Subquery in WHERE clause - python

I am currently working on getting the following query run with sqlalchemy.
SELECT *
FROM
qsreport_test.JiraFehlerFreigabeReleaseTestData a
WHERE a.dtime = (
SELECT
max(dtime)
FROM
qsreport_test.JiraFehlerFreigabeReleaseTestData b
WHERE
b.key=a.key
AND
b.testplanname=a.testplanname
AND
b.priority=a.priority
)
GROUP BY
a.testplanname, a.priority
I searched here and found
SQLAlchemy - subquery in a WHERE clause
and I also know
http://docs.sqlalchemy.org/en/rel_0_9/orm/tutorial.html#using-subqueries
but I still do not get my query to work.
Here are two of my tries. The first one can be performed but does not give me the same result as the manual executed query. The second one can be performed and is now working (was missing a parenthesis before).
Try 1:
subq = session.query(JiraFehlerFreigabeReleaseTestData.key, JiraFehlerFreigabeReleaseTestData.testplanname, JiraFehlerFreigabeReleaseTestData.priority, func.max(JiraFehlerFreigabeReleaseTestData.dtime).label('max_dtime')).filter(JiraFehlerFreigabeReleaseTestData.key == key).filter(JiraFehlerFreigabeReleaseTestData.dtime <= dtime).subquery()
res = session.query(JiraFehlerFreigabeReleaseTestData).filter(and_(JiraFehlerFreigabeReleaseTestData.key == subq.c.key, JiraFehlerFreigabeReleaseTestData.testplanname == subq.c.testplanname, JiraFehlerFreigabeReleaseTestData.priority == subq.c.priority, JiraFehlerFreigabeReleaseTestData.dtime == subq.c.max_dtime)).group_by(JiraFehlerFreigabeReleaseTestData.testplanname).order_by(JiraFehlerFreigabeReleaseTestData.dtime.desc()).all()
Try 2:
b = aliased(JiraFehlerFreigabeReleaseTestData, name='b')
res = session.query(JiraFehlerFreigabeReleaseTestData).filter(JiraFehlerFreigabeReleaseTestData.dtime == (session.query(func.max(b.dtime)).filter(b.key == JiraFehlerFreigabeReleaseTestData.key).filter(b.testplanname == JiraFehlerFreigabeReleaseTestData.testplanname).filter(b.priority == JiraFehlerFreigabeReleaseTestData.priority))).group_by(JiraFehlerFreigabeReleaseTestData.testplanname,JiraFehlerFreigabeReleaseTestData.priority).order_by(JiraFehlerFreigabeReleaseTestData.dtime.desc()).all()
Please show me what I'm doing wrong. My problem is that I do not completely understand how to reference the table of the main query in the subquery. Especially as the two tables are the same in both queries. Perhaps there is strait what to convert my manual sql query to sqlalchemy orm syntax?
Edit:
My second version is working. I was just missing one paranthesis. I fixed the code above.
But why is my first try giving me a different result?

I've not worked with SQLAlchemy yet, but perhaps you can solve your problem, if your arn't using a subquery. Take a look of the query below. Not tested an free translated from yours ;)
SELECT a.*
FROM qsreport_test.JiraFehlerFreigabeReleaseTestData a
INNER JOIN qsreport_test.JiraFehlerFreigabeReleaseTestData b
ON b.key = a.key
AND b.testplanname=a.testplanname
AND b.priority=a.priority
GROUP BY a.testplanname, a.priority
HAVING a.dtime = max(b.dtime)

Related

Make SQLAlchemy "or_" operator return only the 1st match

guys! I have this one query, written with SQLAlchemy:
query = query.filter(
CollectionCardLink.collection_id == card_id,
CollectionCardLink.position.in_([0, count-1]),
or_(CardStyle.user_id == referrer_id, CardStyle.user_id == ANYONE_USER_ID)
).all()
My issue is: this => or_(PostStyle.user_id == current_user_id, PostStyle.user_id == DEFAULT_USER_ID) returns me two records, cause there are records for current_user_id and for DEFAULT_USER_ID in table too.
But I want or_ to work like it works in Python: stop checking after 1st True. So, I need to get PostStyle for DEFAULT_USER_ID only if there's no PostStyle for current_user_id.
Can someone help, please, with that issue?
Also I tried to code: (PostStyle.user_id == current_user_id or PostStyle.user_id == DEFAULT_USER_ID) but it gives me a record for DEFAULT_USER_ID firstly and that's incorrect

how to do a subquery or filter in a condition met by a previous query correctly

Im using python, flask and sqlalchemy have the query below:
query = db.session.query(model.Foo1, model.Foo2).join(model.Foo1, model.Foo1.id == model.Foo2.name_id).first()
Now let's assume that i want to search for the condition Foo1.id != 2 but still make sure that i meet the condition of the query above, what would be the best way to achieve that (i am trying to learn how to do subquery or filter in another query properly)
Thank for the help
For Foo1.id != 2 You don't need a subquery. Filter is enough.
result = db.session.query(model.Foo1, model.Foo2)\
.join(model.Foo1, model.Foo1.id == model.Foo2.name_id)\
.filter(model.Foo1.id != 2).all()
This will return a list of all rows that have ids not equal to 2

