![](/img/trans.png)
[英]SQL: Must appear in the GROUP BY clause or be used in an aggregate function
[英]Must appear in the GROUP BY clause or be used in an aggregate function (PostgreSQL)
我有这样的查询:
select u.fullname ,s.schedate,
sum(case when s.isvisiting =1 then s.isvisiting else 0 end) as visit ,
count(s.custid) as cust,
sum(case when s.isclosing = 1 then s.isclosing else 0 end)as orders
from vmstrschedule s JOIN vmsmsuser u ON u.usercode = s.salesmanid
where u.branchid = 'BLL' group by u.fullname
但它显示如下错误:
必须出现在GROUP BY子句中或在聚合函数中使用
它在MySQL中工作,但是当我在PostgreSQL中尝试时不起作用。 我想每月显示这样的数据:
但是如果我使用此查询:
select u.fullname,s.schedate,
sum(case when s.isvisiting =1 then s.isvisiting else 0 end) as visit ,
count(s.custid) as cust,
sum(case when s.isclosing = 1 then s.isclosing else 0 end)as orders
from vmstrschedule s JOIN vmsmsuser u ON u.usercode = s.salesmanid
where u.branchid = 'BLL' group by u.fullname,s.schedate
order by u.fullname
它将显示如下数据:
如果按“工作”的意思是它为每个组任意选择了一个可能的schedate
值,那是正确的。 但是,postgresql和大多数其他SQL产品希望您明确选择应选择的值
如果每个组中的所有值都相同,则将其放在GROUP BY
子句或任意聚合中(例如MAX(s.schedate)
)应该起作用。 如果值不同,请告诉服务器应该选择哪一个MIN
或MAX
通常比较合适。
select u.fullname ,s.schedate,
sum(case when s.isvisiting =1 then s.isvisiting else 0 end) as visit ,
count(s.custid) as cust,
sum(case when s.isclosing = 1 then s.isclosing else 0 end)as orders
from vmstrschedule s JOIN vmsmsuser u ON u.usercode = s.salesmanid
where u.branchid = 'BLL' group by u.fullname, s.schedate
应该管用
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.