I have a nested arrays as below in one of my collections,
how to retrieve the value from specific sub array & i don't need any of the other fields in that document.
I have tried using following query but it didn't get me the expected output,
query:
db.getCollection('my_collection').find({"_id":"08d4608a-e6c4-1bd8-80e6-8d1ac448c34b"},{"_id":0,"customProperties.0.1":1})
You can try
db.getCollection.find({
"someArray.someNestedArray.name": "value"
})
Related
I want to avoid returning documents from Firestore that I do not need. I have a collection of documents with the following properties:
membership {array}
0 {map}
contactGroupId: "1da9a73f89f8ca4c"
contactGroupResourceName: "contactGroup/1da9a73f89f8ca4c"
1 {map}
contactGroupId: "1da9a73f89f8ca4c"
contactGroupResourceName: "contactGroup/1da9a73f89f8ca4c"
I want to create a compound query that only returns documents if the contactGroupId is equal to 1da9a73f89f8ca4c. Something like ref.where(membership.{key}.contactGroupId, "==", "1da9a73f89f8ca4c"). Is this possible?
It's not possible to filter documents using the contents of a map field inside an array. You will need to restructure your data if you want to perform this query. You have two options.
Put all of the contactGroupId values into their own array field, and use an array-contains query to find them.
Instead of using an array, use the contactGroupId value as the name of a map key that you can use in a query, like where(membership.1da9a73f89f8ca4c.present, '==', true). Your membership field would be structure like this:
membership (map)
- 1da9a73f89f8ca4c
- present (boolean)
- contactGroupResourceName = "contactGroup/1da9a73f89f8ca4c"
I am trying to use $setIntersection to compare a string data type (heroName) and 2 arrays (heroOptions and heroChoosen) to get the object _id of the same values of those three fields. I do not want to use the $expr operator but only the $setIntersection. The problem I am running into is that $setIntersection only compares arrays, and game is a string data type.
Document 1:
_id:ObjectId("111111111111111111112")
heroName:"Batman"
heroOptions:"Batman"
"Superman"
"Robin"
heroChoosen:"Batman"
"The Flash"
Document 1 contains similar value (Batman) in heroName, heroOptions, and heroChoosen so then the _id would be displayed of document 1.
I need to query documents with mongoDb that contain nested arrays. I see a lot of examples using the simple $in operator. The only problem is that I strictly need to check for proper subsets.
Consider the following document.
{data: [[1,2,3], [4,5,6]]}
The query needs to be able to get documents with all of [1,2,3] where 1,2,3 can be in any order, which rules out the following query, because it will only match in the correct order.
{data:{$elemMatch:{$all:[[1,2,3]]}}}
I've also tried nested $elemMatch operators with no success, because the $in operator will return the document even if only one element matches such as the following.
{data:{$elemMatch:{$elemMatch:{$in:[1,4]}}}}
Not sure what your actual query looks like, but this should do what you need:
db.documentDto.find({"some_field":{"$elemMatch":{"$in":[1,2,3]}} })
I haven't got a complete answer (and not much time as its late here) but I would consider
Using aggregation pipeline instead of a query if your not already
Use $unwind operator to deconstruct your nested arrays
Use $sort to sort the contents of the arrays - so you can now compare
Use $match to filter out the arrays which don't fit the array subset values as you can now check based on order.
Use $group to group the result back together based on the _id value
Ref:
http://docs.mongodb.org/manual/reference/operator/aggregation-pipeline/ will give you info on each of the above.
From a quick search I came up with a similar question/example that might be helpful: Mongodb sort inner array
I have two type of documents in a mongodb collection:
one where key sessions has a simple value:
{"sessions": NumberLong("10000000000001")}
one where key sessions has an array of values.
{"sessions": [NumberLong("10000000000001")]}
Is there any way to retrieve all documents from the second category, ie. only documents whose value is an arary and not a simple value?
You can use this kind of query for that:
db.collectionName.find( { $where : "Array.isArray(this.sessions)" } );
but you'd better convert all the records to one type to keep the things consistent.
This code can be simple like this:
db.c.find({sessions:{$gte:[]}});
Explanation:
Because you only want to retrieve documents whose sessions data type is array, and by the feature of $gte (if data types are different between tow operands, it returns false; Double, Integer32, Integer64 are considered as same data type.), giving an empty array as the opposite operand will help to retrieve all results by required.
Also , $gt, $lt, $lte for standard query (attention: different behaviors to operaors with same name in expression of aggregation pipeline) have the same feature. I proved this by practice on MongoDB V2.4.8, V2.6.4.
What I'm trying to do:
Filter a field of a collection that matches a given condition. Instead of returning every item in the field (which is an array of items), I only want to see matched items.
Similar to
select items from test where items.histPrices=[10,12]
It is also similar to what's found on the mongodb website here: http://www.mongodb.org/display/DOCS/Retrieving+a+Subset+of+Fields
Here's what I have been trying:
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"}]})
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"},
{"histPrices":[12,13],"name":"stuff"},{"histPrices":[11,14],"name":"stuff"}]})
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
It will return all the objects that have a match for items.histPrices:[10,12], including ALL of the items in items[]. But I don't want the ones that don't match the condition.
From the comments left on Mongodb two years ago, the solution to get only the items with that histPrices[10,12] is to do it with javascript code, namely, loop through the result set and filter out the other items.
I wonder if there's a way to do that with just the query.
Your find query is wrong
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
Your condition statement should be in the first part of the find statement.In your query {} means fetch all documents similar to this sql
select items from test (no where clause)
you have to change your mongodb find to
db.test.find({"items.histPrices":[10, 12]},{"name":1})
make it work
since your items is an array and if you wanted to return only the matching sub item, you have to use positional operator
db.test.find({"items.histPrices":[10, 12]},{"name":1,'items.$':1})
When working with arrays Embedded to the Document, the best approach is the one suggested by Chien-Wei Huang.
I would just add another aggregation, with the $group (in cases the document is very long, you may not want to retrieve all its content, only the array elements) Operator.
Now the command would look like:
db.test.aggregate({$match:{name:"record"}},
{$unwind:"$items"},
{$match {"items.histPrices":[10, 12]}},
{$group: {_id: "$_id",items: {$push: "$items"}}});)
If you are interested to return only one element from the array in each collection, then you should use projection instead
The same kind of issue solved here:
MongoDB Retrieve a subset of an array in a collection by specifying two fields which should match
db.test.aggregate({$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
But I don't know whether the performance would be OK. You have to verify it with your data.
The usage of $unwind
If you want add some filter condition like name="record", just add another $march at first, ex:
db.test.aggregate({$match:{name:"record"}}, {$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
https://jira.mongodb.org/browse/SERVER-828
Get particular element from mongoDB array
MongoDB query to retrieve one array value by a value in the array