In this case How do I keep documents in aggregation with $unwind aggregation can be ommited but what if you have a more complex case where you need aggregation framework? How can you keep these documents and still unwind those who have non-empty lists?
My case involves aggregation pipeline with multiple stages (match, unwind, project, match, group,..), i simply cannot fallback to plain "match" because those nested arrays need to be presented in a form of report (JasperReports). I first thought that this $unwind empty array will solve my problem but it won't because this array needs to be empty since I'm later grouping these values in JasperReport and showing them as they are.
I hope I was clear enough.
Any ideas?
Starting from MongoDb 3.2, $unwind operator supports preserveNullAndEmptyArrays:<boolean>.
So when preserveNullAndEmptyArrays:true, it will also include value which doesn't have any data or empty data.
for more info, visit - https://docs.mongodb.com/manual/reference/operator/aggregation/unwind/#document-operand-with-options
Related
Is there a way to filter array of subdocuments in the $match stage?
This question is an extension of the above.
Considering the answer of the question, what if in addition, if there are multiple subdocuments that match but you only want a specific element, such as the element with the maximum of a field, etc.
How would you modify the query
I'm new to mongodb and aggregation. Specifically have a question about https://docs.mongodb.com/manual/reference/operator/aggregation-pipeline/
I am trying to create a pipeline where I have a need to create temporary variables to carry values from one stage to the next. I believe the way you do this is to use the 'addFields' operator. Does the addFields operator add these temporary fields to the collection? If so, is there a way around it, as I don't want to store any derivative fields into the database. I have no use for the temporary inter-stage fields after the aggregation operation is complete.
As the $addfields documentation points out, The added fields only apply to the document in the context of the pipeline.
That means the original document is not modified
You can add $addfields at any point in the pipeline, deriving fields from the data in previous stages
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'm querying a mongo collection that has a field that is an array of arrays. I want to find a record with a projection of one deep value out of the array of arrays. Conceptually, this is a $slice of a $slice. Is there a way to do this in Mongo?
For example - I have a record:
{
name: "foo",
text: [["part 1:1", "part 1:2"],["part 2:1","part 2:2"]]
}
and want to select the record with projection "part 2:2".
db.collection.find({"name":"foo"},{text: {$slice: [1,1]}}
gives me the array with both "part 2:1" and "part 2:2". How do I get just "part 2:2"?
You need to use the aggregation pipeline to achieve a $slice chain, due to the limitations in the project statement being part of the find query.
The below query is invalid because the first $slice would return an array, instead of an index, and the execution of the outer scoped $slice fails.
db.collection.find({"name":"foo"},{text: {$slice:[{$slice: [1,1]}]}})
Moreover there is no way to work on an projected field in the same project statement, if possible, we could have modified the text further by applying a $slice to it.
The way to go would be:
Match the record with the name as foo.
Unwind the text array to get to the first level.
Unwind again to get to the level that we want.
Group the records together by name.
Project the last record in the group which is also the last element
of the last nested array.
The Code:
db.collection.aggregate([
{$match:{"name":"foo"}},
{$unwind:"$text"},
{$unwind:"$text"},
{$group:{"_id":"$name","text":{$last:"$text"}}},
{$project:{"name":"$_id","text":1}}
])
or if you would want to project an element appearing in a particular order, then you could use the $skip and $limit operations to achieve this.
var orderOfElement = 2;
db.collection.aggregate([
{$match:{"name":"foo"}},
{$unwind:"$text"},
{$unwind:"$text"},
{$skip:orderOfElement -1},
{$limit:1}
])
Which projects the second element in order in the nested arrays.
If you want the specific part 2:2 the below query will help.
db.user.find({"name":"foo"},{_id:0,name:0})[0].text[1][1];
part 2:2
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