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How to resolve SQLAlchemy Union Throwing Error

I'm using SQL Alchemy(Python, SQLServer) Union on two queries. It throws me the below error. Please help me in resolving it.
query1 = db.query(Employee.LastName).filter(Employee.Age == 30).all()
query2 = db.query(Employee.LastName).filter(Employee.Salary > 25000).all()
query3 = union(query1, query2).all()
**"SELECT construct for inclusion in UNION or other set construct expected, got [('Joseph',),('Alan',),('Joseph',)]."**
Also tried the below query and it throws the below error
query1 = db.query(Employee.LastName).filter(Employee.Age == 30).all()
query2 = db.query(Employee.LastName).filter(Employee.Salary > 25000).all()
query3 = query1.union(query2).all()
**"'list' object has no attribute 'union'"**

Remove the .all() from the first two queries, it turns the queries into lists, but you want to pass Query instances to union.
query1 = db.query(Employee.LastName).filter(Employee.Age == 30) # <- Query
query2 = db.query(Employee.LastName).filter(Employee.Salary > 25000) # <- Query
result = query1.union(query2).all() # <- List

SQLAlchemy subquery access outer tables

I'm having trouble converting this SQL into a valid SQLAlchemy query:
select *
from A
join B on B.Id = (
select top 1 Id
from B
where B.name = A.name
order by B.date
)
I've tried using the subquery but it fails:
query = session.query(A, B)
sub_query = session.query(B)
sub_query = sub_query.filter(B.name == A.name)
sub_query = sub_query.order_by(B.date.desc()).limit(1)
sub_query = sub_query.subquery()
query = query.join(B, B.id == sub_query.c.Id)
By accessing the A in the subquery, SqLAlchemy will add it to the subquery from clause and doesn't use the A from the outer query.
I've seen many SQLAlchemy subquery examples but none of them uses the outer fields.

By using correlate(A) in the subquery we tell the SQLAlchemy that reuses A from the outer query.
For making the join work we should access the Id of the subquery, so we should return only Id and use scalar_subquery() to convert the subquery to a scalar subquery:
query = session.query(A, B)
sub_query = session.query(B.Id)
sub_query = sub_query.filter(B.name == A.name)
sub_query = sub_query.order_by(B.date.desc()).limit(1)
sub_query = sub_query.correlate(A)
query = query.join(B, B.id == sub_query.scalar_subquery())

MySQL - select table name if it contains record for list of tables

I am interested in finding the most efficient manner to query the following:
For a list of table names, return the table name if it contains at least one record that meet the conditions
Essentially, something similar to the following Python code in a single query:
dfs = [pd.read_sql('SELECT name FROM {} WHERE a=1 AND b=2'.format(table), engine) for table in tables]
tables = [table for table, df in zip(tables, dfs) if not df.empty]
Is this possible in MySQL?

Assuming you trust the table names in tables not to contain any surprises leading to SQL injection, you could device something like:
from sqlalchemy import text
selects = [f'SELECT :table_{i} FROM {table} WHERE a = 1 AND b = 2'
for i, table in enumerate(tables)]
stmt = ' UNION '.join(selects)
stmt = text(stmt)
results = engine.execute(
stmt, {f'table_{i}': table for i, table in enumerate(tables)})
or you could use SQLAlchemy constructs to build the same query safely:
from sqlalchemy import table, column, union, and_, select, Integer, literal
tbls = [table(name,
column('a', Integer),
column('b', Integer)) for name in tables]
stmt = union(*[select([literal(name).label('name')]).
select_from(tbl).
where(and_(tbl.c.a == 1, tbl.c.b == 2))
for tbl, name in zip(tbls, tables)])
results = engine.execute(stmt)

You can use a UNION of queries that search each table.
(SELECT 'table1' AS table_name
FROM table1
WHERE a = 1 AND b = 2
LIMIT 1)
UNION
(SELECT 'table2' AS table_name
FROM table2
WHERE a = 1 AND b = 2
LIMIT 1)
UNION
(SELECT 'table3' AS table_name
FROM table3
WHERE a = 1 AND b = 2
LIMIT 1)
...

