如何在此表中找到重复的连续值？

Question

说我有一个表,我查询如下:

select date, value from mytable order by date

这给了我结果:

date                  value
02/26/2009 14:03:39   1                
02/26/2009 14:10:52   2          (a)
02/26/2009 14:27:49   2          (b)
02/26/2009 14:34:33   3
02/26/2009 14:48:29   2          (c)
02/26/2009 14:55:17   3
02/26/2009 14:59:28   4

我对此结果集的行感兴趣,其中值与上一行或下一行中的值相同,如行b,其值= 2与行a相同.我不关心像行c这样的行,它有值= 2但不会直接在值为2的行之后.我怎样才能查询表格,只给出a和b之类的所有行？如果重要的话,这是在Oracle上.

Answer 1

使用超前和滞后分析功能.

create table t3 (d number, v number);
insert into t3(d, v) values(1, 1);
insert into t3(d, v) values(2, 2);
insert into t3(d, v) values(3, 2);
insert into t3(d, v) values(4, 3);
insert into t3(d, v) values(5, 2);
insert into t3(d, v) values(6, 3);
insert into t3(d, v) values(7, 4);

select d, v, case when v in (prev, next) then '*' end match, prev, next from (
  select
    d,
    v,
    lag(v, 1) over (order by d) prev,
    lead(v, 1) over (order by d) next
  from
    t3
)
order by
  d
;

匹配的邻居在匹配列中标有*,

替代文字http://i28.tinypic.com/2drrojt.png