Question

我试图在SQLAlchemy中实现以下查询来处理嵌套集（请参阅here）。我正在努力解决的问题是如何在depth子句中的主SELECT查询（取决于子SELECT查询）中使用带标签的HAVING计算最后。

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;

我觉得我在使用时非常接近：

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，我最终得到了错误：

NameError: global name 'depth' is not defined

哪种有道理。如果我用having(depth <= 1)替换having(func.count('depth') <= 1，我最终会生成以下HAVING子句，该子句不返回任何结果（％s占位符在哪里（＆＃39;深度＆＃39;，1））：

HAVING count(%s) <= %s

我真正需要阅读的内容完全如下：

HAVING depth = 1

有人有什么想法吗？

我最后的办法是实际执行原始查询而不是通过ORM层，但我真的不愿意，因为我离这么近......

提前致谢。

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＆＃39; depth＆＃39;标签始终为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()

从此生成的HAVING子句看起来像（来自原始查询中的categories_2 == parent）：

HAVING count(categories_2.name) - ((count(categories_2.name) - 1) + 1) <= 1

编辑：

我认为包含生成的SQL可能会有所帮助。

SQLAlchemy的

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)

Answer 1

您的SQL查询使用隐式连接，在SQLAlchemy中您需要显式定义它们。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()

如果生成的SQL中的HAVING子句将重复完整表达式而不是其别名，请不要感到惊讶。那是因为那里不允许别名，它只是MySQL扩展，而SQLAlchemy努力生成在大多数情况下都有效的SQL。

在SQLAlchemy中使用HAVING（）子句中的标签

1 个答案: