一次检查多个 case 表达式结果答案

【问题标题】：Check multiple case expressions result at once一次检查多个 case 表达式结果
【发布时间】：2020-08-05 06:11:24
【问题描述】：

我在 select 语句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，我如何检查是否所有案例都返回 0.0？这样我就可以在结果集中排除它？

【问题讨论】：

它可能“看起来”像这样，但它也必须非常不同，因为您提供的内容有语法错误，甚至无法编译。而且您可能只是复制了上一个问题中提供给您的代码。只是没有尝试！

标签： sql sql-server database group-by sql-server-2012

【解决方案1】：

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 table
where status in ('approved', 'cancelled', 'forwarded')
group by person_id;

作为奖励，如果您有很多具有其他状态的行，这可能会提高性能。

或者，您可以在查询中添加having 子句：

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

【讨论】：

【解决方案2】：

最简单的方法可能是将您的查询变成子查询，然后在外部查询中进行过滤：

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 group by 子句，我添加了这一点。

替代方法是使用冗长的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

旁注：这称为case 表达式，而不是case 语句。后者是流控结构，前者是条件逻辑。

【讨论】：