Oracle SQL比较不到两天的日期
嗨，伙计们，我需要你们的支持如何执行这个逻辑。 我现在很结巴，我真的不知道如何进一步处理 目标：比较参考编号、条目日期和状态=1 与Oracle SQL比较不到两天的日期,sql,oracle,Sql,Oracle,嗨，伙计们，我需要你们的支持如何执行这个逻辑。 我现在很结巴，我真的不知道如何进一步处理 目标：比较参考编号、条目日期和状态=1 与 ref\u num，change\u status\u date，status 0（总是最后两行） 比较，如果这些日期change\u status\u date和entry\u date少于2天，则将状态值从status=1更新为status=2，否则如果超过3天，则更改为status=0 知道如何执行正确的select sql和update sql吗 +---
ref\u num，change\u status\u date，status 0（总是最后两行）change\u status\u dateentry\u datestatus=1status=2status=0+--------------+-----------------------+---------------------+----------+
| ref_num | entry_date | change_status_date | status
+--------------+-----------------------+---------------------+----------+
| x326585 | 28/04/2020 16:54:14 | | 1 |
| x326585 | 25/04/2020 13:14:00 | 27/04/2020 23:44:00 | 0 |
| x326585 | 20/04/2020 11:15:02 | 20/04/2020 23:52:01 | 0 |
| A142585 | 28/04/2020 16:55:14 | | 1 |
| A142585 | 26/04/2020 11:54:04 | 27/04/2020 22:54:51 | 0 |
| A142585 | 24/04/2020 10:44:14 | 25/04/2020 13:17:23 | 0 |
| B188532 | 29/04/2020 11:34:41 | | 1 |
| B188532 | 14/04/2020 11:44:24 | 15/05/2020 23:11:10 | 0 |
| B188532 | 11/04/2020 08:34:10 | 13/05/2020 11:44:41 | 0 |
+--------------+-----------------------+---------------------+----------+
END RESULTS:
+--------------+-----------------------+---------------------+----------+
| ref_num | entry_date | change_status_date | status
+--------------+-----------------------+---------------------+----------+
| x326585 | 28/04/2020 16:54:14 | 27/07/2020 23:47:31 | 2 | is less than 3 days (28/04/2020 16:54:14 - 27/04/2020 23:44:00) -> status 2
| x326585 | 25/04/2020 13:14:00 | 27/04/2020 23:44:00 | 0 |
| x326585 | 20/04/2020 11:15:02 | 20/04/2020 23:52:01 | 0 |
| A142585 | 28/04/2020 16:35:58 | 27/07/2020 23:47:31 | 2 | is less than 3 days (28/04/2020 16:35:58 - 27/04/2020 22:54:51) -> status 2
| A142585 | 26/04/2020 11:54:04 | 27/04/2020 22:54:51 | 0 |
| A142585 | 24/04/2020 10:44:14 | 25/04/2020 13:17:23 | 0 |
| B188532 | 29/04/2020 11:34:41 | 27/07/2020 23:47:31 | 0 | is more than 3 days (29/04/2020 11:34:41 - 15/05/2020 23:11:10) -> status 0
| B188532 | 14/04/2020 11:44:24 | 15/05/2020 23:11:10 | 0 |
| B188532 | 11/04/2020 08:34:10 | 13/05/2020 11:44:41 | 0 |
+--------------+-----------------------+---------------------+----------+
感谢您的支持和建议您的解决方案将如下所示
update [tablename] set status=2 where DATEDIFF(day, [tablename].entry_date, [tablename].change_status_date) < 2
update [tablename] set status=0 where DATEDIFF(day, [tablename].entry_date, [tablename].change_status_date) > 3
update[tablename]set status=2，其中DATEDIFF（day，[tablename]。entry\u date，[tablename]。change\u status\u date）<2
更新[tablename]设置状态=0，其中DATEDIFF（day，[tablename]。entry_date，[tablename]。change_status_date）>3
谢谢我就是这样理解这个问题的 样本数据：
SQL> with test (ref_num, entry_date, change_status_date, status) as
2 (select 'x3', to_date('28.04.2020 16:54', 'dd.mm.yyyy hh24:mi'), null , 1 from dual union all
3 select 'x3', to_date('25.04.2020 13:14', 'dd.mm.yyyy hh24:mi'), to_date('27.04.2020 23:44', 'dd.mm.yyyy hh24:mi'), 0 from dual union all
4 select 'x3', to_date('20.04.2020 11:15', 'dd.mm.yyyy hh24:mi'), to_date('20.04.2020 23:52', 'dd.mm.yyyy hh24:mi'), 0 from dual union all
5 --
6 select 'b1', to_date('29.04.2020 11:34', 'dd.mm.yyyy hh24:mi'), null , 1 from dual union all
7 select 'b1', to_date('14.04.2020 11:44', 'dd.mm.yyyy hh24:mi'), to_date('15.05.2020 23:11', 'dd.mm.yyyy hh24:mi'), 0 from dual union all
8 select 'b1', to_date('11.04.2020 08:34', 'dd.mm.yyyy hh24:mi'), to_date('13.05.2020 11:44', 'dd.mm.yyyy hh24:mi'), 0 from dual
9 ),
ref\u numstatus=0的change\u status\u date
；它将与entry\u date
10 temp as
11 (select
12 a.ref_num,
13 a.entry_date,
14 a.change_status_date,
15 --
16 (select max(b.change_status_date)
17 from test b
18 where b.ref_num = a.ref_num
19 and b.status = 0
20 ) compare_change_status_date,
21 a.status
22 from test a
23 )
最后：我假定change\u status\u date
（即NULL
）应替换为sysdate
。这些日期之间的差值应为ABS
，以消除负数
24 select
25 t.ref_num,
26 t.entry_date,
27 --
28 nvl(t.change_status_date, sysdate) change_status_date,
29 --
30 case when t.status = 1 then
31 case when abs(t.entry_date - t.compare_change_status_date) < 2 then 2
32 when abs(t.entry_date - t.compare_change_status_date) > 3 then 0
33 end
34 else t.status
35 end status
36 from temp t
37 order by t.ref_num desc, t.entry_date desc;
RE ENTRY_DATE CHANGE_STATUS_DA STATUS
-- ---------------- ---------------- ----------
x3 28.04.2020 16:54 28.07.2020 08:21 2
x3 25.04.2020 13:14 27.04.2020 23:44 0
x3 20.04.2020 11:15 20.04.2020 23:52 0
b1 29.04.2020 11:34 28.07.2020 08:21 0
b1 14.04.2020 11:44 15.05.2020 23:11 0
b1 11.04.2020 08:34 13.05.2020 11:44 0
6 rows selected.
SQL>
在每一行上设置change\u status\u date
为27/07/2020 23:47:31
的逻辑是什么？我认为是sysdate
，GMB。不太可能。首先，这是Oracle，而您的代码不是。第二，您不应该在行内比较日期。@usama谢谢，但正如littlefoot提到的，这在oracle上不起作用。@littlefoot您有什么建议吗？@James我很抱歉。@UsamaTahir没问题，无论如何谢谢您回答了所有关于直接更新它的最佳建议，可以使用相同的脚本，还是您提出其他建议？谢谢。感谢您的大力支持。是的，它可以重新用于更新或合并。我编辑了我的答案，看一看。thx littlefoot我会检查并尽快返回给你thx。
merge into test a
using (with temp
as (select a.ref_num,
a.entry_date,
a.change_status_date,
--
(select max (b.change_status_date)
from test b
where b.ref_num = a.ref_num
and b.status = 0)
compare_change_status_date,
a.status
from test a)
select t.ref_num,
t.entry_date,
--
nvl (t.change_status_date, sysdate) change_status_date,
--
case
when t.status = 1
then
case
when abs (
t.entry_date - t.compare_change_status_date) <
2
then
2
when abs (
t.entry_date
- t.compare_change_status_date) > 3
then
0
end
else
t.status
end
status
from temp t) x
on ( a.ref_num = x.ref_num
and a.entry_date = x.entry_date)
when matched
then
update set a.status = x.status
where a.status = 1;