http://docs.mongodb.org/manual/reference/operator/aggregation/match/
The documentation says absolutely nothing about what type of object is returned.
I know however from testing that it returns an array if I get multiple documents back.
Must I use $limit if I only want to get back one document even though $match only matches one document (it seems to return an array in any case)?
Related
Such request as db.collection.findOne({$or: [{email: 'email#example.com'},{'linkedIn.id': 'profile.id'}]}); may return an array with two records.
Is it possible to specify to return only the first occurrence so that I always have a model as a response, not an array?
E.g. if there is a record with a specified email, return it and do not return another record, if any, matching profile.id?
Another question is if the order of the params 'email' and 'linkedIn.id' matters.
All this hazel is about LinkedIn strategy, which never returns an email (at least for me) but I have to cater for case when it may return an email. So I construct my query depending on email presence and if it is present, the query is with $or operator. But I would like to avoid checking for whether the response is an object or an array and then perform additional operation on array values to figure out which of the values to use.
According to documentation of mongo DB
findOne()
always returns a single document irrespective of matches it found.
And regarding order of retrieval it will always return the first match except capped collection which maintains order of insertion of documents into collection.
For more detailed description about findOne please refer the documentation as mentioned in following URL
https://docs.mongodb.org/manual/reference/method/db.collection.findOne/
According to the MongoDB docs for db.collection.findOne():
Returns one document that satisfies the specified query criteria. If multiple documents satisfy the query, this method returns the first document according to the natural order which reflects the order of documents on the disk. In capped collections, natural order is the same as insertion order. If no document satisfies the query, the method returns null.
You can't recieve multiple records from db.collection.findOne(). Are you sure you're not using db.collection.find()?
I would like to only return a subdocument if its expire field is in the future. My problem is that I would like to get the root document anyway.
I am currently using:
sub.expire: {"$gte": now}
it only returns the root document if sub.expire is true.
how would I return the root document regardless if sub gets returned or not?
Thanks
This isn't quite how MongoDB is designed. You're phrasing it as, you're trying to query a bunch of subdocuments and return them conditionally while returning the parent documents unconditionally; but as far as MongoDB is concerned there isn't a collection of subdocuments, there's a collection of documents, which each have fields, some of which may be subdocuments. When you say you're "using" this condition on a subdocument -- presumably for find() -- what you're really asking the collection for is documents which satisfy this condition in a subdocument. So documents that don't satisfy that condition aren't returned.
From MongoDB's point of view, what you actually want is closer to, return every document in the collection, but with a certain field (a subdocument, as it happens) stripped out if it's expired, which falls under "projection". There are two ways projection happens in MongoDB, and I'm afraid neither is exactly what you want:
find() lets you project the fields to return, but that's limited to: 1 (do), 0 (don't), or a few special operators if the field is an array, which a subdocument is not.
Aggregation gives you a $project operator to transform the results, but doesn't let you disappear this field (subdocument) entirely: if it has an expiration date in the past, you can replace it with null or (probably better) an empty document, but that's as close as you can get.
This is my meteor code to search whether user exist in the array or not
var u_exist=Polls_Coll.findOne( {option1:{$elemMatch:{ids:"xxx"}}} );
My question is, How to know whether the statement returning something or not(user exist or not)
$elemMatch will return only where one of the conditions supplied actually finds a match in the array. So if you don't get a document back then there was no match.
Also findOne is a single document. Modifiers such as .count() will not work on that. If you have more documents to be expected use find intstead. Also findOne not not make much sense without applying a unique identifier such as _id in the query. Without that you are almost certainly not getting what you want.
While useful for your purpose, findOne is not a good match with the $elemMatch operator. The reasoning is you can possibly get multiple results of the same document having the same set of array elements that matched the condition that you gave.
Buyer beware.
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.
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