Question

说我们有一个mysql数据透视查询（取自pivto表问题的答案 MySQL pivot table）与内部联接。

select projects.org,  SUM(ivt_incidents.severity = 'SEV3') as 
sev3_count, 
SUM(ivt_incidents.severity = 'SEV2') as sev2_count, 
SUM(ivt_incidents.severity = 'SEV1') as sev1_count, 
SUM(ivt_incidents.severity = 'SEV0') as sev0_count, 
SUM(ivt_incidents.severity like '%SEV%') as total_count  from 
ivt_incidents inner join projects on ivt_incidents.jira_key = 
projects.jira_key group by projects.org order by total_count;

在构建sqlalchemy查询时寻求帮助：

response = db.session.query(
        func.sum(Incidents.severity).filter(Incidents.severity == 'SEV3'),
        Projects.org).\
    join(Projects, Projects.jira_key == Incidents.jira_key).\
    group_by(Projects.org)

如果我运行它，它将接受查询，但是如果我调用

response.all()

ProgrammingError：（mysql.connector.errors.ProgrammingError）1064（42000）：SQL语法有错误；请在第1行[SQL：u'SELECT sum（ivt_incidents。）中检查与您的MySQL服务器版本相对应的手册，以在'（WHERE ivt_incidents.severity ='SEV3'）AS anon_1，projects.org AS projects_org'附近使用正确的语法。严重性）（FILTER（WHERE ivt_incidents.severity =％（severity_1）s）AS匿名_1，projects.org AS projects_org \ nFROM ivt_incidents INNER JOIN项目位于projects.jira_key = ivt_incidents.jira_key GROUP BY projects.org'] [参数：{u'严重性_1'：'SEV3'}]（此错误的背景位于：http://sqlalche.me/e/f405）

Answer 1

MySQL不支持您尝试在func.sum(...).filter(...)中使用的SQL标准aggregate FILTER clause中的^✝️，该标准已编译为

SUM(...) FILTER (WHERE ...)

查看原始查询，您应该将谓词表达式作为SUM的参数传递：

response = db.session.query(
        func.sum(Incidents.severity == 'SEV3'),
        Projects.org).\
    join(Projects, Projects.jira_key == Incidents.jira_key).\
    group_by(Projects.org)

仅在调用Query.all()之后才得到错误的原因是，查询实际上是在数据库中发出的。在此之前，您已经构建了Query对象。

^✝️：有一个plugin通过重写查询为MySQL添加了FILTER支持。

具有多个筛选聚合的SQLAlchemy查询

1 个答案: