Question

我想创建一个表来查找缺少的序列号。到达70000后，序号在0到70000之间变为0.在特定的时间段内，我需要找到那些丢失的记录。

Answer 1

此解决方案基于语句，该语句生成从您设置的1到某个限制的所有自然数：

SELECT ROWNUM N FROM dual CONNECT BY LEVEL <= 7000

此解决方案的第二部分是Oracle MINUS运算符（通常称为EXCEPT），用于减去集合。

换句话说，最终查询是：

SELECT ROWNUM id FROM dual CONNECT BY LEVEL <= 7000
MINUS
SELECT id FROM mytable

SQLFiddle demo for 20 numbers

Answer 2

您可以使用Lead and lag functions来检测序列中的间隙解决方案不会限制您使用特定的上限数字，如70000.

的检测：

SELECT * FROM (SELECT lag(c.id) over(ORDER BY id) last_id, c.id curr_id, lead(c.id) over(ORDER BY id) next_id FROM mytable c order by id) WHERE nvl(last_id, curr_id) + 1 <> curr_id AND last_id IS NOT NULL

Sqlfiddle demo。

的遍历：

begin FOR x IN (SELECT * FROM (SELECT lag(c.id) over(ORDER BY id) last_id, c.id curr_id, lead(c.id) over(ORDER BY id) next_id FROM mytable c order by id) WHERE nvl(last_id, curr_id) + 1 <> curr_id AND last_id IS NOT NULL ) LOOP dbms_output.put_line('last_id :' || x.last_id); dbms_output.put_line('curr_id :' || x.curr_id); dbms_output.put_line('next_id :' || x.next_id); dbms_output.put('gaps found: '); for j in x.last_id + 1 .. nvl(x.next_id,x.curr_id) - 1 loop if j != x.curr_id then dbms_output.put(j || ', '); end if; end loop; dbms_output.put_line(''); dbms_output.put_line('*****'); end loop; end;

Answer 3

我刚刚从Tom Kyte那里偷了这个：

select  id, one_before, Diff, dense_rank() over (order by Diff desc) rank from (
select  id, one_before,
            case when (id - one_before) > 1 then (id - one_before)
           else 1
           end Diff
 from (
       select id, lag(id) over(order by id) one_before
      from table_name order by id) )

原始讨论位于http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:8146178058075。

找到丢失的序列号

3 个答案: