找到一条记录，最后一条id_hash相同的记录，并在列中记录两条记录的时间差

Question

我有一个关于 oracle 的查询，我将在其中查询表。我想重新处理这个请求，我有一个列状态。状态中有2个事件：START和Stop，我还有一个hash列，这个任务的key正在执行。这个哈希可以重复

select id, status, date_cr,hesh from SEC_ROUTE

我的答案：

id   status           date_cr                hesh
9    STOP             26.03.2020 14:45:41    7A8E93CA359A39133B6AAC93123    
8    START            26.03.2020 12:23:47    7A8E93CA359A39133B6AAC93127
7    STOP             26.03.2020 11:12:41    7A8E93CA359A39133B6AAC93111  
6    START            26.03.2020 10:56:40    7A8E93CA359A39133B6AAC93111
5    STOP             26.03.2020 10:23:41    7A8E93CA359A39133B6AAC93B09
4    START            26.03.2020 08:13:40    7A8E93CA359A39133B6AAC93B09
3    START            26.03.2020 07:23:41    7A8E93CA359A39133B6AAC93123
2    STOP             26.03.2020 06:18:56    7A8E93CA359A39133B6AAC93123
1    START            26.03.2020 04:12:23    7A8E93CA359A39133B6AAC93123

我要STATUS栏，这是stop的状态，用来记录工作开始和结束的区别 示例（记录差异的格式不是基本的）：

    id  status             date_cr                hesh
    9   STOP(02:21:54)     26.03.2020 14:45:41    7A8E93CA359A39133B6AAC93123 
    8   START              26.03.2020 12:23:47    7A8E93CA359A39133B6AAC93127
    7   STOP(00:16:01)     26.03.2020 11:12:41    7A8E93CA359A39133B6AAC93111
    6   START              26.03.2020 10:56:40    7A8E93CA359A39133B6AAC93111
    5   STOP(02:10:01)     26.03.2020 10:23:41    7A8E93CA359A39133B6AAC93B09
    4   START              26.03.2020 08:13:40    7A8E93CA359A39133B6AAC93B09
    3   START              26.03.2020 07:23:41    7A8E93CA359A39133B6AAC93123
    2   STOP(02:06:33)     26.03.2020 06:18:56    7A8E93CA359A39133B6AAC93123
    1   START              26.03.2020 04:12:23    7A8E93CA359A39133B6AAC93123

0

我在可编程数据库方面还没有太多经验，所以我什至不知道是否可以做到。如果有人愿意提供帮助，我将不胜感激）

Answer 1

假设您不能连续有两个 STOP 并且您想要最后一个 START 值，请使用 lag 查找最后一个开始时间。并从当前时间减去：

create table t (
  id int, status varchar2(5), date_cr timestamp, hash varchar2(32)
);
insert into t values (1, 'STOP', to_date ( '26.03.2020 14:45:41', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93123' );    
insert into t values (2, 'START', to_date ( '26.03.2020 12:23:47', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93127' );
insert into t values (3, 'STOP', to_date ( '26.03.2020 11:12:41', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93111' );  
insert into t values (4, 'START', to_date ( '26.03.2020 10:56:40', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93111' );
insert into t values (5, 'STOP', to_date ( '26.03.2020 10:23:41', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93B09' );
insert into t values (6, 'START', to_date ( '26.03.2020 08:13:40', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93B09' );
insert into t values (7, 'START', to_date ( '26.03.2020 07:23:41', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93123' );
insert into t values (8, 'STOP', to_date ( '26.03.2020 06:18:56', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93123' );
insert into t values (9, 'START', to_date ( '26.03.2020 04:12:23', 'dd.mm.yyyy hh24:mi:ss' ), '7A8E93CA359A39133B6AAC93123' );
commit;

select id,
       status,
       date_cr,
       case 
         when status = 'STOP' then
           date_cr - lag ( date_cr ) over (
             order by date_cr 
           )
       end tm
from   t
order  by date_cr desc;

ID    STATUS    DATE_CR                           TM                    
    1 STOP      26-MAR-2020 14.45.41.000000000    +00 02:21:54.000000    
    2 START     26-MAR-2020 12.23.47.000000000    <null>                 
    3 STOP      26-MAR-2020 11.12.41.000000000    +00 00:16:01.000000    
    4 START     26-MAR-2020 10.56.40.000000000    <null>                 
    5 STOP      26-MAR-2020 10.23.41.000000000    +00 02:10:01.000000    
    6 START     26-MAR-2020 08.13.40.000000000    <null>                 
    7 START     26-MAR-2020 07.23.41.000000000    <null>                 
    8 STOP      26-MAR-2020 06.18.56.000000000    +00 02:06:33.000000    
    9 START     26-MAR-2020 04.12.23.000000000    <null>