我正在使用sqlaclhemy在我的python项目中查询我的数据库,我是sqlalchemy的新手,但是像目前的概念一样,我正在做一些非常简单的事情,比如,
sel = select([staff.c.name]) \
.select_from(staff) \
.where(staff.c.workbase != "") \
.where((staff.c.status != 'Left') & (staff.c.status != 'Name Changed'))我的项目的一部分需要更复杂的sql查询,但我希望将其保留在sqlalchemy中,我的原始sql如下所示,
SELECT A.a_allowance, B.b_allowance, C.c_allowance, A.name, A.leave_allowance
FROM
(SELECT ROUND(leave_allowance * 0.32, 2) as a_allowance, name, leave_allowance FROM staff_list) A
INNER JOIN
(SELECT ROUND(leave_allowance * 0.40, 2) as b_allowance, name FROM staff_list) B
ON A.name = B.name
INNER JOIN
(SELECT ROUND(leave_allowance * 0.28, 2) as c_allowance, name FROM staff_list) C
ON A.name = C.name
WHERE A.name = 'Jones Jones';我不确定如何在sqlalchemy中执行嵌套的select等操作。
发布于 2016-12-06 06:04:46
您可以使用.join()生成连接
>>> print(foo.join(bar, foo.c.bar_id == bar.c.id))
foo JOIN bar ON foo.bar_id = bar.id您可以用子查询替换表达式中的表名:
>>> left = select([foo.c.bar_id]).select_from(foo).where(foo.c.baz > 0).alias("left")
>>> print(left.join(bar, left.c.bar_id == bar.c.id))
(SELECT foo.bar_id AS bar_id
FROM foo
WHERE foo.baz > :baz_1) AS "left" JOIN bar ON "left".bar_id = bar.id在大多数情况下,子查询的行为与表完全相同。
https://stackoverflow.com/questions/40980335
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