I'm trying to find records in MongoDB that are created outside a specific period. The query to search for records inside a specific period is pretty straightforward:
db.test.find({"Published":{'$gt':"2011-08-02", '$lt':"2011-08-06"}})
So naturally, I tried this for "outside" a specific range:
db.test.find({'$not':{"Published":{'$gt':"2011-08-02", '$lt':"2011-08-06"}}})
But this returns an empty result, while there are definately records published then.
What query should I use instead? Can anyone help me? I'm using raw mongo queries.
Thanks in advance
--- UPDATE ---
I found that the following query works, but it doesn't look like the perfect solution:
db.test.find(
{'$or': [
{"Published":{'$lt':"2011-02-02"}},
{"Published":{'$gt':"2011-08-06"}}
]}
)
Is there a cleaner way to do it?
You are putting the $not in the wrong place. Try this:
db.test.find({"Published":{ $not:{$gt:"2011-08-02", $lt:"2011-08-06"} } })
For details, see the MongoDB docs about the $notoperator.
Edit as because of the comment this solution would not work:
> db.dates.find()
{ "_id" : ObjectId("5492d46ef6226b581c80c0a2"), "a" : 1, "date" : "2011-08-04" }
{ "_id" : ObjectId("5492d4e2f6226b581c80c0a3"), "a" : 2, "date" : "2011-08-07" }
> db.dates.find({date:{$not:{$gt:"2011-08-02",$lt:"2011-08-06"}}})
{ "_id" : ObjectId("5492d4e2f6226b581c80c0a3"), "a" : 2, "date" : "2011-08-07" }
Gurus - I'm stuck in a situation that I can't figure out how I can query from the following collection "spouse", which has embedded document "surname" and check for equality with "surname" of this document:
{
"_id" : ObjectId("50bd2bb4fcfc6066b7ef090d"),
"name" : "Gwendolyn",
"surname" : "Davis",
"birthyear" : 1978,
"spouse" : {
"name" : "Dennis",
"surname" : "Evans",
"birthyear" : 1969
},
I need to query:
Output data for all spouses with the same surnames (if the surname of
one of the spouses is not specified, assume that it coincides with the
name of another)
I tried something like this:
db.task.find( {"surname" : { "spouse.surname" : 1 }} )
but it failed)
PLEASE PLEASE Guide me how I can achieve this any example/sample? based on this will be really helpful :-)
Thanks a lot!
You have three options.
Use $where modifier:
db.task.find({$where: 'this.spouse.surname === this.surname'})
Update all your documents and add special flag. After that you will be able to query documents by this flag. It's faster then $where, but requires altering your data.
Use MapReduce. It's quite complicated, but it allows you to do nearly anything.
I want to know the most recent record in a collection. How to do that?
Note: I know the following command line queries works:
1. db.test.find().sort({"idate":-1}).limit(1).forEach(printjson);
2. db.test.find().skip(db.test.count()-1).forEach(printjson)
where idate has the timestamp added.
The problem is longer the collection is the time to get back the data and my 'test' collection is really really huge. I need a query with constant time response.
If there is any better mongodb command line query, do let me know.
This is a rehash of the previous answer but it's more likely to work on different mongodb versions.
db.collection.find().limit(1).sort({$natural:-1})
This will give you one last document for a collection
db.collectionName.findOne({}, {sort:{$natural:-1}})
$natural:-1 means order opposite of the one that records are inserted in.
Edit: For all the downvoters, above is a Mongoose syntax,
mongo CLI syntax is: db.collectionName.find({}).sort({$natural:-1}).limit(1)
Yet another way of getting the last item from a MongoDB Collection (don't mind about the examples):
> db.collection.find().sort({'_id':-1}).limit(1)
Normal Projection
> db.Sports.find()
{ "_id" : ObjectId("5bfb5f82dea65504b456ab12"), "Type" : "NFL", "Head" : "Patriots Won SuperBowl 2017", "Body" : "Again, the Pats won the Super Bowl." }
{ "_id" : ObjectId("5bfb6011dea65504b456ab13"), "Type" : "World Cup 2018", "Head" : "Brazil Qualified for Round of 16", "Body" : "The Brazilians are happy today, due to the qualification of the Brazilian Team for the Round of 16 for the World Cup 2018." }
{ "_id" : ObjectId("5bfb60b1dea65504b456ab14"), "Type" : "F1", "Head" : "Ferrari Lost Championship", "Body" : "By two positions, Ferrari loses the F1 Championship, leaving the Italians in tears." }
Sorted Projection ( _id: reverse order )
> db.Sports.find().sort({'_id':-1})
{ "_id" : ObjectId("5bfb60b1dea65504b456ab14"), "Type" : "F1", "Head" : "Ferrari Lost Championship", "Body" : "By two positions, Ferrari loses the F1 Championship, leaving the Italians in tears." }
{ "_id" : ObjectId("5bfb6011dea65504b456ab13"), "Type" : "World Cup 2018", "Head" : "Brazil Qualified for Round of 16", "Body" : "The Brazilians are happy today, due to the qualification of the Brazilian Team for the Round of 16 for the World Cup 2018." }
{ "_id" : ObjectId("5bfb5f82dea65504b456ab12"), "Type" : "NFL", "Head" : "Patriots Won SuperBowl 2018", "Body" : "Again, the Pats won the Super Bowl" }
sort({'_id':-1}), defines a projection in descending order of all documents, based on their _ids.
