Oracle 获取相邻两条记录的值 lead over 和 lag over（案例：计算相邻两条记录的差值）

有如下原始表数据，现需要根据Id列，将相邻两条记录差值小于等于4的记录查询出来。

数据库：Oracle

使用Oracle的分析函数 lead over 和 lag over，最终 SQL 如下：

select t2.*
  from (select t1.id,
               t1.firstname,
               t1.lastname,
               (case t1.prev_val
                 when null then
                  null
                 else
                  t1.id - t1.prev_val
               end) as prev_diff,
               (case t1.next_val
                 when null then
                  null
                 else
                  t1.next_val - t1.id
               end) as next_diff
          from (select t.id,
                       t.firstname,
                       t.lastname,
                       lag(t.id) over(order by t.id asc) as prev_val,
                       lead(t.id) over(order by t.id asc) as next_val
                  from SHANHY_REACTOR_EXAMPLE_CUSTOMER t) t1) t2
                  where t2.prev_diff <= 4 or t2.next_diff <= 4;

SQL 执行结果如下：

（END）