Sql 查找缺失的序列号
我想创建一个表来查找缺少的序列号。序列号在0到70000之间达到70000后变为0。在特定的时间段内,我需要找到那些丢失的记录。此解决方案基于一条语句,该语句生成从1到您设置的某个限制的所有自然数:Sql 查找缺失的序列号,sql,oracle,oracle11g,sequence,Sql,Oracle,Oracle11g,Sequence,我想创建一个表来查找缺少的序列号。序列号在0到70000之间达到70000后变为0。在特定的时间段内,我需要找到那些丢失的记录。此解决方案基于一条语句,该语句生成从1到您设置的某个限制的所有自然数: SELECT ROWNUM N FROM dual CONNECT BY LEVEL <= 7000 .我刚才从Tom Kyte那里偷了这个: select id, one_before, Diff, dense_rank() over (order by Diff desc) rank
SELECT ROWNUM N FROM dual CONNECT BY LEVEL <= 7000
.我刚才从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。您可以使用它来检测序列中的间隙。
该解决方案不会将您限制为特定的上限数字,如70000。
检测:
。
穿越:
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) )
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
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;