I searched for similar questions but couldn't find any. Feel free to point me in their direction.
Say I have this data:
{ "_id" : ObjectId("5694c9eed4c65e923780f28e"), "name" : "foo1", "attr" : "foo" }
{ "_id" : ObjectId("5694ca3ad4c65e923780f290"), "name" : "foo2", "attr" : "foo" }
{ "_id" : ObjectId("5694ca47d4c65e923780f294"), "name" : "bar1", "attr" : "bar" }
{ "_id" : ObjectId("5694ca53d4c65e923780f296"), "name" : "bar2", "attr" : "bar" }
If I want to get the latest record for each attribute group, I can do this:
> db.content.aggregate({$group: {_id: '$attr', name: {$last: '$name'}}})
{ "_id" : "bar", "name" : "bar2" }
{ "_id" : "foo", "name" : "foo2" }
I would like to have my data grouped by attr and then sorted by _id so that only the latest record remains in each group, and that's how I can achieve this. BUT I need a way to avoid naming all the fields that I want in the result (in this example "name") because in my real use case they are not known ahead.
So, is there a way to achieve this, but without having to explicitly name each field using $last and just taking all fields instead? Of course, I would sort my data prior to grouping and I just need to somehow tell Mongo "take all values from the latest one".
See some possible options here:
Do multiple find().sort() queries for each of the attr values you
want to search.
Grab the original _id of the $last doc, then do a findOne() for each of those values (this is the more extensible option).
Use the $$ROOT system variable as shown here.
This wouldn't be the quickest operation, but I assume you're using this more for analytics, not in response to a user behavior.
Edited to add slouc's example posted in comments:
db.content.aggregate({$group: {_id: '$attr', lastItem: { $last: "$$ROOT" }}}).
Related
I have a simple order table in mongoDB
{
"_id" : NumberInt(2),
"bar" : "Maggie Choos Bar"
},
{
"_id" : NumberInt(3),
"bar" : "Corona Bar"
{
I want to find the BIGGEST "_id" number in the table
db.getCollection("order").find({}).sort({"$_id":-1}).limit(1);
But no matter if I sort 1 or -1 I keep getting the result with _id : 2
Any ideas?
The field name prefixed with a dollar sign, like $_id, is for references in an aggregation pipeline. For a sort document, use just the field name:
db.getCollection("order").find({}).sort({"_id":-1}).limit(1);
I did this in my mongodb:
db.teams.insert({name:"Alpha team",employees:[{name:"john"},{name:"david"}]});
db.teams.insert({name:"True team",employees:[{name:"oliver"},{name:"sam"}]});
db.teams.insert({name:"Blue team",employees:[{name:"jane"},{name:"raji"}]});
db.teams.find({"employees.name":/.*o.*/});
But what I got was:
{ "_id" : ObjectId("5ddf3ca83c182cc5354a15dd"), "name" : "Alpha team", "employees" : [ { "name" : "john" }, { "name" : "david" } ] }
{ "_id" : ObjectId("5ddf3ca93c182cc5354a15de"), "name" : "True team", "employees" : [ { "name" : "oliver" }, { "name" : "sam" } ] }
But what I really want is
[{"name":"john"},{"name":"oliver"}]
I'm having a hard time finding examples of this without using some kind of programmatic iterator/loop. Or examples I find return the parent document, which means I'd have to parse out the embedded array employees and do some kind of UNION statement?
Eg.
How to get embedded document in mongodb?
Retrieve only the queried element in an object array in MongoDB collection
Can someone point me in the right direction?