Sorted Projection ( _id: reverse order ): getting the latest (last) document from a collection.
> db.Sports.find().sort({'_id':-1}).limit(1)
{ "_id" : ObjectId("5bfb60b1dea65504b456ab14"), "Type" : "F1", "Head" : "Ferrari Lost Championship", "Body" : "By two positions, Ferrari loses the F1 Championship, leaving the Italians in tears." }
I need a query with constant time response
By default, the indexes in MongoDB are B-Trees. Searching a B-Tree is a O(logN) operation, so even find({_id:...}) will not provide constant time, O(1) responses.
That stated, you can also sort by the _id if you are using ObjectId for you IDs. See here for details. Of course, even that is only good to the last second.
You may to resort to "writing twice". Write once to the main collection and write again to a "last updated" collection. Without transactions this will not be perfect, but with only one item in the "last updated" collection it will always be fast.
php7.1 mongoDB:
$data = $collection->findOne([],['sort' => ['_id' => -1],'projection' => ['_id' => 1]]);
My Solution :
db.collection("name of collection").find({}, {limit: 1}).sort({$natural: -1})
If you are using auto-generated Mongo Object Ids in your document, it contains timestamp in it as first 4 bytes using which latest doc inserted into the collection could be found out. I understand this is an old question, but if someone is still ending up here looking for one more alternative.
db.collectionName.aggregate(
[{$group: {_id: null, latestDocId: { $max: "$_id"}}}, {$project: {_id: 0, latestDocId: 1}}])
Above query would give the _id for the latest doc inserted into the collection
This is how to get the last record from all MongoDB documents from the "foo" collection.(change foo,x,y.. etc.)
db.foo.aggregate([{$sort:{ x : 1, date : 1 } },{$group: { _id: "$x" ,y: {$last:"$y"},yz: {$last:"$yz"},date: { $last : "$date" }}} ],{ allowDiskUse:true })
you can add or remove from the group
help articles: https://docs.mongodb.com/manual/reference/operator/aggregation/group/#pipe._S_group
https://docs.mongodb.com/manual/reference/operator/aggregation/last/
Mongo CLI syntax:
db.collectionName.find({}).sort({$natural:-1}).limit(1)
Let Mongo create the ID, it is an auto-incremented hash
mymongo:
self._collection.find().sort("_id",-1).limit(1)
I need help incrementing value of all keys in participants without having to know name of the keys inside of it.
> db.conversations.findOne()
{
"_id" : ObjectId("4faf74b238ba278704000000"),
"participants" : {
"4f81eab338ba27c011000001" : NumberLong(2),
"4f78497938ba27bf11000002" : NumberLong(2)
}
}
I've tried with something like
$mongodb->conversations->update(array('_id' => new \MongoId($objectId)), array('$inc' => array('participants' => 1)));
to no avail...
You need to redesign your schema. It is never a good idea to have "random key names". Even though MongoDB is schemaless, it still means you need to have defined key names. You should change your schema to:
{
"_id" : ObjectId("4faf74b238ba278704000000"),
"participants" : [
{ _id: "4f81eab338ba27c011000001", count: NumberLong(2) },
{ _id: "4f78497938ba27bf11000002", count: NumberLong(2) }
]
}
Sadly, even with that, you can't update all embedded counts in one command. There is currently an open feature request for that: https://jira.mongodb.org/browse/SERVER-1243
In order to still update everything, you should:
query the document
update all the counts on the client side
store the document again
In order to prevent race conditions with that, have a look at "Compare and Swap" and following paragraphs.
It is not possible to update all nested elements in one single move in current version of MongoDB. So I can advice to use "foreach {}".
Read realted topic: How to Update Multiple Array Elements in mongodb
I hope this feature will be implemented in next version.
I have a basic structure like this:
> db.users.findOne()
{
"_id" : ObjectId("4f384903cd087c6f720066d7"),
"current_sign_in_at" : ISODate("2012-02-12T23:19:31Z"),
"current_sign_in_ip" : "127.0.0.1",
"email" : "something#gmail.com",
"encrypted_password" : "$2a$10$fu9B3M/.Gmi8qe7pXtVCPu94mBVC.gn5DzmQXH.g5snHT4AJSZYCu",
"last_sign_in_at" : ISODate("2012-02-12T23:19:31Z"),
"last_sign_in_ip" : "127.0.0.1",
"name" : "Trip Jameson",
"sign_in_count" : 100,
"usertimes" : [
...thousands and thousands of records like this one....
{
"enddate" : 348268392.115282,
"idle" : 0,
"startdate" : 348268382.116728,
"title" : "My Awesome Title"
},
]
}
So I want to find only usertimes for a single user where the title was "My Awesome Title", and then I want to see what the value for "idle" was in that record(s)
So far all I can figure out is that I can find the entire user record with a search like:
> db.users.find({'usertimes.title':"My Awesome Title"})
This just returns the entire User record though, which is useless for my purposes. Am I misunderstanding something?
Return only partial embedded documents is currently not supported by MongoDB
The matching User record will always be returned (at least with the current MongoDB version).
see this question for similar reference
Filtering embedded documents in MongoDB
This is the correspondent Jira on MongoDB space
http://jira.mongodb.org/browse/SERVER-142
Use:
db.users.find({'usertimes.title': "My Awesome Title"}, {'idle': 1});
May I suggest you take a more detailed look at http://www.mongodb.org/display/DOCS/Querying, it'll explain things for you.