SQL：另一个日期/时间24小时内的最小日期/时间值

我正在尝试查找在另一个日期/时间值的24小时内出现的第一个日期/时间值

遭遇表：

ENC_ID    ADMSN_TIME         DISCH_TIME
16755387  03/19/13 11:10:00  03/22/13 12:55:00

问题列表：

ENC_ID      REVIEWED_TIME       USER_ID
16755387    03/19/13 13:02:00   FOOBAR
16755387    03/19/13 13:03:00   FOOBAR
16755387    03/20/13 07:52:00   FOOBAR
16755387    03/22/13 08:35:00   FOOBAR
16755387    03/22/13 08:35:00   FOOBAR
16755387    03/22/13 08:35:00   FOOBAR

期望结果集：

ENC_ID    ADMSN_TIME         ADMSN_REVIEW_TIME  ADMSN_REVIEWER  DISCH_REVIEW_TIME  DISCH_REVIEWER
16755387  03/19/13 11:10:00  03/19/13 13:02:00  FOOBAR          03/22/13 08:35:00  FOOBAR

规则：

---

ABS（revieved_TIME-ADMSN_TIME）我提出了以下内容，它适用于一些测试。使用LEAD分析函数和一个并集

SELECT enc_id,
       admsn_time,
       CASE
          WHEN TO_CHAR (admsn_time, 'DDMMYYYY') =
                  TO_CHAR (admsn_review_time, 'DDMMYYYY')
          THEN
             admsn_review_time
          ELSE
             NULL
       END
          AS admsn_review_time,
       CASE
          WHEN TO_CHAR (admsn_time, 'DDMMYYYY') =
                  TO_CHAR (admsn_review_time, 'DDMMYYYY')
          THEN
             admsn_user
          ELSE
             NULL
       END
          AS admsn_review_user,
       disch_time,
       CASE
          WHEN TO_CHAR (disch_time, 'DDMMYYYY') =
                  TO_CHAR (disch_review_time, 'DDMMYYYY')
          THEN
             disch_review_time
          ELSE
             NULL
       END
          AS disch_review_time,
       CASE
          WHEN TO_CHAR (disch_time, 'DDMMYYYY') =
                  TO_CHAR (disch_review_time, 'DDMMYYYY')
          THEN
             disch_user
          ELSE
             NULL
       END
          AS disch_review_user
  FROM (SELECT enc_id,
               admsn_time,
               LEAD (admsn_time, 1)
                  OVER (PARTITION BY enc_id ORDER BY admsn_time)
                  admsn_review_time,
               LEAD (user_id, 1)
                  OVER (PARTITION BY enc_id ORDER BY admsn_time)
                  admsn_user,
               disch_time,
               LEAD (disch_time, 1)
                  OVER (PARTITION BY enc_id ORDER BY disch_time)
                  disch_review_time,
               LEAD (user_id, 1)
                  OVER (PARTITION BY enc_id ORDER BY disch_time)
                  disch_user,
               ch
          FROM (SELECT e.enc_id,
                       admsn_time,
                       disch_time,
                       'E' AS ch,
                       NULL AS user_id
                  FROM encounter e
                UNION
                SELECT enc_id,
                       reviewed_time,
                       reviewed_time,
                       'P',
                       user_id
                  FROM problem_list pl))
 WHERE ch = 'E';

