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})
Related
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
I Have An Array With 4 Object Items I want query to my collection and return 4 items that have this uid's...
myArray = > [{uid : 'test'},{uid : 'test2'},{uid : 'test3'},{uid : 'test4'}]
ProductCollection.find({????},(err,result)=>{})
NOTE : I dont want use any loop
I dont want use any loop
I will assume that's related to query the DB several times, one for each uid value.
Anyway, you can go to the database once to filter elements that match an array of values, like your case, using MongoDB's $in operator.
But you would have to format the uid values to an array of the values themselves instead of the array of objets, this can be accomplished with a simple .map call (don't know if you will consider this a loop) to get the filter value in the correct format.
var uids = myArray.map((item) => item.uid })
// ['test', 'test2', 'test3', 'test4']
And after that you can query your DB with this uids values
ProductCollection.find({'uid': {'$in': uids} },(err,result)=>{})
(Assuming 'uid' it the property you have in your ProductCollection that you are trying to filter by)
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'm looking for a way to get special fields from mongoskin find function. in other words in SQL lang we say select column1,column2,column3 from mytable rather than select *
currently my query is like below and i want to get specify the fields that I'm looking for rather the the whole json object.
db.collection('collacta').find().toArray(function(err, result) {
if (result) {
...
} else {
...
};
});
thanks
For getting the projection of fields, you should pass the DBObject for projection,
DBCursor cursor = collection.find(query, projectionQuery);
The projection is The DBObject in form of key-value pair. where,
key is the name of field you want to project. value can be either 0 or 1.
0 - means exclude the particular column from result set.
1 - means include the particular column in result set.
For more info, see here.
I'm using Flask-SQLAlchemy with PostgreSQL. I have the following two models:
class Course(db.Model):
id = db.Column(db.Integer, primary_key = True )
course_name =db.Column(db.String(120))
course_description = db.Column(db.Text)
course_reviews = db.relationship('Review', backref ='course', lazy ='dynamic')
class Review(db.Model):
__table_args__ = ( db.UniqueConstraint('course_id', 'user_id'), { } )
id = db.Column(db.Integer, primary_key = True )
review_date = db.Column(db.DateTime)#default=db.func.now()
review_comment = db.Column(db.Text)
rating = db.Column(db.SmallInteger)
course_id = db.Column(db.Integer, db.ForeignKey('course.id') )
user_id = db.Column(db.Integer, db.ForeignKey('user.id') )
I want to select the courses that are most reviewed starting with at least two reviews. The following SQLAlchemy query worked fine with SQlite:
most_rated_courses = db.session.query(models.Review, func.count(models.Review.course_id)).group_by(models.Review.course_id).\
having(func.count(models.Review.course_id) >1) \ .order_by(func.count(models.Review.course_id).desc()).all()
But when I switched to PostgreSQL in production it gives me the following error:
ProgrammingError: (ProgrammingError) column "review.id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT review.id AS review_id, review.review_date AS review_...
^
'SELECT review.id AS review_id, review.review_date AS review_review_date, review.review_comment AS review_review_comment, review.rating AS review_rating, review.course_id AS review_course_id, review.user_id AS review_user_id, count(review.course_id) AS count_1 \nFROM review GROUP BY review.course_id \nHAVING count(review.course_id) > %(count_2)s ORDER BY count(review.course_id) DESC' {'count_2': 1}
I tried to fix the query by adding models.Review in the GROUP BY clause but it did not work:
most_rated_courses = db.session.query(models.Review, func.count(models.Review.course_id)).group_by(models.Review.course_id).\
having(func.count(models.Review.course_id) >1) \.order_by(func.count(models.Review.course_id).desc()).all()
Can anyone please help me with this issue. Thanks a lot
SQLite and MySQL both have the behavior that they allow a query that has aggregates (like count()) without applying GROUP BY to all other columns - which in terms of standard SQL is invalid, because if more than one row is present in that aggregated group, it has to pick the first one it sees for return, which is essentially random.
So your query for Review basically returns to you the first "Review" row for each distinct course id - like for course id 3, if you had seven "Review" rows, it's just choosing an essentially random "Review" row within the group of "course_id=3". I gather the answer you really want, "Course", is available here because you can take that semi-randomly selected Review object and just call ".course" on it, giving you the correct Course, but this is a backwards way to go.
But once you get on a proper database like Postgresql you need to use correct SQL. The data you need from the "review" table is just the course_id and the count, nothing else, so query just for that (first assume we don't actually need to display the counts, that's in a minute):
most_rated_course_ids = session.query(
Review.course_id,
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
order_by(func.count(Review.course_id).desc()).\
all()
but that's not your Course object - you want to take that list of ids and apply it to the course table. We first need to keep our list of course ids as a SQL construct, instead of loading the data - that is, turn it into a derived table by converting the query into a subquery (change the word .all() to .subquery()):
most_rated_course_id_subquery = session.query(
Review.course_id,
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
order_by(func.count(Review.course_id).desc()).\
subquery()
one simple way to link that to Course is to use an IN:
courses = session.query(Course).filter(
Course.id.in_(most_rated_course_id_subquery)).all()
but that's essentially going to throw away the "ORDER BY" you're looking for and also doesn't give us any nice way of actually reporting on those counts along with the course results. We need to have that count along with our Course so that we can report it and also order by it. For this we use a JOIN from the "course" table to our derived table. SQLAlchemy is smart enough to know to join on the "course_id" foreign key if we just call join():
courses = session.query(Course).join(most_rated_course_id_subquery).all()
then to get at the count, we need to add that to the columns returned by our subquery along with a label so we can refer to it:
most_rated_course_id_subquery = session.query(
Review.course_id,
func.count(Review.course_id).label("count")
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
subquery()
courses = session.query(
Course, most_rated_course_id_subquery.c.count
).join(
most_rated_course_id_subquery
).order_by(
most_rated_course_id_subquery.c.count.desc()
).all()
A great article I like to point out to people about GROUP BY and this kind of query is SQL GROUP BY techniques which points out the common need for the "select from A join to (subquery of B with aggregate/GROUP BY)" pattern.