Please add projections to filter out the fields you don't need. Please refer the project link mongodb projections
Your find query should be constructed with the projection parameters like below:
db.teams.find({"employees.name":/.*o.*/}, {_id:0, "employees.name": 1});
This will return you:
[{"name":"john"},{"name":"oliver"}]
Can be solved with a simple aggregation pipeline.
db.teams.aggregate([
{$unwind : "$employees"},
{$match : {"employees.name":/.*o.*/}},
])
EDIT:
OP Wants to skip the parent fields. Modified query:
db.teams.aggregate([
{$unwind : "$employees"},
{$match : {"employees.name":/.*o.*/}},
{$project : {"name":"$employees.name",_id:0}}
])
Output:
{ "name" : "john" }
{ "name" : "oliver" }
This my code:
db.test.find() {
"_id" : ObjectId("4d3ed089fb60ab534684b7e9"),
"title" : "Sir",
"name" : {
"_id" : ObjectId("4d3ed089fb60ab534684b7ff"),
"first_name" : "Farid"
},
"addresses" : [
{
"city" : "Baku",
"country" : "Azerbaijan"
},{
"city" : "Susha",
"country" : "Azerbaijan"
},{
"city" : "Istanbul",
"country" : "Turkey"
}
]
}
I want get output only all city. Or I want get output only all country. How can i do it?
I'm not 100% about your code example, because if your 'find' by ID there's no need to search by anything else... but I wonder whether the following can help:
db.test.insert({name:'farid', addresses:[
{"city":"Baku", "country":"Azerbaijan"},
{"city":"Susha", "country":"Azerbaijan"},
{"city" : "Istanbul","country" : "Turkey"}
]});
db.test.insert({name:'elena', addresses:[
{"city" : "Ankara","country" : "Turkey"},
{"city":"Baku", "country":"Azerbaijan"}
]});
Then the following will show all countries:
db.test.aggregate(
{$unwind: "$addresses"},
{$group: {_id:"$country", countries:{$addToSet:"$addresses.country"}}}
);
result will be
{ "result" : [
{ "_id" : null,
"countries" : [ "Turkey", "Azerbaijan"]
}
],
"ok" : 1
}
Maybe there are other ways, but that's one I know.
With 'cities' you might want to take more care (because I know cities with the same name in different countries...).
Based on your question, there may be two underlying issues here:
First, it looks like you are trying to query a Collection called "test". Often times, "test" is the name of an actual database you are using. My concern, then, is that you are trying to query the database "test" to find any collections that have the key "city" or "country" on any of the internal documents. If this is the case, what you actually need to do is identify all of the collections in your database, and search them individually to see if any of these collections contain documents that include the keys you are looking for.
(For more information on how the db.collection.find() method works, check the MongoDB documentation here: http://docs.mongodb.org/manual/reference/method/db.collection.find/#db.collection.find)
Second, if this is actually what you are trying to do, all you need to for each collection is define a query that only returns the key of the document you are looking for. If you get more than 0 results from the query, you know documents have the "city" key. If they don't return results, you can ignore these collections. One caveat here is if data about "city" is in embedded documents within a collection. If this is the case, you may actually need to have some idea of which embedded documents may contain the key you are looking for.
Is there a way to use db.collection.find() to query for a specific value in a sub-document and find those documents that match. For example:
{
{ 'Joe' : {eyecolor : 'brown'},
{ 'Mary' : {eyecolor : 'blue'},
....
}
I want to return the names of all people whose eyecolor is blue.
You need to specify the full path to a value for search to work:
db.people.find({ "Joe.eyecolor" : "brown" })
You can't switch to an array of people instead of an associative array style you're using now, as there is no way to return only array elements that match conditions. You can use $elemMatch to return the first match, but that's not likely what you'd want. Or, you could still use arrays, but you'd need to filter the array further within your client code (not the database).
You might be able to use the Aggregation framework, but it wouldn't use indexes efficiently, as you'd need to $unwind the entire array, and then do filtering, brute force. And if the data contained is more complex, the fact that projections when using the AF require you to manually specify all fields, it becomes a bit cumbersome.
