SQL - 根据日期比较行并转置差异答案

【问题标题】：SQL - Compare rows based on date and transpose differenceSQL - 根据日期比较行并转置差异
【发布时间】：2021-05-12 14:38:47
【问题描述】：

场景：

表'HIST'

RID	VALUE	HIST_DATE
1	V111	2019-01-01
1	V112	2020-02-11
1	V112	2020-03-08
1	V113	2020-04-11
1	V114	2021-03-15
2	V211	2020-04-11
2	V211	2021-03-16
3	V311	2019-05-01
3	V312	2020-01-01

预期输出：

RID	VALUE_OLD	VALUE_NEW
1	V113	V114

我想显示按 RID 分组的“VALUE”列中的行之间的差异，其中 HIST_DATE 在 '2020-03-31' 和 '2021-04-01' 之间强>

然后转置“VALUE_OLD”和“VALUE_NEW”列中的不同值。

注意：假设在这些日期之间只能出现 2 个不同的值

Oracle SQL

【问题讨论】：

请解释逻辑。表中的其余值发生了什么变化？
我只想知道两个时间检查点之间的值差异。其余差异不需要

标签： sql oracle

【解决方案1】：

另一种选择是使用 LAG/LEAD 分析功能：

LAG 从表中的前一行返回值。
LEAD 从表的下一行返回值。

select rid
    , LAG(hist_date) OVER(partition by rid order by hist_date ) as HIST_DATE_OLD 
    , hist_date  as HIST_DATE
    , LAG(value) OVER(partition by rid order by hist_date ) as VALUE_OLD 
    , value as VALUE_NEW
from hist

select rid
    , lead(hist_date) OVER(partition by rid order by hist_date desc) as HIST_DATE_OLD 
    , hist_date  as HIST_DATE
    , lead(value) OVER(partition by rid order by hist_date desc) as VALUE_OLD 
    , value as VALUE_NEW
from hist
order by rid, hist_date

然后您可以根据需要进行过滤和格式化

select rid, VALUE_OLD, VALUE_NEW from (
    select rid
       , TO_CHAR(lead(hist_date) OVER(partition by rid order by hist_date desc), 'YYYY-MM-DD') as HIST_DATE_OLD 
       , TO_CHAR(hist_date , 'YYYY-MM-DD') as HIST_DATE
       , lead(value) OVER(partition by rid order by hist_date desc) as VALUE_OLD 
       , value as VALUE_NEW
    from hist
    where HIST_DATE between '2020-03-31' and '2021-04-01'
    and   rid = 1  
order by rid, hist_date
)
where HIST_DATE_OLD IS NOT NULL

RID VALUE_OLD VALUE_NEW 
--- --------- --------- 
  1 V113      V114

select * from (
    select rid
       , TO_CHAR(lead(hist_date) OVER(partition by rid order by hist_date desc), 'YYYY-MM-DD') as HIST_DATE_OLD 
       , TO_CHAR(hist_date , 'YYYY-MM-DD') as HIST_DATE
       , lead(value) OVER(partition by rid order by hist_date desc) as VALUE_OLD 
       , value as VALUE_NEW
    from hist
order by rid, hist_date
)
where HIST_DATE between '2020-03-31' and '2021-04-01'
    and   rid = 1  

RID HIST_DATE_OLD HIST_DATE  VALUE_OLD VALUE_NEW 
--- ------------- ---------- --------- --------- 
  1 2020-03-08    2020-04-11 V112      V113      
  1 2020-04-11    2021-03-15 V113      V114

注意主SELECT内外的过滤器

在外面，您在执行 LAG/LEAD 操作后进行过滤（上图）
在内部，您在执行 LAG/LEAD 操作之前进行过滤（如下）

RID HIST_DATE_OLD HIST_DATE  VALUE_OLD VALUE_NEW 
--- ------------- ---------- --------- --------- 
  1               2020-04-11           V113      
  1 2020-04-11    2021-03-15 V113      V114

【讨论】：

【解决方案2】：

假设在这些日期之间只能出现 2 个不同的值

 select distinct RID, 
        first_value(VALUE) over(order by HIST_DATE) VALUE_OLD,
        first_value(VALUE) over(order by HIST_DATE desc) VALUE_NEW
 from HIST
 where HIST_DATE between  date '2020-03-31' and date '2021-04-01'

【讨论】：

【解决方案3】：

你必须：

查找每个 rid (newvals) 的最后 hist_date
查找最后一个 (newvals) 之前的每个 rid (oldvals) 的 last hist_date
读取历史表以查找每个消除的新值
读取历史表以查找每个消除的先前 val

无论是否有其他较旧的值，您总是显示每个消除的最后两个值。

SELECT
    n.rid,
    o.value AS value_old,
    n.value AS value_new
FROM
    (
        SELECT
            rid,
            MAX(hist_date) AS DATE
        FROM
            hist
        GROUP BY
            rid
    )
    newvals,
    (
        SELECT
            hist.rid,
            MAX(hist_date) AS DATE
        FROM
            hist,
            (
                SELECT
                    rid,
                    MAX(hist_date) AS DATE
                FROM
                    hist
                GROUP BY
                    rid
            )
            newvals
        WHERE
            hist.rid=newvals.rid
        AND hist.hist_date<newvals.date
        GROUP BY
            hist.rid
    )
    oldvals,
    hist n,
    hist o
WHERE
    n.rid=o.rid
and n.rid=newvals.rid
and n.rid=oldvals.rid
AND n.hist_date=newvals.date
AND o.hist_date=oldvals.date

【讨论】：