在SQLAlchemy中使用HAVING()子句中的标签
我正在尝试在SQLAlchemy中实现一个处理嵌套集合的查询(可以在这里查看)。我遇到的问题是,如何在主SELECT
查询中使用标记的depth
计算(这个计算依赖于子SELECT
查询),并在最后的HAVING
子句中使用它。
SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
nested_category AS parent,
nested_category AS sub_parent,
(
SELECT node.name, (COUNT(parent.name) - 1) AS depth
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.name = 'PORTABLE ELECTRONICS'
GROUP BY node.name
ORDER BY node.lft
)AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth <= 1
ORDER BY node.lft;
我觉得我快成功了,使用了:
node = aliased(Category)
parent = aliased(Category)
sub_parent = aliased(Category)
sub_tree = DBSession.query(node.name,
(func.count(parent.name) - 1).label('depth')).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.name == category_name).\
group_by(node.name).\
order_by(node.lft).subquery()
children = DBSession.query(node.name,
(func.count(parent.name) - (sub_tree.c.depth + 1)).label('depth')).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.lft.between(sub_parent.lft, sub_parent.rgt)).\
filter(sub_parent.name == sub_tree.c.name).\
group_by(node.name).having(depth <= 1).\
order_by(node.lft).all()
但是,我最终得到了一个错误:
NameError: global name 'depth' is not defined
这有点道理。如果我把having(depth <= 1)
替换成having(func.count('depth') <= 1)
,我得到的HAVING
子句是这样的,结果却没有返回(其中的%s占位符是('depth', 1)):
HAVING count(%s) <= %s
我真正需要的格式是这样的:
HAVING depth = 1
有人有什么想法吗?
我最后的选择是直接执行原始查询,而不是通过ORM层,但我真的不想这样,因为我已经很接近了……
提前谢谢大家。
编辑:
我还尝试了以下代码,但它没有返回正确的结果(就好像'depth'标签总是为0):
node = aliased(Category)
parent = aliased(Category)
sub_parent = aliased(Category)
sub_tree_depth = (func.count(parent.name) - 1).label('depth')
depth = (func.count(parent.name) - (sub_tree_depth + 1)).label('depth')
sub_tree = DBSession.query(node.name,
sub_tree_depth).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.name == category_name).\
group_by(node.name).\
order_by(node.lft).subquery()
children = DBSession.query(node.name,
depth).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.lft.between(sub_parent.lft, sub_parent.rgt)).\
filter(sub_parent.name == sub_tree.c.name).\
group_by(node.name).having(depth <= 1).\
order_by(node.lft).all()
从这个生成的HAVING
子句看起来像这样(categories_2在原始查询中是parent):
HAVING count(categories_2.name) - ((count(categories_2.name) - 1) + 1) <= 1
编辑:
我觉得提供生成的SQL可能会有帮助。
SQLAlchemy
node = aliased(Category)
parent = aliased(Category)
sub_parent = aliased(Category)
sub_tree = DBSession.query(node.name,
(func.count(parent.name) - 1).label('depth')).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.name == category_name).\
group_by(node.name).\
order_by(node.lft).subquery()
depth = (func.count(parent.name) - (sub_tree.c.depth + 1)).label('depth')
children = DBSession.query(node.name, depth).\
filter(node.lft.between(parent.lft, parent.rgt)).\
filter(node.lft.between(sub_parent.lft, sub_parent.rgt)).\
filter(sub_parent.name == sub_tree.c.name).\
group_by(node.name).having(depth <= 1).\
order_by(node.lft)
生成的SQL
'SELECT categories_1.name AS categories_1_name, count(categories_2.name) - (anon_1.depth + %s) AS depth
FROM categories AS categories_1, categories AS categories_2, (SELECT categories_1.name AS name, count(categories_2.name) - %s AS depth
FROM categories AS categories_1, categories AS categories_2
WHERE categories_1.lft BETWEEN categories_2.lft AND categories_2.rgt AND categories_1.name = %s GROUP BY categories_1.name ORDER BY categories_1.lft) AS anon_1, categories AS categories_3
WHERE categories_1.lft BETWEEN categories_2.lft AND categories_2.rgt AND categories_1.lft BETWEEN categories_3.lft AND categories_3.rgt AND categories_3.name = anon_1.name GROUP BY categories_1.name
HAVING count(categories_2.name) - (anon_1.depth + %s) <= %s ORDER BY categories_1.lft' (1, 1, u'Institutional', 1, 1)
1 个回答
17
你的SQL查询使用了隐式连接,但在SQLAlchemy中,你需要明确地定义这些连接。除此之外,你的第二次尝试几乎是正确的:
node = aliased(Category)
parent = aliased(Category)
sub_parent = aliased(Category)
sub_tree = DBSession.query(node.name,
(func.count(parent.name) - 1).label('depth')).\
join(parent, node.lft.between(parent.lft, parent.rgt)).\
filter(node.name == category_name).\
group_by(node.name).\
order_by(node.lft).subquery()
depth = (func.count(parent.name) - (sub_tree.c.depth + 1)).label('depth')
children = DBSession.query(node.name, depth).\
join(parent, node.lft.between(parent.lft, parent.rgt)).\
join(sub_parent, node.lft.between(sub_parent.lft, sub_parent.rgt)).\
join(sub_tree, sub_parent.name == sub_tree.c.name).\
group_by(node.name, sub_tree.c.depth).\
having(depth <= 1).\
order_by(node.lft).all()
不要惊讶,如果生成的SQL中的HAVING
子句会重复整个表达式,而不是使用它的别名。这是因为在这里不允许使用别名,这只是MySQL的一个扩展,而SQLAlchemy则努力生成在大多数情况下都能正常工作的SQL。