To most efficiently do the query you're showing, you'd need to not use subdocuments, and instead place the people as individual documents:
{
name: "Joe",
eyecolor: "brown"
}
Then, you could just do a simple search like:
db.people.find({eyecolor: "brown"})
Yes and no. You can query for all documents that have a matching person, but you can't query for all persons directly. In other words, subdocuments are not virtual collections, you'll always have the 'parent' document returned.
The example you posted comes with the additional complexity that you're using the name as a field key, which prevents you from using the dot notation.
In general, if you have a number of similar things, it's best to put them in a list, e.g.
{
"_id" : 132,
"ppl" : [ { "Name" : "John", "eyecolor" : "blue" },
{ "Name" : "Mary", "eyecolor" : "brown" },
...
]
}
Then, you can query using the aggregation framework:
db.collection.aggregate([
// only match documents that have a person w/ blue eyes (can use indexing)
{$match : { "ppl.eyecolor" : "blue" } },
// unwind the array of people
{$unwind : "$ppl" },
// match only those with blue eyes
{$match : { "ppl.eyecolor" : "blue" }},
// optional projection to make the result a list of people
{$project : { Name : "$ppl.Name", EyeColor: "$ppl.eyecolor" }} ]);
Which gives a result like
"result" : [
{
"_id" : 132,
"Name" : "John",
"EyeColor" : "blue"
},
{
"_id" : 12,
"Name" : "Jimmy",
"EyeColor" : "blue"
},
{
"_id" : 4312,
"Name" : "Jimmy",
"EyeColor" : "blue"
},
{
"_id" : 4312,
"Name" : "Marc",
"EyeColor" : "blue"
}
],
"ok" : 1
So, what I'm trying to do is query all documents that have a City of 'Paris' and a State of 'France'. I need to do some kind of join, but I haven't been able to figure out how to construct it.
I'm using the c# driver, but I'll gladly accept help using any method.
{
"_id" : ObjectId("519b407f3c22a73a7c29269f"),
"DocumentID" : "1",
"Meta" : [{
"Name" : "City",
"Value" : "Paris",
}, {
"Name" : "State",
"Value" : "France",
}
}]
}
{
"_id" : ObjectId("519b407f3c22a73a7c29269g"),
"DocumentID" : "2",
"Meta" : [{
"Name" : "City",
"Value" : "Paris",
}, {
"Name" : "State",
"Value" : "Texas",
}
}]
}
The $elemMatch operator is used to indicate that all the conditions within it must be matched by the same array element. So (to switch to shell syntax) to match all documents which have meta city Paris you would do
db.collection.find( {Meta:{$elemMatch:{Name:"City",Value:"Paris"}}} )
This assures you won't match something which has Name: "somethingelse", Value: "Paris" somewhere in its array with a different array element matching the Name:"City".
Now, default combination for combining query conditions is "and" so you can continue adding attributes:
db.collection.find( {Meta: {
$elemMatch:{Name:"City",Value:"Paris"},
$elemMatch:{Name:"State",Value:"France"}
}
}
)
Now if you want to add another condition you keep adding it but if you want a NOT then you do it like this:
db.collection.find( {Meta: {
$elemMatch:{Name:"City",Value:"Paris"},
$elemMatch:{Name:"State",Value:"France"},
$not: {$elemMatch:{Name:"Arrondissement",Value:"Louvre"}}
}
}
)
I might be answering my own question here, but I'm new to MongoDB, so while this appears to give me the results I'm after, it might not be the optimum approach.
var result = collection.Find(
Query.And(
Query.ElemMatch("Meta", Query.EQ("Name", "City")),
Query.ElemMatch("Meta", Query.EQ("Value", "Paris")),
Query.ElemMatch("Meta", Query.EQ("Name", "State")),
Query.ElemMatch("Meta", Query.EQ("Value", "France")))
);
Which leads to a follow up - how would I get all of the documents whose 'City' is 'Paris' and 'State' is 'France' but whose 'Arrondissement' is not 'Louvre'?