如何比较Oracle表中的行?

您必须有另一列来建立具有相同排名的行的顺序,否则,“前一行”的概念是没有意义的。假设您这样做:

seq rank continuationofrow
1   1    row 
2   2    row 
3   3    row 
4   4    row 
5   4    row  
6   4    row

不,您可以使用分析函数:

select seq, rank, continuationofrow, lag(rank) over (order by seq) as prev_rank
from mytable;


seq rank continuationofrow prev_rank
1   1    row  
2   2    row               1
3   3    row               2
4   4    row               3
5   4    row               4
6   4    row               4

select
  ...
 lag(rank, 1) over (order by ordering-columns)
from
  ..