Get total record count for SqlAlchemy query result, which uses pagination(limit, offset)

I am using sqlalchemy query with limit and offset. Now I need to get the total count of the query result.
For now I am using count query and limit query separately and getting the results. Is there any efficient way to get the count in a single sqlalchemy query.
Here is the sample of currently what I am using,
# getting total count
docs_count = DB_SESSION.query(Documents).count()
# using limit and offset
docs_list = DB_SESSION.query(Documents).limit(10).offset(0)
Now is it possible to combine this into a single query, or there any efficient method to do this. Thanks all.
I was facing the same issue.
What I did is to add the count operation into the query through an 'over' clause as follow:
from sqlalchemy.sql import func
docs_query = DB_SESSION.query(Documents,func.count(Documents.id).over().label("total"))
docs_query = docs_query.limit(10).offset(0)
docs_results = docs_query.all()
if len(docs_results) > 0 :
docs_list= docs_results[0]
docs_count= docs_results[0][1]
SQLAlchemy info :
count:
https://docs.sqlalchemy.org/en/14/orm/query.html#sqlalchemy.orm.Query.count
over:
https://docs.sqlalchemy.org/en/14/core/sqlelement.html#sqlalchemy.sql.expression.over
I don't know if it is the best way to do it but it is a bit faster than what you have done. Hope it will help you or someone else.

