问题描述
我正在努力在sqlAlchemy中编写一个WITH AS VALUES
子句。
让我们假设下表
CREATE TABLE Example ("name" varchar(5),"level" varchar(5));
INSERT INTO Example ("name","level") VALUES
('John','one'),('Alice','two'),('Bob','three')
;
WITH matched_levels (level_name,level_score) as (
values ('one',1.0),('two',2.0),('three',3.0)
)
select e.name,m.level_score
from Example e
join matched_levels m on e.level = m.level_name;
-- name level_score
-- John 1
-- Alice 2
-- Bob 3
另请参阅this SQL fiddle。
在我发现其他SO问题([1],[2],[3])之后,我提出了以下建议
matching_levels = sa.select([sa.column('level_name'),sa.column('level_score')]).select_from(
sa.text("values ('one',3.0)")) \
.cte(name='matched_levels')
result = session.query(Example).join(
matching_levels,matching_levels.c.level_name == Example.level
).all()
转换为该无效查询
WITH matched_levels AS
(SELECT level_name,level_score
FROM values ('one',3.0))
SELECT "Example".id AS "Example_id","Example".name AS "Example_name","Example".level AS "Example_level"
FROM "Example" JOIN matched_levels ON matched_levels.level_name = "Example".level
- [1] https://stackoverflow.com/a/60551492/1821901
- [2] Join on a CTE in SQLAlchemy
- [3] https://stackoverflow.com/a/42128983/1821901
解决方法
根据此answer:
您可以尝试以这种方式重写您的 matching_levels
查询:
matching_levels = select(Values(
column('level_name',String),column('level_score',Float),name='temp_table').data([('one',1.0),('two',2.0),('three',3.0)])
).cte('matched_levels')
result = session.query(Example).join(
matching_levels,matching_levels.c.level_name == Example.level
).all()