SQL，在查询期间根据数据条件创建新列

Question

我已经设法拼凑出一个有效的 SQL 查询，它使用了表的联合和连接的组合，为我提供了我需要的中间结果。

SELECT  n.study_id AS StudyId,
        n.practice_id AS PracticeId,
    n.FluVaxCode,
    n.Date,
    date(p.BaseStart / 1000, 'unixepoch') AS BaseStart, 
    date(p.BaseEnd / 1000, 'unixepoch') AS BaseEnd,
    date(p.OutcomeStart / 1000, 'unixepoch') AS OutcomeStart, 
    date(p.OutcomeEnd / 1000, 'unixepoch') AS OutcomeEnd,
    CASE WHEN Date BETWEEN BaseStart AND BaseEnd THEN 'Y' ELSE 'N' END AS BaseVax,
    CASE WHEN Date BETWEEN OutcomeStart AND OutcomeEnd THEN 'Y' ELSE 'N' END AS OutcomeVax
FROM toypractice p INNER JOIN
     (SELECT  t.study_id, t.practice_id,
     date(t.event_date / 1000, 'unixepoch') AS Date, 
     t.code_id AS FluVaxCode
     FROM toytherapy t
     WHERE t.code_id IN ('dher.', 'a6b1.', 'bk31.')
     UNION 
     SELECT  c.study_id, c.practice_id, 
     DATE(c.event_date / 1000, 'unixepoch') AS Date, 
     c.code_id AS FluVaxCode
     FROM toyclinical c
     WHERE c.code_id IN ('1383.', '229..', 'X77RW')
     ORDER BY FluVaxCode DESC
     ) n 
     ON p.practice_id = n.practice_id;

我最终得到多个列，包括三个：

Date, BaseStart, BaseEnd

所有这些都是 YYYY-MM-DD 格式（我认为）。有没有办法，在查询期间，我可以在查询结束时添加一些代码，以便它将在输出中创建一个新列，这样如果 Date 介于 BaseStart 和 BaseEnd 之间，则新列中的值将是' Y'，否则新列中的值将是'N'？在这种情况下，我使用 sqlite 进行开发/测试，但它最终必须在 mssqlserver 2019 中工作。谢谢。 J

Answer 1

尝试将以下内容添加到您当前的选择列表中：

    , CASE WHEN n.Date BETWEEN date(p.BaseStart / 1000, 'unixepoch') AND date(p.BaseEnd / 1000, 'unixepoch') THEN 'a' ELSE 'b' END AS is_between

这可以通过不同的方式完成。但以上应该可以工作。

这是另一个应该起作用的调整。我已将原始 event_date 添加到 n 派生表的结果中，并将相应的 CASE 表达式添加到外部 SELECT 列表中。其他 CASE 表达式可能需要进行类似的调整。

SELECT n.study_id AS StudyId
     , n.practice_id AS PracticeId
     , n.FluVaxCode
     , n.Date
     , date(p.BaseStart / 1000, 'unixepoch') AS BaseStart
     , date(p.BaseEnd / 1000, 'unixepoch') AS BaseEnd
     , date(p.OutcomeStart / 1000, 'unixepoch') AS OutcomeStart
     , date(p.OutcomeEnd / 1000, 'unixepoch') AS OutcomeEnd
     , CASE WHEN Date BETWEEN BaseStart AND BaseEnd THEN 'Y' ELSE 'N' END AS BaseVax
     , CASE WHEN Date BETWEEN OutcomeStart AND OutcomeEnd THEN 'Y' ELSE 'N' END AS OutcomeVax
     , CASE WHEN n.event_date BETWEEN p.BaseStart AND p.BaseEnd THEN 'a' ELSE 'b' END AS is_between
  FROM toypractice p
  JOIN (
         SELECT t.study_id, t.practice_id
              , t.event_date
              , date(t.event_date / 1000, 'unixepoch') AS Date
              , t.code_id AS FluVaxCode
           FROM toytherapy t
          WHERE t.code_id IN ('dher.', 'a6b1.', 'bk31.')
          UNION 
         SELECT c.study_id, c.practice_id
              , c.event_date
              , DATE(c.event_date / 1000, 'unixepoch') AS Date
              , c.code_id AS FluVaxCode
           FROM toyclinical c
          WHERE c.code_id IN ('1383.', '229..', 'X77RW')
          ORDER BY FluVaxCode DESC
       ) n 
    ON p.practice_id = n.practice_id
;

The fiddle

类似这样的：

sqlite 的特定测试用例，作为起点。

• cte - 提供几行代表您当前的查询结果
• cte2 - 展示我们如何将日期 (conv_date) 转换为与基本值兼容
• 最终查询表达式 - 显示如何使用从 cte2 派生的 conv_date 正确计算 BETWEEN。

注意，我尝试直接使用日期，这会产生错误的is_between 结果。 is_between2 显示正确的结果。

WITH cte (date, BaseStart, BaseEnd) AS (
         SELECT 1467241200000, '2016-06-20', '2016-06-30' UNION
         SELECT 1467241200000, '2017-06-20', '2017-06-30'
     )
SELECT date(date / 1000, 'unixepoch'), BaseStart, BaseEnd
     , CASE WHEN Date BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between
     , CASE WHEN date(date / 1000, 'unixepoch') BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between2
  FROM cte
;

and also:

WITH cte (date, BaseStart, BaseEnd) AS (
         SELECT 1467241200000, '2016-06-20', '2016-06-30' UNION
         SELECT 1467241200000, '2017-06-20', '2017-06-30'
     )
   , cte2 AS (
         SELECT *
              , date(date / 1000, 'unixepoch') AS conv_date
           FROM cte
     )
SELECT date(date / 1000, 'unixepoch'), BaseStart, BaseEnd
     , CASE WHEN Date      BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between
     , CASE WHEN conv_date BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between2
  FROM cte2
;

结果：

date(date / 1000, 'unixepoch')	BaseStart	BaseEnd	is_between	is_between2
2016-06-29	2016-06-20	2016-06-30	b	a
2016-06-29	2017-06-20	2017-06-30	b	b

一般答案：

SELECT Date, BaseStart, BaseEnd
     , CASE WHEN Date BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between
  FROM ( your SQL ) AS derived_table
;

or

WITH cte AS (
       your sql
     )
SELECT Date, BaseStart, BaseEnd
     , CASE WHEN Date BETWEEN BaseStart AND BaseEnd THEN 'a' ELSE 'b' END AS is_between
  FROM cte
;

如果您选择的标识符与数据库支持的 SQL 语言中的保留或关键字冲突，则可能需要添加引号。