Sql 一次检查多个大小写表达式结果

我在select语句中的聚合函数中有一个case表达式，看起来像这样

select person_id,
    sum(case status = 'approved' then hours else 0.0 end) as hours
    sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
    sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
from table

---

现在，如何检查所有案例是否都返回0.0？因此，我可以在结果集中排除它？

最简单的方法可能是将查询转换为子查询，然后在外部查询中进行筛选：

select *
from (
    select person_id,
        sum(case status = 'approved' then hours else 0.0 end) as hours
        sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
        sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
    from mytable
    group by person_id
) t
where hours + void_hrs + fwd_hrs > 0

请注意，您的原始查询缺少一个

GROUPBY

子句，我添加了该子句

另一种方法是使用冗长的

having

子句：

select person_id,
    sum(case status = 'approved' then hours else 0.0 end) as hours
    sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
    sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
from mytable
group by person_id
having sum(
    case status 
        when 'approved' then hours
        when 'cancelled' then void_hrs
        when 'forwarded' then fwd_hrs
        else 0.0
    end
) > 0

select person_id,
       sum(case status = 'approved' then hours else 0.0 end) as hours
       sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
       sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
from table
where status in ('approved', 'cancelled', 'forwarded')
group by person_id;

---

旁注：这称为

case

表达式，而不是case语句。后者是一个流控制结构，而前者是条件逻辑。

我只想添加一个

where

子句：

select person_id,
    sum(case status = 'approved' then hours else 0.0 end) as hours
    sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
    sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
from mytable
group by person_id
having sum(
    case status 
        when 'approved' then hours
        when 'cancelled' then void_hrs
        when 'forwarded' then fwd_hrs
        else 0.0
    end
) > 0

select person_id,
       sum(case status = 'approved' then hours else 0.0 end) as hours
       sum(case status = 'cancelled' then void_hrs else 0.0 end) as void_hrs
       sum(case status = 'forwarded' then fwd_hrs else 0.0 end) as fwd_hrs
from table
where status in ('approved', 'cancelled', 'forwarded')
group by person_id;

作为奖励，如果您有许多其他状态的行，这可能会提高性能

或者，您可以将

having

子句添加到查询中：

having sum(case when status in ('approved', 'cancelled', 'forwarded') then 1 else 0 end) > 0

它可能“看起来”像这样，但也必须非常不同，因为您提供的内容有语法错误，甚至无法编译。你可以复制上一个问题中给出的代码。只是不尝试！