【问题标题】:How to get around PARTITION BY syntax error in MySql如何解决 MySql 中的 PARTITION BY 语法错误
【发布时间】:2026-02-20 03:25:01
【问题描述】:

MySql 引发ProgrammingError 运行此sql

SELECT parents.uuid AS parents_uuid, children.uuid AS children_uuid,
children.parent_uuid AS children_parent_uuid, 
count(*) OVER (PARTITION BY parents.uuid) AS children_count
FROM parents, children
WHERE children.parent_uuid = parents.uuid ORDER BY children_count DESC

错误:

sqlalchemy.exc.ProgrammingError: (_mysql_exceptions.ProgrammingError) (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(PARTITION BY parents.uuid)' at line 1") [SQL: 'SELECT parents.uuid AS parents_uuid, children.uuid AS children_uuid, children.parent_uuid AS children_parent_uuid, count(*) OVER (PARTITION BY parents.uuid)']

对于Parent 可以有多个children 的架构,而Child 有单个parent

Base = declarative_base()
class Parent(Base):
    __tablename__ = 'parents'
    uuid = Column(String(64), primary_key=True, unique=True)
    def __init__(self):  
        self.uuid = uuid.uuid4()   

class Child(Base):
    __tablename__ = 'children'
    uuid = Column(String(64), primary_key=True, unique=True)
    parent_uuid = Column(String(64), ForeignKey('parents.uuid'))
    def __init__(self, parent_uuid=None):  
        self.uuid = uuid.uuid4()   
        self.parent_uuid = parent_uuid

【问题讨论】:

  • MySQL 不支持窗口函数。

标签: python mysql sql sqlalchemy


【解决方案1】:

MySQL 不支持窗口函数。要解决此问题,请使用相关子查询来获取计数。

SELECT 
p.uuid AS parents_uuid, 
c.uuid AS children_uuid,
c.parent_uuid AS children_parent_uuid, 
(SELECT COUNT(*) FROM children WHERE parent_uuid = p.uuid) AS children_count
FROM parents p
JOIN children c ON c.parent_uuid = p.uuid 
ORDER BY children_count DESC

或者使用派生表来获取计数并将join它提供给现有查询。

SELECT 
p.uuid AS parents_uuid, 
c.uuid AS children_uuid,
c.parent_uuid AS children_parent_uuid, 
counts.children_count
FROM parents p
JOIN (SELECT parent_uuid,COUNT(*) as children_count
      FROM children 
      GROUP BY parent_uuid) counts ON counts.parent_uuid = p.uuid 
JOIN children c ON c.parent_uuid = p.uuid 
ORDER BY counts.children_count DESC

【讨论】:

    【解决方案2】:

    【讨论】:

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