使用投射功能

Question

这就是我的桌子的样子；

              select * into temp3
      from (values('306B','123456','24-25','Erin, Fromans','1','City','2024-06-03 00:00:00.000', NULL),
                  ('306B','123456','24-25','Erin, Fromans','2','City',NULL, '2024-07-12 00:00:00.000'),
          ('306B','123456','24-25','Erin, Fromans','4','City','2024-08-26 00:00:00.000', NULL),
              ('306B','123456','24-25','Erin, Fromans','5','City',NULL, NULL),
          ('306B','123456','24-25','Erin, Fromans','6','City',NULL, '2024-11-15 00:00:00.000'),
      ('0208A','321654','24-25','Guerrazzi, Kiana','4','Ocala','2024-08-26 00:00:00.000', NULL),
      ('0208A','321654','24-25','Guerrazzi, Kiana','5','Ocala',NULL, NULL),
      ('0208A','321654','24-25','Guerrazzi, Kiana','6','Ocala',NULL, '2024-11-15 00:00:00.000'),
      ('108B','4399899','24-25','Obeid, Tarek','2','City','2024-10-07 00:00:00.000',NULL),
      ('108B','4399899','24-25','Obeid, Tarek','3','City',NULL, '2024-11-15 00:00:00.000'),
       ('416A','4580487','24-25','Castellano, Isabella','1','City','2024-06-03 00:00:00.000', NULL),
           ('416A','4580487','24-25','Castellano, Isabella','2','City',NULL, '2024-07-12 00:00:00.000'),
           ('416A','4580487','24-25','Castellano, Isabella','4','City','2024-08-26 00:00:00.000', NULL),
       ('416A','4580487','24-25','Castellano, Isabella','5','City',NULL, NULL),
           ('416A','4580487','24-25','Castellano, Isabella','6','City',NULL, '2024-11-15 00:00:00.000'))
                     as c(apt,emplid,ay,name,block,site,indate,outdate)

我的代码如下；

       select t.apt, t.emplid,t.ay,t.name, t.block,t.site,
           case when t.indate is not null then t.indate else (
         select min(t1.indate) 
           from temp3 t1 where t1.apt = t.apt and t1.name=t.name and t1.ay=t.ay and t1.site=t.site   
               ) end indate,
             case when t.outdate is not null then t.outdate else (
              select max(t1.outdate) from temp3 t1 
                  where t1.apt = t.apt and t1.name=t.name and t1.ay=t.ay and t1.site=t.site
                 ) end outdate
                            from temp3 t
                    where block='5' and t.ay='24-25'

我的代码没有显示出需要。我的代码中错误的部分是“indate”和“outdate”值。它不应该是最小值或最大值。

需要的是，如果您拉动 block=5 并且“indate”为 null，那么它应该检查第一个 block=4，如果为 null，则检查 block=3，依此类推。

“过时”也是如此，如果“过时”对于 block=5 为 null，则检查 block=6，如果为 null，则检查 block=7，依此类推..

所以数据应该看起来像；

('306B','123456','24-25','Erin, Fromans','5','City','2024-08-26 00:00:00.000', '2024-11-15 00:00:00.000'),
('0208A','321654','24-25','Guerrazzi, Kiana','5','Ocala','2024-08-26 00:00:00.000', '2024-11-15 00:00:00.000'),
('416A','4580487','24-25','Castellano, Isabella','5','City','2024-08-26 00:00:00.000', '2024-11-15 00:00:00.000'),

Answer 1

• 子查询选择最近的较低/较高

  block

  ，其中

  indate

  /

  outdate

  不是

  NULL

  。

• ORDER BY

  t1.block

  DESC

  因此首先选择

  lower

  /

  higher

  块。

• LIMIT 1

  或

  TOP 1

  (SQL Server) 仅顶部结果。

SELECT t.apt, t.emplid, t.ay, t.name, t.block, t.site,
  COALESCE(
    t.indate,
    (  SELECT t1.indate FROM temp3 t1 WHERE t1.apt = t.apt AND t1.name = t.name AND t1.ay = t.ay AND t1.site = t.site AND t1.block < t.block AND t1.indate IS NOT NULL ORDER BY t1.block DESC LIMIT 1 )
  ) AS indate,
  COALESCE(
    t.outdate,
    (  SELECT t2.outdate FROM temp3 t2 WHERE t2.apt = t.apt AND t2.name = t.name AND t2.ay = t.ay AND t2.site = t.site AND t2.block > t.block AND t2.outdate IS NOT NULL ORDER BY t2.block ASC LIMIT 1 )
  ) AS outdate
FROM temp3 t
WHERE t.block = 5 AND t.ay = '24-25'

将给出以下输出：

适合	明确	好	名字	块	网站	注明	过时了
306B	123456	24-25	艾琳，弗罗曼斯	5	城市	2024-08-26 00:00:000	2024-11-15 00:00:000
0208A	321654	24-25	格拉兹、琪亚娜	5	奥卡拉	2024-08-26 00:00:000	2024-11-15 00:00:000
416A	4580487	24-25	伊莎贝拉卡斯特拉诺	5	城市	2024-08-26 00:00:000	2024-11-15 00:00:000