处理 CASE 语句中的重叠条件

Question

SELECT (CASE WHEN ymd BETWEEN CURRENT_DATE -4 AND CURRENT_DATE -1 THEN '3day total' 
WHEN ymd BETWEEN CURRENT_DATE -11 AND CURRENT_DATE -1 THEN '10day total' 
WHEN ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1 THEN '30day total' END) AS 'Period',
SUM(cost) cost
FROM table
WHERE ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1
GROUP BY 1

结果实际上似乎给了我第 1-3 天、第 4-10 天和第 11-30 天的时段。我相信这是因为存在重叠的条件，一旦满足第一个条件，SQL 就会停止处理 CASE 语句。

我想要的是每个桶的总数（即 3 天总和、10 天总和和 30 天总和）。

有没有办法在不添加额外字段的情况下做到这一点？

PS - 语法与传统 sql 有点不同，因为它是 vsql (vertica)。

Answer 1

制作三个总计而不是一个总计，并使用条件来确定对每条记录进行计数的位置，而不是尝试确定单个总计的期间：

select
  sum(case when ymd between CURRENT_DATE - 4 and CURRENT_DATE - 1 then cost else 0 end) as '3day total',
  sum(case when ymd BETWEEN CURRENT_DATE - 11 and CURRENT_DATE - 5 then cost else 0 end) as '10day total',
  sum(case when ymd BETWEEN CURRENT_DATE - 31 and CURRENT_DATE - 12 then cost else 0 end) as '30day total'
from
  table
and
  ymd between CURRENT_DATE -31 and CURRENT_DATE -1
group by
  1

注意：我不确定您想在哪里计算哪些日期范围，所以我制作了它们以便它们不会重叠，因为如果您使用 between，这最有意义。如果您仍然希望它们重叠，那么您可以进行比较，而不是使用 between，因为任何晚于 CURRENT_DATE - 1 的值都已被过滤掉。

编辑：

要获得行中的结果，您可以在选择之间进行联合：

select '3day total' as period, sum(cost) as cost
from table
where ymd between CURRENT_DATE - 4 and CURRENT_DATE - 1
union all
select '10day total', sum(cost)
from table
where ymd BETWEEN CURRENT_DATE - 11 and CURRENT_DATE - 5
union all
select '30day total', sum(cost)
from table
where ymd BETWEEN CURRENT_DATE - 31 and CURRENT_DATE - 12

Answer 2

如果你想保持三行一列，这里有一种方法：

SELECT which,
       sum(CASE WHEN ymd BETWEEN CURRENT_DATE - diff AND CURRENT_DATE -1 THEN cost
                else 0
           end) as cost
FROM table t cross join
     (select '3day total' as which, 4 as diff union all
      select '10day total', 11 union all
      select '30day total', 31
     ) w
WHERE ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1
GROUP BY which;

编辑：

cross join 的工作原理是为原始表中的每一行创建三行。每个组都考虑这些行中的每一行——因此重叠不是问题。这些组由它们的名称和diff 值定义。

换句话说，如果数据以 7 天前的一行开始：

ymd                  cost
CURRENT_DATE - 7      $10

然后将 cross join 乘以 3：

which         diff       ymd                 cost
3day total      4        CURRENT_DATE - 7     $10
10day total    11        CURRENT_DATE - 7     $10
30day total    31        CURRENT_DATE - 7     $10

然后，当这些由which 聚合时，会计算出正确的值：

which         Total
3day total      $0
10day total    $10
30day total    $10

Answer 3

尝试为每个存储桶设置不同的字段。我假设您想对每个存储桶的成本求和，如果您想对不同的字段求和，您可能需要更改 THEN cost。

您的查询中还缺少WHERE 子句。条件以AND 语句开头，您可能需要更正此问题。

SELECT SUM(CASE WHEN ymd BETWEEN CURRENT_DATE -4 AND CURRENT_DATE -1 THEN cost ELSE 0 END) as '3day total',
SUM(CASE WHEN ymd BETWEEN CURRENT_DATE -11 AND CURRENT_DATE -1 THEN cost ELSE 0 END) as '10day total', 
SUM(CASE WHEN ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1 THEN cost ELSE 0 END) as  '30day total' END) AS 'Period'
FROM table
WHERE ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1
GROUP BY 1

Answer 4

尝试扭转你的条件：

SELECT (CASE WHEN ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1 THEN '30day total' 
WHEN ymd BETWEEN CURRENT_DATE -11 AND CURRENT_DATE -1 THEN '10day total' 
WHEN ymd BETWEEN CURRENT_DATE -4 AND CURRENT_DATE -1 THEN '3day total' 
END) AS 'Period',
SUM(cost) cost
FROM table
WHERE ymd BETWEEN CURRENT_DATE -31 AND CURRENT_DATE -1
GROUP BY 1