I have a mongodb query where i want to get documents if a field has particular value.
db.collection.find({key:{$in:['value1','value2']}}) if i run above command i get documents containing either 'value1' or 'value2'. but lets just say there are no values. and i search db.collection.find({key:{$in:[]}}), nothing is displayed. and db.collection.find({key:{$in:[*]}}) gives unexpected token* which wild card do i use in $in to show all results.?
I think this is logically consistent behavior for $in. The query
db.collection.find({ "key" : { "$in" : [] } })
could be translated as "find all the documents where the value of key is one of the values contained in the array []". Since there are no values in the array [], there are no matching documents. If you want to find all of the extant values for key, use .distinct to return them as an array:
db.collection.distinct("key")
.distinct will use an index if possible.
If you want a query to match all documents, omit the query selector from .find:
db.collection.find()
as suggested in the comments.
Related
Currently I could filter with this MongoDb query
db.getCollection("entity").find(
{
"NameDetails.Name.0.NameValue.0.EntityName" : /ABC/
}
);
How do I loop through all the Name and then all the NameValue to search for /ABC/? If any of it matches, it returns as result.
You need to either use elemMatch or unwind.
If you know that it would match one element always, use elemMatch
or
If you want all the matching elements in the array, go for unwind then group.
I have a mongo DB collection that looks something like this:
{
{
_id: objectId('aabbccddeeff'),
objectName: 'MyFirstObject',
objectLength: 0xDEADBEEF,
objectSource: 'Source1',
accessCounter: {
'firstLocationCode' : 283,
'secondLocationCode' : 543,
'ThirdLocationCode' : 564,
'FourthLocationCode' : 12,
}
}
...
}
Now, assuming that this is not the only record in the collection and that most/all of the documents contain the accessCounter subdocument/field how will I go with selecting the x first documents where I have the most access from a specific location.
A sample "query" will be something like:
"Select the first 10 documents From myCollection where the accessCounter.firstLocationCode are the highest"
So a sample result will be X documents where the accessCounter. will be the greatest is the database.
Thank your for taking the time to read my question.
No need for an aggregation, that is a basic query:
db.collection.find().sort({"accessCounter.firstLocation":-1}).limit(10)
In order to speed this up, you should create a subdocument index on accessCounter first:
db.collection.ensureIndex({'accessCounter':-1})
assuming the you want to do the same query for all locations. In case you only want to query firstLocation, create the index on accessCounter.firstLocation.
You can speed this up further in case you only need the accessCounter value by making this a so called covered query, a query of which the values to return come from the index itself. For example, when you have the subdocument indexed and you query for the top secondLocations, you should be able to do a covered query with:
db.collection.find({},{_id:0,"accessCounter.secondLocation":1})
.sort("accessCounter.secondLocation":-1).limit(10)
which translates to "Get all documents ('{}'), don't return the _id field as you do by default ('_id:0'), get only the 'accessCounter.secondLocation' field ('accessCounter.secondLocation:1'). Sort the returned values in descending order and give me the first ten."
All,
I have a mangodb collection with below fields.
_ID
Title
Description
Tags , array
I have created 2 index on _id and tags field. I have created index for people to search the content with help of keywords.
I have created the index with tags:-1 to show the latest inserted records to show first. But even after that it is showing in the ascending order of _id.
How to create the index on tags field to show the last inserted to show first at the same time it should allow me to search on tags field faster .
If the _id field is the default ObjectId which reflects the insertion order, and you want to query all the documents with a specific Tags by descending insertion order, you can use the query as below:
find({ Tags : $value }).sort({ _id : -1 })
For this query, you can create a compound index on { Tags : 1, _id : -1 }. All the documents with the same Tags will be sorted in descending insertion order and this index should work well for this query.
Please note that if you are doing range query on Tags, like:
find({ Tags : { $in : [ $value1, $value2 ] }}).sort({ _id : -1 })
find({ Tags : { $gt : $value}}).sort({ _id : -1})
It wouldn’t be able to use the index to sort the result documents, and will need to sort the results in memory. You can run the query with .explain(true) to check the query plan. If scanAndOrder is true, it means the query cannot use the order of documents in the index for returning sorted results.
There are also some documents and blogs relate to indexes and that I'd recommend reading:
http://docs.mongodb.org/manual/tutorial/sort-results-with-indexes/
http://emptysqua.re/blog/optimizing-mongodb-compound-indexes/
http://blog.mongolab.com/2012/06/cardinal-ins/
I'm trying to use the aggregation framework to group a lot of strings together to indentify the unique ones. I must also keep some information about the rest of the fields. This would be analogous to me using the * operator in mysql with a group by statement.
SELECT *
FROM my_table
GROUP BY field1
I have tried using the aggregation framework, and it works fine just to get unique fields.
db.mycollection.aggregate({
$group : { _id : "$field1"}
})
What if I want the other fields that went with that. MySQL would only give me the first one that appeared in the group (which I'm fine with). Thats what I thought the $first operator did.
db.mycollection.aggregate({
$group : {
_id : "$field1",
another_field : {$first : "$field2"}
}})
This way it groups by field1 but still gives me back the other fields attached to document. When I try this I get:
exception: aggregation result exceeds maximum document size (16MB)
Which I have a feeling is because it is returning the whole aggregation back as one document. Can I return it as another json array?
thanks in advance
You're doing the aggregation correctly, but as the error message indicates, the full result of the aggregate call cannot be larger than 16 MB.
Work-arounds would be to either add a filter to reduce the size of the result or use map-reduce instead and output the result to another collection.
If you unique values of the result does not exceed 2000 you could use group() function like
db.mycollection.group( {key : {field1 : 1, field2 : 1}}, reduce: function(curr, result){}, initial{} })
Last option would be map reduce:
db.mycollection.mapReduce( function() { emit( {field1 :1, field2: 1}, 1); }, function(key, values) { return 1;}, {out: {replace: "unique_field1_field2"}})
and your result would be in "unique_field1_field2" collection
Another alternative is use the distinct function:
db.mycollection.distinct('field1')
This functions accepts a second argument, a query, where you can filter the documents.
This is my document:
{
title:"Happy thanksgiving",
body: "come over for dinner",
blocked:[
{user:333, name:'john'},
{user:994, name:'jessica'},
{user:11, name: 'matt'},
]
}
What is the query to find all documents that do not have user 11 in "blocked"?
You can use $in or $nin for "not in"
Example ...
> db.people.find({ crowd : { $nin: ["cool"] }});
I put a bunch more examples here: http://learnmongo.com/posts/being-part-of-the-in-crowd/
Since you are comparing against a single value, your example actually doesn't need a NOT IN operation. This is because Mongo will apply its search criteria to every element of an array subdocument. You can use the NOT EQUALS operator, $ne, to get what you want as it takes the value that cannot turn up in the search:
db.myCollection.find({'blocked.user': {$ne: 11}});
However if you have many things that it cannot equal, that is when you would use the NOT IN operator, which is $nin. It takes an array of values that cannot turn up in the search:
db.myCollection.find({'blocked.user': {$nin: [11, 12, 13]}});
Try the following:
db.stack.find({"blocked.user":{$nin:[11]}})
This worked for me.
See http://docs.mongodb.org/manual/reference/operator/query/nin/#op._S_nin
db.inventory.find( { qty: { $nin: [ 5, 15 ] } } )
This query will
select all documents in the inventory collection where the qty field
value does not equal 5 nor 15. The selected documents will include
those documents that do not contain the qty field.
If the field holds an array, then the $nin operator selects the
documents whose field holds an array with no element equal to a value
in the specified array (e.g. , , etc.).