Develop Reference

How to access a CTE column in sqlalchemy filter()?

I have a query object with two columns: one is a real column from a table, the other is aggregate.
The query object is returned by a function and used later in the code.
def thequery():
cte1 = (
db.query(
UserFile.user_id,
func.min(UserFile.id).label("min_id")
).group_by(UserFile.user_id)
).cte("cte1")
return db.query(User.name, cte1.c.min_id).join(cte1, cte1.c.user_id == User.id)
Since the query is not yet executed, I can add more filters.
q = thequery()
q.filter(User.name == "John") # This works because `User` is a model.
However, I can't find a way to filter by my CTE column.
q.filter(min_id == 200) # How to refer to min_id?

from sqlalchemy import literal_column
q.filter(literal_column("cte1.min_id") == 200)
You can use expression above

How to iteratively create UNION ALL SQL statement using Python?

I am connecting to Snowflake to query row count data of view table from Snowflake. I am also querying metadata related to View table. My Query looks like below. I was wondering if I can iterate through UNION ALL statement using python ? When I try to run my below query I received an error that says "view_table_3" does not exist.
Thanks in advance for your time and efforts!
Query to get row count for Snowflake view table (with metadata)
view_tables=['view_table1','view_table2','view_table3','view_table4']
print(f""" SELECT * FROM (SELECT TABLE_SCHEMA,TABLE_NAME,CREATED,LAST_ALTERED FROM SCHEMA='INFORMATION_SCHEMA.VIEWS' WHERE TABLE_SCHEMA='MY_SCHEMA' AND TABLE_NAME IN ({','.join("'" +x+ "'" for x in view_tables)})) t1
LEFT JOIN
(SELECT 'view_table1' table_name2, count(*) as view_row_count from MY_DB.SCHEMA.view_table1
UNION ALL SELECT {','.join("'" +x+ "'" for x in view_tables[1:])},count(*) as view_row_count from MY_DB.SCHEMA.{','.join("" +x+ "" for x.replace("'"," ") in view_tables)})t2
on t1.TABLE_NAME =t2.table_name2 """)

If you want to make a union dynamically, put the entire SELECT query inside the generator, and then join them with ' UNION '.
sql = f'''SELECT * FROM INFORMATION_SCHEMA.VIEWS AS v
LEFT JOIN (
{' UNION '.join(f"SELECT '{table}' AS table_name2, COUNT(*) AS view_row_count FROM MY_SCHEMA.{table}" for table in view_tables)}
) AS t2 ON v.TABLE_NAME = t2.table_name2
WHERE v.TABLE_NAME IN ({','.join(f"'{table}'" for table in view_tables)})
'''
print(sql);

Django SQL Windows Function with Group By

I have this MariaDB Query
SELECT
DAYOFWEEK(date) AS `week_day`,
SUM(revenue)/SUM(SUM(revenue)) OVER () AS `rev_share` FROM orders
GROUP BY DAYOFWEEK(completed)
It will result in a table which show the revenue share.
My goal is to archive the same with the help of Django's ORM layer.
I tried the following by using RawSQL:
share = Orders.objects.values(week_day=ExtractIsoWeekDay('date')) \
.annotate(revenue_share=RawSQL('SUM(revenue)/SUM(SUM(revenue)) over ()'))
This results in a single value without a group by. The query which is executed:
SELECT
WEEKDAY(`orders`.`date`) + 1 AS `week_day`,
(SUM(revenue)/SUM(SUM(revenue)) over ()) AS `revenue_share`
FROM `orders`
And also this by using the Window function:
share = Orders.objects.values(week_day=ExtractIsoWeekDay('date')) \
.annotate(revenue_share=Sum('revenue')/Window(Sum('revenue')))
Which results into the following query
SELECT
WEEKDAY(`order`.`date`) + 1 AS `week_day`,
(SUM(`order`.`revenue`) / SUM(`order`.`revenue`) OVER ()) AS `rev_share`
FROM `order` GROUP BY WEEKDAY(`order`.`date`) + 1 ORDER BY NULL
But the data is completely wrong. It looks like the Window is not using the whole table.
Thanks for your help in advance.

How to query set as ORDER BY and GROUP BY in django?

This my is query:
SELECT kategoriharga,ongkoskirim,diskon,ratingproduk,ratingtoko,label
FROM
(SELECT *
FROM pohonkeputusan
where perdaerah='Kabupaten Toba Samosir'
order by label desc
) AS sub
GROUP BY
kategoriharga,ongkoskirim,diskon,ratingproduk,ratingtoko
How to make to be query set in Django?

I don't understand why you want to group by all fields. Try to use distinct:
Pohonkeputusan.objects.filter(perdaerah='Kabupaten Toba Samosir').order_by('-label').values_list('kategoriharga', 'ongkoskirim', 'diskon', 'ratingproduk', 'ratingtoko').distinct()

SQLAlchemy subquery in from clause without join

i need a little help.
I have following query and i'm, curious about how to represent it in terms of sqlalchemy.orm. Currently i'm executing it by session.execute. Its not critical for me, but i'm just curious. The thing that i'm actually don't know is how to put subquery in FROM clause (nested view) without doing any join.
select g_o.group_ from (
select distinct regexp_split_to_table(g.group_name, E',') group_
from (
select array_to_string(groups, ',') group_name
from company
where status='active'
and array_to_string(groups, ',') like :term
limit :limit
) g
) g_o
where g_o.group_ like :term
order by 1
limit :limit
I need this subquery thing because of speed issue - without limit in the most inner query function regexp_split_to_table starts to parse all data and does limit only after that. But my table is huge and i cannot afford that.
If something is not very clear, please, ask, i'll do my best)

I presume this is PostgreSQL.
To create a subquery, use subquery() method. The resulting object can be used as if it were Table object. Here's how your query would look like in SQLAlchemy:
subq1 = session.query(
func.array_to_string(Company.groups, ',').label('group_name')
).filter(
(Company.status == 'active') &
(func.array_to_string(Company.groups, ',').like(term))
).limit(limit).subquery()
subq2 = session.query(
func.regexp_split_to_table(subq1.c.group_name, ',')
.distinct()
.label('group')
).subquery()
q = session.query(subq2.c.group).\
filter(subq2.c.group.like(term)).\
order_by(subq2.c.group).\
limit(limit)
However, you could avoid one subquery by using unnest function instead of converting array to string with arrayt_to_string and then splitting it with regexp_split_to_table:
subq = session.query(
func.unnest(Company.groups).label('group')
).filter(
(Company.status == 'active') &
(func.array_to_string(Company.groups, ',').like(term))
).limit(limit).subquery()
q = session.query(subq.c.group.distinct()).\
filter(subq.c.group.like(term)).\
order_by(subq.c.group).\
limit(limit)