Oracle SQL - 如何按时间包含相关记录

Question

给出下面列出的数据，如何select只记录，其中：

a) 至少有一张相同 client_id 的先前门票存在并且

b) 每张先行票的最大时间差不得超过14天。换句话说，如果工单有 a) 中描述的后继者，并且此后继者是在 >14 天后创建的，则不得考虑。

create table tickets (
ticket_id number,
client_id number,
start_time date);

insert into tickets values (1,1,to_date('201601011330','yyyymmddhh24mi'));
insert into tickets values (2,1,to_date('201601021320','yyyymmddhh24mi'));
insert into tickets values (3,1,to_date('201601101330','yyyymmddhh24mi'));
insert into tickets values (4,1,to_date('201603101330','yyyymmddhh24mi'));
insert into tickets values (5,2,to_date('201601011630','yyyymmddhh24mi'));
insert into tickets values (6,2,to_date('201601201330','yyyymmddhh24mi'));
insert into tickets values (7,3,to_date('201602011330','yyyymmddhh24mi'));
insert into tickets values (8,4,to_date('201602290000','yyyymmddhh24mi'));
insert into tickets values (9,4,to_date('201603011630','yyyymmddhh24mi'));
insert into tickets values (10,4,to_date('201604011120','yyyymmddhh24mi'));
insert into tickets values(11,4,to_date('201604101030','yyyymmddhh24mi'));
commit;

Answer 1

你可以用解析函数做你想做的事。我认为是这样的逻辑：

select t.*
from (select t.*,
             row_number() over (partition by client_id order by start_time) as seqnum,
             lag(start_time) over (partition by client_id order by start_time) as prev_st
      from tickets t
     ) t
where (start_time - prev_st) < 14 and seqnum >= 2;

我意识到我不知道 (b) 中的 "it" 指的是什么 -- 相关记录的后继者。正如所写，seqnum >= 2 是多余的，因为每个客户端的第一条记录都没有满足第一个条件（prev_st 是 NULL）。

如果这不是您所需要的，那么 row_number()、lag() 和 lead() 的某种组合似乎是正确的。

Answer 2

没有解析函数也可以。

select * from tickets t1
  where exists (
    select 1 from tickets t2
      where t1.client_id = t2.client_id
        and t1.start_time>t2.start_time
        and t1.start_time<=t2.start_time+14
   );

Oracle SQL - 如何按时间包含相关记录

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