SQLAlchemy Joining with subquery issue

I am trying to translate SQL into SQLAlchemy. The SQL version of the query I want is as follows:
SELECT * from calendarEventAttendee
JOIN calendarEventAttendanceActual ON calendarEventAttendanceActual.id = calendarEventAttendee.attendanceActualId
LEFT JOIN
(SELECT bill.id, bill.personId, billToEvent.eventId FROM bill JOIN billToEvent ON bill.id = billToEvent.billId) b
ON b.eventId = calendarEventAttendee.eventId AND b.personId = calendarEventAttendee.personId
WHERE b.id is NULL
My SQLAlchemy query is as follows:
query = db.session.query(CalendarEventAttendee).join(CalendarEventAttendanceActual)
sub_query = db.session.query(Bill, BillToEvent).join(BillToEvent, BillToEvent.billId == Bill.id).subquery()
query = query.outerjoin(sub_query, and_(sub_query.Bill.personId == CalendarEventAttendee.personId, Bill.eventId == CalendarEventAttendee.eventId))
results = query.all()
I am getting an error AttributeError: 'Alias' object has no attribute 'Bill'
If I adjust the SQLAlchemy query to the following:
sub_query = db.session.query(Bill, BillToEvent).join(BillToEvent, BillToEvent.billId == Bill.id).subquery()
query = query.outerjoin(sub_query, and_(sub_query.Bill.personId == CalendarEventAttendee.personId, sub_query.BillToEvent.eventId == CalendarEventAttendee.eventId))
results = query.all()
I get an error AttributeError: Bill
Any help would be appreciated, thanks!

Once you call subquery(), there is no access to objects, but only to columns via .c.{column_name} accessor.
Do the following for sub_query instead: load only the columns you need in order to avoid any name collisions:
sub_query = db.session.query(
Bill.id, Bill.personId, BillToEvent.eventId
).join(BillToEvent, BillToEvent.billId == Bill.id).subquery()
Then in your query use column names with .c.column_name:
query = query.outerjoin(
sub_query, and_(
sub_query.c.personId == CalendarEventAttendee.personId,
sub_query.c.eventId == CalendarEventAttendee.eventId)
)
results = query.all()

How can I get the youngest objects from SQLAlchemy?

Each row in my table has a date. The date is not unique. The same date is present more than one time.
I want to get all objects with the youngest date.
My solution work but I am not sure if this is a elegent SQLAlchemy way.
query = _session.query(Table._date) \
.order_by(Table._date.desc()) \
.group_by(Table._date)
# this is the younges date (type is date.datetime)
young = query.first()
query = _session.query(Table).filter(Table._date==young)
result = query.all()
Isn't there a way to put all this in one query object or something like that?

You need a having clause, and you need to import the max function
then your query will be:
from sqlalchemy import func
stmt = _session.query(Table) \
.group_by(Table._date) \
.having(Table._date == func.max(Table._date)
This produces a sql statement like the following.
SELECT my_table.*
FROM my_table
GROUP BY my_table._date
HAVING my_table._date = MAX(my_table._date)
If you construct your sql statement with a select, you can examine the sql produced in your case using. *I'm not sure if this would work with statements query
str(stmt)

Two ways of doing this using a sub-query:
# #note: do not need to alias, but do in order to specify `name`
T1 = aliased(MyTable, name="T1")
# version-1:
subquery = (session.query(func.max(T1._date).label("max_date"))
.as_scalar()
)
# version-2:
subquery = (session.query(T1._date.label("max_date"))
.order_by(T1._date.desc())
.limit(1)
.as_scalar()
)
qry = session.query(MyTable).filter(MyTable._date == subquery)
results = qry.all()
The output should be similar to:
# version-1
SELECT my_table.id AS my_table_id, my_table.name AS my_table_name, my_table._date AS my_table__date
FROM my_table
WHERE my_table._date = (
SELECT max("T1"._date) AS max_date
FROM my_table AS "T1")
# version-2
SELECT my_table.id AS my_table_id, my_table.name AS my_table_name, my_table._date AS my_table__date
FROM my_table
WHERE my_table._date = (
SELECT "T1"._date AS max_date
FROM my_table AS "T1"
ORDER BY "T1"._date DESC LIMIT ? OFFSET ?
)