Support i have a document structure as follows
var order = {
order_id : "1234",
plan_name: "Basic plan",
app_id:["app-id-1","app-id-2"]
}
It has one to many relationship structure.
i inserted it in mongo like this
db.orders.insert(order)
I also created index on order_id and app_id.
Now , i have an app_id , i want to find its order_id.
How to write query to find
I know i need to use
db.orders.find({"app_id":""}) , but app_id is an array.
Or is there a other technique i am missing ?
You can query on arrays in several ways.
If you want to find all orders with app-id-1 as a value in the app_id array you can simply query
app_id: "app-1-id"
This will return all orders which have that value in their array.
if you want to find all orders which have a set of app ids you can query
app_id: {$all: <array_of_app_ids>}
To find all orders which contain only a specific set of ids and no others you can pair $all with $size
app_id: {$all: <array_of_app_ids>, $size: <array_of_app_ids.length>}
but if you pass a single value, mongoDB will return all entries which have that value in their array
Related
I have an array of 10 unique Object IDs named Arr
I have 10,000 documents in a collection named xyz.
How can I find documents using Object IDs in the array Arr from the collection xyz with only one request?
There are $all and $in operators but are used to query fields with an array.
Or do I need to make requests equal to the length of Arr and get individual document using findOne?
EDIT:
I'm expecting something like this:
db.getCollection("xyz").find({"_id" : [array containing 10 unique IDs]})
....for which the result callback will contain an array of all the matched IDs of query array.
According to the documentation here: https://docs.mongodb.com/manual/reference/operator/query/in/
You should use the following query:
db.getCollection("xyz").find({"Arr" : { $in: [123, 456, 789 ] }});
I have a mongodb query where i want to get documents if a field has particular value.
db.collection.find({key:{$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.
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."
I am using mongo's shell and want to do what is basically equivalent to "SQL's select col INTO var" and then use the value of var to look up other rows in the same table or others (Joins). For example, in PL/SQL I will declare a variable called V_Dno. I also have a table called Emp(EID, Name, Sal, Dno). I can access the value of Dno for employee 100 as, "Select Dno into V_Dno from Emp where EID = 100). In MongoDB, when I find the needed employee (using its _id), I end up with a document and not a value (a field). In a sense, I get equivalent to the entire row in SQL and not just a column. I am doing the following to find the given emp:
VAR V_Dno = db.emp.find ({Eid : 100}, {Dno : 1});
The reason I want to do this to traverse from one document into the other using the value of a field. I know I can do it using the DBRef, but I wanted to see if I could tie documents together using this method.
Can someone please shed some light on this?
Thanks.
find returns a cursor that lets you iterate over the matching documents. In this case you'd want to use findOne instead as it directly returns the first matching doc, and then use dot notation to access the single field.
var V_Dno = db.emp.findOne({Eid : 100}, {Dno : 1}).Dno;
Using your query as a starting point:
var vdno = db.emp.findOne({Eid: 100, Dno :1})
This returns a document from the emp collection where the Eid = 100 and the Dno = 1. Now that I have this document in the vdno variable I can "join" it to another collection. Lets say you have a Department collection, a document in the department collection has a manual reference to the _id field in the emp collection. You can use the following to filter results from the department collection based on the value in your variable.
db.department.find({"employee._id":vdno._id})
I have Document
class Store(Document):
store_id = IntField(required=True)
items = ListField(ReferenceField(Item, required=True))
meta = {
'indexes': [
{
'fields': ['campaign_id'],
'unique': True
},
{
'fields': ['items']
}
]
}
And want to set up indexes in items and store_id, does my configuration right?
Your second index declaration looks like it should do what you want. But to make sure that the index is really effective, you should use explain. Connect to your database with the mongo shell and perform a find-query which should use that index followed by .explain(). Example:
db.yourCollection.find({items:"someItem"}).explain();
The output will be a document with lots of fields. The documentation explains what exactly each field means. Pay special attention to these fields:
millis Time in milliseconds the query required
indexOnly (self-explaining)
n number of returned documents
nscannedObjects the number of objects which had to be examined without using an index. For an index-only query this should be equal to n. When it is higher, it means that some documents could not be excluded by an index and had to be scanned manually.