计算 Oracle 中每个 ID 列的两个最近日期之间的天数答案

【问题标题】：Calculate no of days between two latest dates for each ID column in Oracle计算 Oracle 中每个 ID 列的两个最近日期之间的天数
【发布时间】：2019-12-27 22:25:57
【问题描述】：

我有一张这样的桌子：

ID Submitted Date
00001 20-JUL-2018
0000117-MAR-2017
00001@ 987654328@
0000220-JUL-2018
0000222-AUG-2018 kbd>
0000223-APR-2019

我想计算每个 ID 的最近两个提交日期之间的天数。

与 ID = 00001 一样，天数应为 20-JUN-2019 - 20-JUL-2018。

我正在使用 Oracle 11G。

【问题讨论】：

请向我们展示您的尝试并指出您遇到的问题。
Oracle 版本？ ...

标签： sql oracle datediff

【解决方案1】：

略有不同；您可以找到每一行的上一个日期，还可以根据日期顺序分配排名：

select id, submitted_date,  
  lag(submitted_date) over (partition by id order by submitted_date) as previous_date,
  dense_rank() over (partition by id order by submitted_date desc) as rnk
from your_table;
ID    SUBMITTED_DATE PREVIOUS_DATE        RNK
----- -------------- ------------- ----------
00001 2017-03-17                            3
00001 2018-07-20     2017-03-17             2
00001 2019-06-20     2018-07-20             1
00002 2018-07-20                            3
00002 2018-08-22     2018-07-20             2
00002 2019-04-23     2018-08-22             1

然后将其用作内联视图以仅获取最新日期 - 排名第一：

select id, submitted_date, previous_date,
  submitted_date - previous_date as diff
from
(
  select id, submitted_date,  
    lag(submitted_date) over (partition by id order by submitted_date) as previous_date,
    dense_rank() over (partition by id order by submitted_date desc) as rnk
  from your_table
)
where rnk = 1;

ID    SUBMITTED_DATE PREVIOUS_DATE       DIFF
----- -------------- ------------- ----------
00001 2019-06-20     2018-07-20           335
00002 2019-04-23     2018-08-22           244

【讨论】：

【解决方案2】：

试试这个：

SELECT ID, 
MAX(SUBMITTED_DATE - PREV_SUBMITTED_DATE) AS DIFF
FROM
(SELECT ID, SUBMITTED_DATE,  
LAG(SUBMITTED_DATE) OVER (PARTITION BY ID ORDER BY SUBMITTED_DATE) AS PREV_SUBMITTED_DATE
FROM
(SELECT ID, SUBMITTED_DATE,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY SUBMITTED_DATE DESC) AS RN
FROM TABLE)
WHERE RN <= 2)
GROUP BY ID;

或者

SELECT ID, MAX(DIFF) AS DIFF FROM
(SELECT ID,
NTH_VALUE(SUBMITTED_DATE,1) OVER (PARTITION BY ID ORDER BY SUBMITTED_DATE DESC) -  
NTH_VALUE(SUBMITTED_DATE,2) OVER (PARTITION BY ID ORDER BY SUBMITTED_DATE DESC) AS DIFF
FROM TABLE)
GROUP BY ID;

干杯！！

【讨论】：

【解决方案3】：

一种方法使用lag()和lead()：

select t.*,
       (submitted_date - prev_sd) as day_diff
from (select t.*,
             lag(submitted_date) over (partition by id order by submitted_date) as prev_sd,
             lead(submitted_date) over (partition by id order by submitted_date) as next_sd
      from t
     ) t
where next_sd is null;

这种方法避免了任何聚合。

【讨论】：