Question

我编写的sql创建了4列：Tier0，Tier1，Tier2和Weight。

重量与等级相关联

当Tier0 ='X'时，权重= 0，
  当Tier1 ='X'时，重量= 0.7，和
  当Tier2 ='X'时，权重= 1。

现在我的SELECT语句的这部分代码是

CASE 
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) > (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 'X' Else '' END AS Tier0,

CASE 
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) = (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 'X' Else '' END AS Tier1,

CASE 
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) < (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 'X' Else '' END AS Tier2,

CASE 
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) > (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 0
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) = (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 0.7
WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) < (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
THEN 1  
END AS Weight

有没有办法在前三个案例陈述中包含最后的长案例？

如果可能的话，我想为每个层都有一个case语句，它们都将X放在正确的层列中，并且还正确地更新了权重列。

任何帮助都将不胜感激。

Answer 1

您可以使用派生表：

SELECT  Tier0,
        Tier1,
        Tier2,
        CASE 
            WHEN Tier0 = 'X' THEN 0
            WHEN Tier1 = 'X' THEN 0.7
            WHEN Tier2 = 'X' THEN 1
        END [Weight]
FROM (Your current query here) AS T;

或CTE：

;WITH CTE AS
(
    Your current query here
)
SELECT  Tier0,
        Tier1,
        Tier2,
        CASE 
            WHEN Tier0 = 'X' THEN 0
            WHEN Tier1 = 'X' THEN 0.7
            WHEN Tier2 = 'X' THEN 1
        END [Weight]
FROM CTE;

Answer 2

这是怎么回事？

CASE 
  WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) > (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
  THEN 0
  WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) = (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
  THEN 0.7 
  WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) < (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
  THEN 1  
END AS Weight

Answer 3

您可以计算CTE中的Tier0，Tier1，Tier2列，然后使用它来计算Weight列：

with cte as (
  select 
    CASE 
    WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) > (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
    THEN 'X' Else '' END AS Tier0,

    CASE 
    WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) = (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
    THEN 'X' Else '' END AS Tier1,

    CASE 
    WHEN (SUM(vst.len_of_stay) / COUNT(vst.len_of_stay)) < (SUM(drg.drg_std_days_stay) / COUNT(drg.drg_std_days_stay))
    THEN 'X' Else '' END AS Tier2

  FROM ...)
SELECT Tier0, Tier1, Tier2,
       case when Tier0 = 'X' then 0
            when Tier1 = 'X' then 0.7
            when Tier2 = 'X' then 1
        end as weight
  from cte

至少这有助于复制逻辑。

使用单个语句更新多个列

3 个答案: