In find operation fields can be excluded, but what if I want to do a find then a sort and just after then the projection. Do you know any trick, operation for it?
doc: fields {Object}, the fields to return in the query. Object of fields to include or exclude (not both), {‘a’:1}
You can run a usual find query with conditions, projections, and sort. I think you want to sort on a field that you don't want to project. But don't worry about that, you can sort on that field even after not projecting it.
If you explicitly select projection of sorting field as "0", then you won't be able to perform that find query.
//This query will work
db.collection.find(
{_id:'someId'},
{'someField':1})
.sort('someOtherField':1)
//This query won't work
db.collection.find(
{_id:'someId'},
{'someField':1,'someOtherField':0})
.sort('someOtherField':1)
However, if you still don't get required results, look into the MongoDB Aggregation Framework!
Here is the sample query for aggregation according to your requirement
db.collection.aggregate([
{$match: {_id:'someId'}},
{$sort: {someField:1}},
{$project: {_id:1,someOtherField:1}},
])
Related
I have an existing collection where I can do some queries on. For further data processing, it would be handy to create some subset collections via query.
I understood that I can use the aggregate function with $match and $expr to e.g. $group some values and at the end use $out to get a new collection with the results.
The thing I am hanging on is not to $group anything, but just put the objects that $match finds into a new collection only. So not the complete objects with all their values. Just the one I am matching. Like when you db[collection].find({$match: {...}}, {"key1": 1, "key2": 0})
Where I get the new matching objects just containing key1: value1 but not key2: value2, which is also in the original collection.
How do I achieve that using aggregate without grouping anything? I read through the documentation and couldn't find any other stage operation that looks good.
As i mentioned in the comments $project is the right operator to use to achieve this.
I need to query documents with mongoDb that contain nested arrays. I see a lot of examples using the simple $in operator. The only problem is that I strictly need to check for proper subsets.
Consider the following document.
{data: [[1,2,3], [4,5,6]]}
The query needs to be able to get documents with all of [1,2,3] where 1,2,3 can be in any order, which rules out the following query, because it will only match in the correct order.
{data:{$elemMatch:{$all:[[1,2,3]]}}}
I've also tried nested $elemMatch operators with no success, because the $in operator will return the document even if only one element matches such as the following.
{data:{$elemMatch:{$elemMatch:{$in:[1,4]}}}}
Not sure what your actual query looks like, but this should do what you need:
db.documentDto.find({"some_field":{"$elemMatch":{"$in":[1,2,3]}} })
I haven't got a complete answer (and not much time as its late here) but I would consider
Using aggregation pipeline instead of a query if your not already
Use $unwind operator to deconstruct your nested arrays
Use $sort to sort the contents of the arrays - so you can now compare
Use $match to filter out the arrays which don't fit the array subset values as you can now check based on order.
Use $group to group the result back together based on the _id value
Ref:
http://docs.mongodb.org/manual/reference/operator/aggregation-pipeline/ will give you info on each of the above.
From a quick search I came up with a similar question/example that might be helpful: Mongodb sort inner array
I'm using MongoDB 2.6.8. According to the $maxTimeMS reference, these two queries should behave identically:
> db.collection.find({}).maxTimeMS(100)
> db.collection.find({}, {$maxTimeMS: 100})
The first query does exactly what I want, but the second query restricts only returns the object IDs of the documents. I tried increasing $maxTimeMS to 100000000 and there was no change in behavior.
Why am I getting different results for these two commands?
You found a bug in the documentation.
The reason that db.collection.find({}, {$maxTimeMS: 100}) returns only the _id of each object is because mongoDB is interpreting the {$maxTimeMS: 100} portion of the query as a projection.
So it thinks you want to see all the documents and you want to see the fields _id and the field $maxTimeMS. Of course, none of your documents have a $maxTimeMS field, so they only show the _id.
The proper way to perform the query you want without the shortcut is:
db.collection.find({ $query: {}, $maxTimeMS: 100 })
I'm trying to get data from one or more subdocuments but I don't know the name of the field that will hold the subdocument. Here are some examples of what the documents look like.
https://github.com/vz-risk/VCDB/blob/master/data/json/0C5DE044-B9B4-408D-9E65-D367EED12AB2.json
https://github.com/vz-risk/VCDB/blob/master/data/json/064F5887-C2DA-4139-B3AA-D55906F8C30A.json
I would like to get the action varieties for these incidents, so in the case of the first one I would like to get action.malware.variety and action.social.variety. In the second example it would be action.hacking.variety and action.malware.variety. So the problem is that I don't know what field is going to hold the subdocument. It could be one of hacking, malware, social, error, misuse, physical, and environmental.
So I would like to $unwind that field and do some stuff with the key name. Is this something that can be done with aggregation or do I need to switch over to mapReduce?
You seem to be talking about a case where you are not sure if all of the hacking, social or malaware parts are there right. I think you want $project first using the $ifNull operator as in:
db.stuff.aggregate([
{$project:
{
hacking: {$ifNull: ["$action.hacking.variety",[null]]},
social: {$ifNull: ["$action.social.variety",[null]]},
malware: {$ifNull: ["$action.malware.variety",[null]]}
}
},
{$unwind: "$hacking"},
{$unwind: "$social"},
{$unwind: "$malware"}
])
That should give you documents with something in each of those values.
Sort of pretty much the same with any of your possible list of values.
What is the difference between the $match operator used inside the aggregate function and the regular find in Mongodb?
Why doesn't the find function allow renaming the field names like the aggregate function?
e.g. In aggregate we can pass the following string:
{ "$project" : { "OrderNumber" : "$PurchaseOrder.OrderNumber" , "ShipDate" : "$PurchaseOrder.ShipDate"}}
Whereas, find does not allow this.
Why does not the aggregate output return as a DBCursor or a List? and also why can't we get a count of the documents that are returned?
Thank you.
Why does not the aggregate output return as a DBCursor or a List?
The aggregation framework was created to solve easy problems that otherwise would require map-reduce.
This framework is commonly used to compute data that requires the full db as input and few document as output.
What is the difference between the $match operator used inside the aggregate function and the regular find in Mongodb?
One of differences, like you stated, is the return type. Find operations output return as a DBCursor.
Other differences:
Aggregation result must be under 16MB. If you are using shards, the full data must be collected in a single point after the first $group or $sort.
$match only purpose is to improve aggregation's power, but it has some other uses, like improve the aggregation performance.
and also why can't we get a count of the documents that are returned?
You can. Just count the number of elements in the resulting array or add the following command to the end of the pipe:
{$group: {_id: null, count: {$sum: 1}}}
Why doesn't the find function allow renaming the field names like the aggregate function?
MongoDB is young and features are still coming. Maybe in a future version we'll be able to do that. Renaming fields is more critical in aggregation than in find.
EDIT (2014/02/26):
MongoDB 2.6 aggregation operations will return a cursor.
EDIT (2014/04/09):
MongoDB 2.6 was released with the predicted aggregation changes.
I investigated a few things about the aggregation and find call:
I did this with a descending sort in a table of 160k documents and limited my output to a few documents.
The Aggregation command is slower than the find command.
If you access to the data like ToList() the aggregation command is faster than the find.
if you watch at the total times (point 1 + 2) the commands seem to be equal
Maybe the aggregation automatically calls the ToList() and does not have to call it again. If you dont call ToList() afterwards the find() call will be much faster.
7 [ms] vs 50 [ms] (5 documents)