Limit child collections in initial query sqlalchemy

I am building an api which can return children of resources if the user requests it. For example, user has messages. I want the query to be able to limit the number of message objects that are returned.
I found a useful tip aboutl imiting the number of objects in child collections here. Basically, it indicates the following flow:
class User(...):
# ...
messages = relationship('Messages', order_by='desc(Messages.date)', lazy='dynamic')
user = User.query.one()
users.messages.limit(10)
My use case involves returning sometimes large numbers of users.
If I were to follow the advice in that link and used .limit() then I would need to iterate over the entire collection of users calling .limit() on each one. This is much less efficient then, say, using LIMIT in the original sql expression which created the collection.
My question is whether it is possible using declarative to efficiently(N+0) load a large collection of objects while limiting the number of children in their child collections using sqlalchemy?
UPDATE
To be clear, the below is what I am trying to avoid.
users = User.query.all()
messages = {}
for user in users:
messages[user.id] = user.messages.limit(10).all()
I want to do something more like:
users = User.query.option(User.messages.limit(10)).all()
This answer comes from Mike Bayer on the sqlalchemy google group. I'm posting it here to help folks:
TLDR:
I used version 1 of Mike's answer to solve my problem because, in this case, I do not have foreign keys involved in this relationship and so cannot make use of LATERAL. Version 1 worked great, but be sure to note the effect of offset. It threw me off during testing for a while because I didn't notice it was set to something other than 0.
Code Block for version 1:
subq = s.query(Messages.date).\
filter(Messages.user_id == User.id).\
order_by(Messages.date.desc()).\
limit(1).offset(10).correlate(User).as_scalar()
q = s.query(User).join(
Messages,
and_(User.id == Messages.user_id, Messages.date > subq)
).options(contains_eager(User.messages))
Mike's Answer
so you should ignore whether or not it uses "declarative", which has nothing to do with querying, and in fact at first ignore Query too, because first and foremost this is a SQL problem. You want one SQL statement that does this. What query in SQL would load lots of rows from the primary table, joined to the first ten rows of the secondary table for each primary?
LIMIT is tricky because it's not actually part of the usual "relational algebra" calculation. It's outside of that because it's an artificial limit on rows. For example, my first thought on how to do this was wrong:
select * from users left outer join (select * from messages limit 10) as anon_1 on users.id = anon_1.user_id
This is wrong because it only gets the first ten messages in the aggregate, disregarding user. We want to get the first ten messages for each user, which means we need to do this "select from messages limit 10" individually for each user. That is, we need to correlate somehow. A correlated subquery though is not usually allowed as a FROM element, and is only allowed as a SQL expression, it can only return a single column and a single row; we can't normally JOIN to a correlated subquery in plain vanilla SQL. We can however, correlate inside the ON clause of the JOIN to make this possible in vanilla SQL.
But first, if we are on a modern Postgresql version, we can break that usual rule of correlation and use a keyword called LATERAL, which allows correlation in a FROM clause. LATERAL is only supported by modern Postgresql versions, and it makes this easy:
select * from users left outer join lateral
(select * from message where message.user_id = users.id order by messages.date desc limit 10) as anon1 on users.id = anon_1.user_id
we support the LATERAL keyword. The query above looks like this:
subq = s.query(Messages).\
filter(Messages.user_id == User.id).\
order_by(Messages.date.desc()).limit(10).subquery().lateral()
q = s.query(User).outerjoin(subq).\
options(contains_eager(User.messages, alias=subq))
Note that above, in order to SELECT both users and messages and produce them into the User.messages collection, the "contains_eager()" option must be used and for that the "dynamic" has to go away. This is not the only option, you can for example build a second relationship for User.messages that doesn't have the "dynamic" or you can just load from query(User, Message) separately and organize the result tuples as needed.
if you aren't using Postgresql, or a version of Postgresql that doesn't support LATERAL, the correlation has to be worked into the ON clause of the join instead. The SQL looks like:
select * from users left outer join messages on
users.id = messages.user_id and messages.date > (select date from messages where messages.user_id = users.id order by date desc limit 1 offset 10)
Here, in order to jam the LIMIT in there, we are actually stepping through the first 10 rows with OFFSET and then doing LIMIT 1 to get the date that represents the lower bound date we want for each user. Then we have to join while comparing on that date, which can be expensive if this column isn't indexed and also can be inaccurate if there are duplicate dates.
This query looks like:
subq = s.query(Messages.date).\
filter(Messages.user_id == User.id).\
order_by(Messages.date.desc()).\
limit(1).offset(10).correlate(User).as_scalar()
q = s.query(User).join(
Messages,
and_(User.id == Messages.user_id, Messages.date >= subq)
).options(contains_eager(User.messages))
These kinds of queries are the kind that I don't trust without a good test, so POC below includes both versions including a sanity check.
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
import datetime
Base = declarative_base()
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
messages = relationship(
'Messages', order_by='desc(Messages.date)')
class Messages(Base):
__tablename__ = 'message'
id = Column(Integer, primary_key=True)
user_id = Column(ForeignKey('user.id'))
date = Column(Date)
e = create_engine("postgresql://scott:tiger#localhost/test", echo=True)
Base.metadata.drop_all(e)
Base.metadata.create_all(e)
s = Session(e)
s.add_all([
User(id=i, messages=[
Messages(id=(i * 20) + j, date=datetime.date(2017, 3, j))
for j in range(1, 20)
]) for i in range(1, 51)
])
s.commit()
top_ten_dates = set(datetime.date(2017, 3, j) for j in range(10, 20))
def run_test(q):
all_u = q.all()
assert len(all_u) == 50
for u in all_u:
messages = u.messages
assert len(messages) == 10
for m in messages:
assert m.user_id == u.id
received = set(m.date for m in messages)
assert received == top_ten_dates
# version 1. no LATERAL
s.close()
subq = s.query(Messages.date).\
filter(Messages.user_id == User.id).\
order_by(Messages.date.desc()).\
limit(1).offset(10).correlate(User).as_scalar()
q = s.query(User).join(
Messages,
and_(User.id == Messages.user_id, Messages.date > subq)
).options(contains_eager(User.messages))
run_test(q)
# version 2. LATERAL
s.close()
subq = s.query(Messages).\
filter(Messages.user_id == User.id).\
order_by(Messages.date.desc()).limit(10).subquery().lateral()
q = s.query(User).outerjoin(subq).\
options(contains_eager(User.messages, alias=subq))
run_test(q)
If you apply limit and then call .all() on it, you will get all objects once and it will not get objects one by one , causing performance issue that you mentioned.
simply apply limit and get all objects.
users = User.query.limit(50).all()
print(len(users))
>>50
Or for child objects / relationships
user = User.query.one()
all_messages = user.messages.limit(10).all()
users = User.query.all()
messages = {}
for user in users:
messages[user.id] = user.messages.limit(10).all()
So, I think you'll need to load the messages in a second query and then later associate with your users somehow.
The following is database dependent; as discussed in this question, mysql does not support in queries with limits, but sqlite at least will parse the query. I didn't look at the plan to see if it did a good job.
The following code will find all the message objects you care about. You then need to associate them with users.
I've tested this to confirm that it produces a query sqlite can parse; I have not confirmed that sqlite or any other database does the right thing with this query.
I had to cheat a bit and use the text primitive to refer to the outer user.id column in the select because SQLAlchemy kept wanting to include an additional join to users in the inner select subjquery.
from sqlalchemy import Column, Integer, String, ForeignKey, alias
from sqlalchemy.sql import text
from sqlalchemy.orm import Session
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class User(Base):
__tablename__ = 'users'
id = Column(Integer, primary_key = True)
name = Column(String)
class Message(Base):
__tablename__ = 'messages'
user_id = Column(Integer, ForeignKey(User.id), nullable = False)
id = Column(Integer, primary_key = True)
s = Session()
m1 = alias(Message.__table__)
user_query = s.query(User) # add any user filtering you want
inner_query = s.query(m1.c.id).filter(m1.c.user_id == text('users.id')).limit(10)
all_messages_you_want = s.query(Message).join(User).filter(Message.id.in_(inner_query))
To associate the messages with users, you could do something like the following assuming your Message has a user relation and your user objects have a got_child_message method that does whatever you like for this
users_resulting = user_query.all() #load objects into session and hold a reference
for m in all_messages_you_want: m.user.got_child_message(m)
Because you already have the users in the session and because the relation is on User's primary key, m.user resolves to query.get against the identity map.
I hope this helps you get somewhere.
#melchoirs answer is the best. I basically putting this here for futureselves
I played around with the above stated answer, and it works, I needed it more so to limit the number of associations returned before passing into a Marshmallow Serializer.
Some issues for clarification:
the subquery runs per association, hence it finds the corresponding date to base off properly
think about the limit/offset as give me 1 (limit) record starting at the next X (offset). Hence what is the Xth oldest record, and then in the main query it gives everything back from that. Its damn smart
It appears that if the association has less than X records, it returns nothing, as the offset is past the records, and henceforth the main query does not return a record.
Using the above as a template, I came up with the below answer. The initial query/count guard is due to the issue that if the associated records are less than the offset, nothing is found. In addition, I needed to add an outerjoin in the event that there are no associations either.
At the end, I found this query to be a bit or ORM voodoo, and didn't want to go that route. I instead exclude the histories from the device serializer, and require a second history lookup using the device ID. That set can be paginated and makes everything a bit cleaner.
Both methods work, it just comes down to the why you'll need to do the one query versus a couple. In the above, there was probably business reasons to get everythng back more efficiently with the single query. For my use case, readability, and convention trumped the voodoo
#classmethod
def get_limited_histories(cls, uuid, limit=10):
count = DeviceHistory.query.filter(DeviceHistory.device_id == uuid).count()
if count > limit:
sq = db.session.query(DeviceHistory.created_at) \
.filter(DeviceHistory.device_id == Device.uuid) \
.order_by(DeviceHistory.created_at.desc()) \
.limit(1).offset(limit).correlate(Device)
return db.session.query(Device).filter(Device.uuid == uuid) \
.outerjoin(DeviceHistory,
and_(DeviceHistory.device_id == Device.uuid, DeviceHistory.created_at > sq)) \
.options(contains_eager(Device.device_histories)).all()[0]
It then behaves similar to a Device.query.get(id) but Device.get_limited_histories(id)
ENJOY

