How to fire find query on sub-documents in MongoDB - mongodb

I am not able to get values in a sub-document like the first query below.
> db.posts.find({'repository': {'language':'Python'}}).count()
0
> db.posts.find({'actor': 'swiftlinux'}).count()
12
Can someone tell me how to get results when the query is based on a sub-document?

Should be
db.posts.count({'repository.language': 'Python'})
Sub-documents are queried with a dot. Also, you apply the count the results of the query, not the result of the find method.

Related

How can Mongo query specify that two conditions in an array are satisfied at the same time?

For example, the doc is
{'id':'1',
'tag':[{'tag_a':'A',
'score':'10'},
{'tag_b':'B',
'score':'0'}
]
}
I want get the doc that satisfied tag_b get 10 points in my collection. Obviously, I should not get this doc. However, when I use the query below
{$and:[{'tag.tag_b':'B','tag.score':'10'}]}
This doc will appear in the results.
How can I avoid this situation. Thanks!

MongoDB $regex with $in clause

I need a mongodb query something like
db.getCollection("xyz").find({"_id" : {$regex : {$in : [xxxx/*]}}})
My Use case is -- I have a list of Strings such as
[xyz/12/poi, abc/98/mnb, ytn/65/tdx, ...]
The ids that are there in the collection(test) are something like
xyz/12/poi/2019061304.
I will get the values like xyz/12/poi from the input list, the other part of the id being yyyymmddhh format.
So, I need to go to the collection and find all the documents matching the input list with the ID of the documents in the test collection.
I can retrieve the documents individually but that does not seem to be a feasible option as the size of the input list is more than 10000.
Can you guys suggest a more feasible solution. Thanks in advance.
I tried using $in with $regex. But it seems mongodb does not support that. I have also tried pattern matching but even that is not feasible for me. Can you please suggest an alternative to using $in with $regex in mongodb.
Expected result could be an aggragate query/a normal query so that we hit the database only once and get the desired output rather than hitting the db for 10000 odd times.

Get a list of records from a collection sorted by count and uniqueness of a field

So I have a bunch of documents in a MongoDB collection and it seems that the collection is growing a little faster than we thought.
Is there a way to get a list from a collection that will count the number of documents that have X as a value in a field.
For example(I'll just make data up)
there are 4 values possible for the field (reference).
/content/public
/content/private
/resource/something
/much/wow
Is there a way to get a list from mongo that says:
1231 Records have /content/public as the value for reference.
21312312 have /content/private
34 have /resource/something
34242 have /much/wow
Use the aggregation tools for this. You haven't listed a language in your question, so here's the mongodb command directly. This assumes your collection is named 'urls'.
db.urls.aggregate([{$group: {_id:'$reference', total:{$sum:1} } }]);

MongoDB $in not only one result in case of repeated elements

I need to get the users whose ids are contained in an array. For this i'm using the $in operator, however being this inside an aggregate operation, i'd like to get back a specific user all the time it's id is present in the array, not just one. For example:
The ids array is A=[a,b,c,b] and U(x) is user with id x
with users.find({_id:{$in:A}}) i get these users as result: U(a),U(b),U(c)
instead i'd like to get back the result: U(a),U(b),U(c),U(b)
so get the user back every time it's id appears.
I understand that $in is working as expected but does anyone have an idea on how can i achieve this?
Thanks
This isn't possible using a MongoDB query.
MongoDB's query engine iterates over the documents in a collection (or over an index if there's a useful one) and returns to you any documents that match your query, in the order it finds them. Whether b appears once, twice, or a hundred times in your query makes no difference: the document with _id of b matches the query and is returned once, when MongoDB finds it.
You can do a post-processing step in your programming language to repeat documents as many times as you want.

In Mongodb, how to retrieve the subset of an object that matches a condition?

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