LAG和LEAD函数ORACLE比较最后两行

Question

我只想选择name1列中更改为name1列的任何值的ID，具体取决于最后2个ID。因此，请参见以下逻辑：

示例1：

ID NAME1 NAME2
102 A    AA
101 A    AA
100 A    AA
102 B    BB
101 B    AA
100 B    CC

结果应为：100, 101 (after comparison of 100 and 101), 102 (after comparison of 101 and 102)

示例2：

ID NAME1 NAME2
102 A    AA
101 A    AA
100 A    AA
102 B    AA
101 B    AA
100 B    CC

结果应为：100, 101 (after comparison of 100 and 101) , 102 is not shown because it hasn't changed when we compare 101 and 102.

仍然应该选择最小的ID。

请参阅下面的SQL Fiddle：

http://sqlfiddle.com/#!4/1735c/1

谢谢！

Answer 1

我不确定逻辑，但据我所知，您可以按以下方式使用lag和lead：

Select * from
(Select t.*, 
       Lag(name2) over (partition by name1 order by id desc) as lg,
       Lead(name2) over (partition by name1 order by id desc) as ld
From your_table t)
Where lg is not null 
  and (lg <> name2
       Or (ld is not null 
           and ld <> name2));

干杯！