How to add a row number column to an SqlAlchemy query?

I would like to have the row number as a column of my queries. Since I am using MySql, I cannot use the built-in func.row_number() of SqlAlchemy. The result of this query is going to be paginated, therefore I would like to keep the row number before the split happen.
session.query(MyModel.id, MyModel.date, "row_number")
I tried to use an hybrid_property to increment a static variable inside MyModel class that I reset before my query, but it didn't work.
#hybrid_property
def row_number(self):
cls = self.__class__
cls.row_index = cls.row_index + 1
return literal(self.row_index)
#row_number.expression
def row_number(cls):
cls.row_index = cls.row_index + 1
return literal(cls.row_index)
I also tried to mix a subquery with this solution :
session.query(myquery.subquery(), literal("#rownum := #rownum + 1 AS row_number"))
But I didn't find a way to make a textual join for (SELECT #rownum := 0) r.
Any suggestions?
EDIT
For the moment, I am looping on the results of the paginated query and I am assigning the calculated number from the current page to each row.
SQLAlchemy allows you to use text() in some places, but not arbitrarily. I especially cannot find an easy/documented way of using it in columns or joins. However, you can write your entire query in SQL and still get ORM objects out of it. Example:
query = session.query(Foobar, "rownum")
query = query.from_statement(
"select foobar.*, cast(#row := #row + 1 as unsigned) as rownum"
" from foobar, (select #row := 0) as init"
)
That being said, I don't really see the problem with something like enumerate(query.all()) either. Note that if you use a LIMIT expression, the row numbers you get from MySQL will be for the final result and will still need to have the page start index added. That is, it's not "before the split" by default. If you want to have the starting row added for you in MySQL you can do something like this:
prevrow = 42
query = session.query(Foobar, "rownum")
query = query.from_statement(sqlalchemy.text(
"select foobar.*, cast(#row := #row + 1 as unsigned) as rownum"
" from foobar, (select #row := :prevrow) as init"
).bindparams(prevrow=prevrow))
In this case the numbers will start at 43 since it's pre-incrementing.

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