{
_id:456,
bizs:[{_id:'',name:'e'},{_id:'',name:'f'}]
}
{
_id:789,
bizs:[{_id:'',name:'x'},{_id:'',name:'y'}]
}
Now, I want to update the bizs subdocument by matching with my array of ids.
That is to say, my array filter for update query is [123,789], which will match against the _id fields of each document.
I have tried using findByIdAndUpdate() but that doesn't allow an array for the update query
How can I update the 2 matching documents (like my example above) without having to put findByIdAndUpdate inside a forloop to match the array element with the _id?
You can not use findByIdAndUpdate when updating multiple documents, findByIdAndUpdate is from mongoose which is a wrapper to native MongoDB's findOneAndUpdate. When you pass a single string as a filter to findByIdAndUpdate like : Collection.findByIdAndUpdate({'5e179dac627ef7823643cd97'}, {}) - then mongoose will internally convert string to ObjectId() & form it as a filter like :_id : ObjectId('5e179dac627ef7823643cd97') to execute findOneAndUpdate. So it means you can only update one document at a time, So if you've multiple documents to be updated use update with option {multi : true} or updateMany.
Assume if you wanted to push a new object to bizs, this is how query looks like :
collection.updateMany({ _id: { $in: [123, 456] } }, {
$push: {
bizs: {
"_id": "",
"name": "new"
}
}
})
Note : Update operations doesn't return the documents in response rather they will return write result which has information about n docs matched & n docs modified.
{
"_id":"12345",
"model":{
"T":0,
"serviceTask":[
{
"_id":"6789",
"obj":{
"params" :{
"action": "getEmployeeData",
"employeeId":"123"
},
"url":"www.test.com",
"var":"test",
},
"opp":100
}
]
}
}
I have similar structured documents in my collection. How do I query the documents that match action value to "getEmployeeData". I tried dot notation with $elemMatchbut couldn't get the results.
Dot notation works fine here, it will return the document if serviceTask contains at least one action set to getEmployeeData
db.collection.find({
"model.serviceTask.obj.params.action": "getEmployeeData"
})
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"]}});
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