总小时数和按小时分组答案

【问题标题】：Sum the total hours and group by hours总小时数和按小时分组
【发布时间】：2016-04-27 15:11:12
【问题描述】：

我想将 0 到 40 之间的所有时间分组为一个总和。 41 - 50 为一个总和，50+ 为另一个总和。

select hours,
       sum(hours)
from   employee
where  hours between 0 and 40
group by hours;

上述查询按小时分组，因此我将结果按小时划分，例如我有 1、2.3、0.5、35.5、30 等。

1       403
2.3     4.6
0.5     53
35.5    284
30      1230

但我想要类似的东西 403+4.6+53+284+1230 = 1974.6 因为他们都在 40 岁以下我该怎么做？

【问题讨论】：

标签： sql oracle

【解决方案1】：

您可以使用条件聚合，按构建小时间隔的值进行分组。通过您的示例，您可以没有整数值，因此您应该使用显式关系运算符来拥有，例如 40-50 组中的 40.1：

select sum(hours),
       case
          when hours <= 40 then '0-40'
          when hours > 40 and hours <= 50 then '41-50'
          when hours > 50 then '50-...'
         end
from employee
group by case
          when hours <= 40 then '0-40'
          when hours > 40 and hours <= 50 then '41-50'
          when hours > 50 then '50-...'
         end

【讨论】：

【解决方案2】：

select sum(case when hours between 0 and 40 then hours else 0 end) hours_1,
       sum(case when hours between 41 and 50 then hours else 0 end) hours_41,
       sum(case when hours > 50 then hours else 0 end) hours_51
from employee

【讨论】：

【解决方案3】：

GROUP-ing 基于CASE

select (case when hours between 0 and 40
              then '0 - 40'
              when hours between 41 and 50
              then '41 - 50'
              else
                   '50+'
          end) as hours_range,
        sum(hours) 
from employee
group by (case when hours between 0 and 40
              then '0 - 40'
              when hours between 41 and 50
              then '41 - 50'
              else
                   '50+'
          end);

【讨论】：

如何将它们作为行而不是列，我是否必须与另一个选择进行联合？

【解决方案4】：

select '1 to 40',sum(hours)

from   employee
where  hours between 0 and 40

union all

  select '41 to 50',sum(hours)

from   employee
where  hours between 41 and 50

union all

  select '50+',sum(hours)

from   employee
where  hours>50

【讨论】：

哪个版本的性能更好，union all 或像其他答案中一样执行case 语句