首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >时,用大小写简化更新查询

时,用大小写简化更新查询
EN

Stack Overflow用户
提问于 2020-02-29 08:21:30
回答 1查看 41关注 0票数 0

我有一个查询,其中有很多重复的条件。我想知道--能不能把这个条件变成变量或者类似的东西,所以不是每次都会重复整个句子吗?

重复的部分:

代码语言:javascript
复制
ISNULL((CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR
     WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END),0)

在查询中,如下所示:

代码语言:javascript
复制
UPDATE @tblALL_HOURS SET REG_WORK_COST = (CASE WHEN MINUTE_FROM >= 0 AND MINUTE_FROM <= @6h AND MINUTE_TO >= @6h AND MINUTE_TO <= @18h
AND IS_ABSENT = 0
THEN (DATEDIFF(MINUTE, DATEADD(MINUTE, @6h, CAST(CAST(DATE_TO AS DATE) AS DATETIME)), date_to)/60.00) * (BRUTO_HOUR + ISNULL((CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END),0))

WHEN MINUTE_FROM >= 0 AND MINUTE_FROM <= @6h AND MINUTE_TO >= @18h AND MINUTE_TO < @24h
AND IS_ABSENT = 0
THEN 12.00 * (BRUTO_HOUR + ISNULL((CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END),0))

WHEN MINUTE_FROM >= @6h AND MINUTE_FROM <= @18h AND MINUTE_TO >= @6h AND MINUTE_TO <= @18h
AND IS_ABSENT = 0
THEN (DATEDIFF(MINUTE, DATE_FROM, DATE_TO) / 60.00) * (BRUTO_HOUR + ISNULL((CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END),0))

WHEN MINUTE_FROM >= @6h AND MINUTE_FROM <= @18h AND MINUTE_TO >= @18h AND MINUTE_TO < @24h
AND IS_ABSENT = 0
THEN (DATEDIFF(MINUTE, DATE_FROM, DATEADD(MINUTE, @18h, CAST(CAST(DATE_FROM AS DATE) AS DATETIME))) / 60.00) * (BRUTO_HOUR + ISNULL((CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END),0))

END)

是否有可能有这样的东西代替:

代码语言:javascript
复制
WHEN MINUTE_FROM >= @6h AND MINUTE_FROM <= @18h AND MINUTE_TO >= @18h AND MINUTE_TO < @24h AND IS_ABSENT = 0 THEN (DATEDIFF(MINUTE, DATE_FROM, DATEADD(MINUTE, @18h, CAST(CAST(DATE_FROM AS DATE) AS DATETIME))) / 60.00) * (BRUTO_HOUR + @myCondition)
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-02-29 08:52:00

使用一个简单的CTE,它返回@tblALL_HOURS的所有列和ISNULL函数返回的值:

代码语言:javascript
复制
ISNULL(CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END,0)

作为一列,然后使用CTE更新,并使用该列代替条件:

代码语言:javascript
复制
WITH cte AS (
  SELECT *, 
    ISNULL(CASE WHEN IS_HOLIDAY = 1 THEN 2.5 * NETO_HOUR WHEN DAY_IN_WEEK = 'Sunday' THEN NETO_HOUR END,0) AS myCondition 
  FROM @tblALL_HOURS 
)
UPDATE cte 
SET REG_WORK_COST = CASE 
    WHEN MINUTE_FROM >= 0 AND MINUTE_FROM <= @6h AND MINUTE_TO >= @6h AND MINUTE_TO <= @18h AND IS_ABSENT = 0
      THEN (DATEDIFF(MINUTE, DATEADD(MINUTE, @6h, CAST(CAST(DATE_TO AS DATE) AS DATETIME)), date_to)/60.00) * (BRUTO_HOUR + myCondition)
    WHEN MINUTE_FROM >= 0 AND MINUTE_FROM <= @6h AND MINUTE_TO >= @18h AND MINUTE_TO < @24h AND IS_ABSENT = 0
      THEN 12.00 * (BRUTO_HOUR + myCondition)
    WHEN MINUTE_FROM >= @6h AND MINUTE_FROM <= @18h AND MINUTE_TO >= @6h AND MINUTE_TO <= @18h AND IS_ABSENT = 0
      THEN (DATEDIFF(MINUTE, DATE_FROM, DATE_TO) / 60.00) * (BRUTO_HOUR + myCondition)
    WHEN MINUTE_FROM >= @6h AND MINUTE_FROM <= @18h AND MINUTE_TO >= @18h AND MINUTE_TO < @24h AND IS_ABSENT = 0
      THEN (DATEDIFF(MINUTE, DATE_FROM, DATEADD(MINUTE, @18h, CAST(CAST(DATE_FROM AS DATE) AS DATETIME))) / 60.00) * (BRUTO_HOUR + myCondition)
END
票数 2
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/60463257

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档