---

不确定这是否比您的第二个解决方案总体上更有效，但它避免了

不同的

，并将这两个问题加入到

问题列表中

。仍然觉得太复杂了-做一个

unpivot

然后再做一个

pivot

看起来像是一个黑客-但这部分是为了我自己的娱乐：

select enc_id, admsn_time, admsn_review_time, admsn_reviewer,
    disch_time, disch_review_time, disch_reviewer
from (
    select enc_id, admsn_time, disch_time, user_id, reviewed_time, review_type
    from (
        select *
        from (
            select e.enc_id, e.admsn_time, e.disch_time,
                pl.user_id, pl.reviewed_time,
                row_number() over (partition by e.enc_id
                    order by abs(pl.reviewed_time - e.admsn_time))
                        as admsn_reviewed_rn,
                row_number() over (partition by e.enc_id
                    order by abs(pl.reviewed_time - e.disch_time))
                        as disch_reviewed_rn
            from encounter e
            left join problem_list pl on pl.enc_id = e.enc_id
        )
        unpivot (rn for review_type in (admsn_reviewed_rn as 'ADMSN',
            disch_reviewed_rn as 'DISCH'))
    )
    where rn = 1
    and (abs(reviewed_time - admsn_time) <= 1
        or abs(reviewed_time - disch_time) <= 1)
)
pivot (min(reviewed_time) as review_time, min(user_id) as reviewer
for review_type in ('ADMSN' as admsn, 'DISCH' as disch))
/

---

你的基本问题是如何在时间T受到约束后找到第一次。从这里开始

SELECT  E.ENC_ID, 
        E.ADMSN_TIME,
        min(REVIEWED_TIME) as ADMSN_REVIEWED_TIME,
FROM    ENCOUNTER as E
JOIN    PROBLEM_LIST as P 
ON      E.ENC_ID = P.ENC_ID
AND     ADMSN_TIME < REVIEWED_TIME
AND     REVIEWED_TIME <= ADMSN_TIME + 1
GROUP BY E.ENC_ID, E.ADMSN_TIME

选择E.ENC\u ID，
E.ADMSN_时间，
最小值（审核时间）作为ADMSN审核时间，
从邂逅到E
将问题列表作为P连接
在E.ENC_ID=P.ENC_ID上
和ADMSN\u时间<审核时间
当你说你想要最小值时，这表明我应该在你的查询中的某个地方看到一个min（）函数。@DanBracuk，在子选择中有一个
min
<代码>选择分钟（查看时间）.

select enc_id, admsn_time, admsn_review_time, admsn_reviewer,
    disch_time, disch_review_time, disch_reviewer
from (
    select enc_id, admsn_time, disch_time, user_id, reviewed_time, review_type
    from (
        select *
        from (
            select e.enc_id, e.admsn_time, e.disch_time,
                pl.user_id, pl.reviewed_time,
                row_number() over (partition by e.enc_id
                    order by abs(pl.reviewed_time - e.admsn_time))
                        as admsn_reviewed_rn,
                row_number() over (partition by e.enc_id
                    order by abs(pl.reviewed_time - e.disch_time))
                        as disch_reviewed_rn
            from encounter e
            left join problem_list pl on pl.enc_id = e.enc_id
        )
        unpivot (rn for review_type in (admsn_reviewed_rn as 'ADMSN',
            disch_reviewed_rn as 'DISCH'))
    )
    where rn = 1
    and (abs(reviewed_time - admsn_time) <= 1
        or abs(reviewed_time - disch_time) <= 1)
)
pivot (min(reviewed_time) as review_time, min(user_id) as reviewer
for review_type in ('ADMSN' as admsn, 'DISCH' as disch))
/

    ENC_ID ADMSN_TIME          ADMSN_REVIEW_TIME   ADMSN_ DISCH_TIME          DISCH_REVIEW_TIME   DISCH_
---------- ------------------- ------------------- ------ ------------------- ------------------- ------
  16755387 03/19/0013 11:10:00 03/19/0013 13:02:00 FOOBAR 03/22/0013 12:55:00 03/22/0013 08:35:00 FOOBAR

SELECT  E.ENC_ID, 
        E.ADMSN_TIME,
        min(REVIEWED_TIME) as ADMSN_REVIEWED_TIME,
FROM    ENCOUNTER as E
JOIN    PROBLEM_LIST as P 
ON      E.ENC_ID = P.ENC_ID
AND     ADMSN_TIME < REVIEWED_TIME
AND     REVIEWED_TIME <= ADMSN_TIME + 1
GROUP BY E.ENC_ID, E.ADMSN_TIME