Is it possible to use Dot Notation when dealing with nested documents?
http://www.mongodb.org/display/DOCS/Dot+Notation+(Reaching+into+Objects)
I'm trying to query the results of a map/reduce and therefore need to
run a query like this:
find({'_id.page' : 'ThisPage', '_id.user' : 'AUser'})
Trying this in Node code returns no rows but the same query works as
expected in mongodb shell.
Dot notation isn't required for reaching inside of documents for queries, you can use document notation instead.
find({'_id.page' : 'ThisPage', '_id.user' : 'AUser'})
could instead be
find({_id: {page: 'ThisPage', user: 'AUser'}})
It is very possible, I've done it before.
Why do you have nested documents under your _id property? Not sure what your use case is but that seems a bit strange. _id is a special property that is always the unique id of the document. So it might be getting treated special by the driver (i.e. doesn't expect there to be sub documents). Maybe try putting your sub-documents under a different property name.
Related
I have a mongodb collection full of 65k+ documents, each one with a properties named site_histories. The value of it is an array that might be empty, or might not be. If it is not empty, it will have one or more objects similar to this:
"site_histories" : "[{\"site_id\":\"129373\",\"accepted\":\"1\",\"rejected\":\"0\",\"pending\":\"0\",\"user_id\":\"12743\"}]"
I need to make a query that will look for every instance in the collection of a document that does not have a given user_id.
I'm pretty new to Mongo, so I was trying to make a query that would find every instance that does have the given user_id, which I was then planning on adding a "$ne" to, but even that didn't work. This is the query I was using that didn't work:
db.test.find({site_histories: { $elemMatch: {user_id: '12743\' }}})
So can anyone tell me why this query didn't work? And can anyone help me format a query that will do what I need the final query to do?
If your site_histories really is an array, it should be as simple as doing:
db.test.find({"site_histories.user_id": "12743"})
That looks in all the elements of the array.
However, I'm a bit scared of all those backslashes. If site_histories is a string, that won't work. It would mean that the schema is poorly designed, you'd maybe try with $regex
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.
I have an use case in which I want to compare each record of two collections in mongodb and after comparing each record I need to find mismatch fields of all record.
Let us take an example, in collection1 I have one record as {id : 1, name : "bks"}
and in collection2 I have a record as {id : 1, name : "abc"}
When I compare above two records with same key, then field name is a mismatch field as name is different.
I am thinking to achieve this use case using mapreduce in mongodb. But I am facing some problems while accessing collection name in map function. When I tried to compare it in map function, I got error as : "errmsg" : "exception: ReferenceError: db is not defined near '
Can anyone give me some thoughts on how to compare records using mapreduce?
I might have helped you to read the documentation:
When upgrading to MongoDB 2.4, you will need to refactor your code if your map-reduce operations, group commands, or $where operator expressions include any global shell functions or properties that are no longer available, such as db.
So from your error fragment, you appear to be referencing db in order to access another collection. You cannot do that.
If indeed you are intending to "compare" items in one collection to those in another, then there is no other approach other than looping code:
db.collection.find().forEach(function(doc) {
var another = db.anothercollection.findOne({ "_id": doc._id });
// Code to compare
})
There is simply no concept of "joins" as such available to MongoDB, and operations such as mapReduce or aggregate or others strictly work with one collection only.
The exception is db.eval(), but as per all of strict warnings in the documentation, this is almost always a very bad idea.
Live with your comparison in looping code.
In MongoDB,
To query for records that contain certain fields you can do:
collection.find({'field_name1': {'$exists': true}})
And that will return any record that has the 'field_name1' field set...
But how do you query mongo to find records that contains ONLY 'field_name1' (and no other fields)? I'd like to be able to do this for, say, a list of fields.
The sad answer, as you'll often find with MongoDB and other NoSQL databases is probably that it would be best to structure your data in a way that allows you to query it as simply as possible.
That said, there are ways of doing this, but as far as I know, it requires you execute JavaScript server side. This will be slow, and cannot possibly take advantage of indexes and other logical features of MongoDB, so use it only if it's absolutely necessary, if performance is at all important.
So, the easiest way to do this, is probably to create a function that returns the number of fields in an object, which we can use with the $where query syntax. This allows you to run arbitrary JavaScript queries against your data, and can be combined with normal syntax queries.
Sadly, my JavaScript-fu is a little weak, so I don't know how (or if) you can get at the count of members of an object in JS in a one-liner, so to do this, I would store a function server side.
From the mongo shell, execute the following:
db.system.js.save(
{
"_id" : "countFields",
"value" : function(x) { i=0; for(p in x) { i++; } return i}
}
)
With that, you have a saved JavaScript function, server side, called countFields that returns the number of elements in an object. Now, you need to execute your find-operation with the $where query:
db.collection.find({
'field_name1': {'$exists': True},
'$where' : 'countFields(this)==2'
})
This would give you only the documents that meet both the $exists condition, and the $where clause. Note that I'm comparing with 2 in the example, since the countFields function counts _id as a field.
I know certain commends need the hashmap / dictionary to be ordered, but does the actual BSON document in MongoDB matter and would the index still work?
E.g.
db.people.ensureIndex({LName:1, FName:1});
Would it work on both:
{LName:"abc", FName:"def"},
{FName:"ghi", LName:"jkl"}
?
Thanks
The order of a document's properties does not affect indexing.
You can see this for yourself by running this query:
db.people.find({LName: "abc"}).explain()
and then this query:
db.people.find({LName: "jkl"}).explain()
you should see that MongoDB will use the index in both cases (the cursor property should be something like "BtreeCursor LName_1_FName_1").