Question

我有一张桌子：

表1

tran_id    user_id    start_date    end_date
1          100        01-06-2018    18-06-2018
2          100        14-06-2018    14-06-2018
4          100        19-07-2018    19-07-2018
7          101        05-01-2018    06-01-2018
9          101        08-01-2018    08-01-2018
3          101        03-01-2018    03-01-2018

演示-Link

这是逻辑：

我需要找到两个trans_id之间的日期差，该成员按start_date排序，而start_date和end_date没有重叠。

我们需要检查用户一次处理一条记录的最大终止日期。

逻辑将是：

对于成员：

按user_id和start_date排序所有记录
trans_id = 1，end_date = 18-06-2018 ，设置max_end_date = 18-06-2018
trans_id = 2，end_date = 14-06-2018，end_date <max_end_date，继续前进
trans_id = 3，end_date = 19-07-2018，end_date> max_end_date，在输出中添加一条记录，其中
- transidfrom = 1（因为这是max_end_date的记录）
- transidto = 4（因为这是end_date> max_end_date的记录）
- transidfrom_end_date = 18-06-2018，从end_date的{{1}}中选择trans_id
- transidfrom = 19-07-2018，从transidto_start_date的{{1}}中选择start_date
- trans_id = transidto-datediff

输出如下：

table2

transidto_start_date

在Oracle SQL中有1种查询方法吗？

Answer 1

如果我正确理解了您的要求，那么也许可以做到：

FSITJA@db01 2019-07-08 12:08:59> with table1(tran_id, user_id, start_date, end_date) as (
  2      select 1, 100, date '2018-06-01', date '2018-06-18' from dual union all
  3      select 2, 100, date '2018-06-14', date '2018-06-14' from dual union all
  4      select 4, 100, date '2018-07-19', date '2018-07-19' from dual union all
  5      select 7, 101, date '2018-01-05', date '2018-01-06' from dual union all
  6      select 9, 101, date '2018-01-08', date '2018-01-08' from dual union all
  7      select 3, 101, date '2018-01-03', date '2018-01-03' from dual )
  8  select rownum        as my_id,
  9         tran_id       as transidfrom,
 10         next_tran_id  as transidto,
 11         end_date      as transidfrom_end_date,
 12         next_end_date as transidto_start_date,
 13         datediff
 14    from (select tran_id,
 15                 user_id,
 16                 start_date,
 17                 end_date,
 18                 lead(tran_id) over (partition by user_id order by end_date) next_tran_id,
 19                 lead(start_date) over (partition by user_id order by end_date) next_end_date,
 20                 lead(start_date) over (partition by user_id order by end_date) - end_date datediff
 21            from table1)
 22   where datediff > 0;

     MY_ID TRANSIDFROM  TRANSIDTO TRANSIDFROM_END_DAT TRANSIDTO_START_DAT   DATEDIFF
---------- ----------- ---------- ------------------- ------------------- ----------
         1           1          4 2018-06-18 00:00:00 2018-07-19 00:00:00         31
         2           3          7 2018-01-03 00:00:00 2018-01-05 00:00:00          2
         3           7          9 2018-01-06 00:00:00 2018-01-08 00:00:00          2

3 rows selected.

SQL Fiddle example

根据条件查找Oracle SQL中的日期差异

1 个答案: