Question

我正在编写一个在指定范围之间生成数字的查询。

我有一张桌子

NUM_RANGES

ID  START_NUMBER  END_NUMBER
--  ------------  ----------
 1             1           5
 2             6          10

我需要得到这个结果：

使用此查询，我可以获得正确的结果，但仅适用于where子句中的指定id：

select   id, start_number + level - 1 next_tag
                        from  (select id, start_number,end_number
                        from NUM_RANGES
                        where id = 1
       ) 
        connect by level <=  end_number -  start_number + 1

没有“where id = 1”我得到62行重复，其中有明显的帮助，但更大的范围1 - 200,200-500它的工作太慢..

感谢您的帮助！

Answer 1

在Oracle 12c上，您可以使用CROSS APPLY：

select *
from num_ranges
cross apply(
   select level - 1 + start_number as my_new_number
   from dual
   connect by level <= end_number - start_number + 1
);

Answer 2

在Oracle 11.2及更早版本中，您可以使用分层查询：

with
     num_ranges ( id, start_number, end_number ) as (
       select 1, 1,  5 from dual union all
       select 2, 9, 12 from dual
     )
-- End of simulated input data (for testing purposes only, not part of the solution).
-- SQL query begins below this line.
select     id, start_number + level - 1 as nmbr
from       num_ranges
connect by level <= end_number - start_number + 1
       and prior id = id
       and prior sys_guid() is not null
order by   id, nmbr  --  If needed
;

        ID       NMBR
---------- ----------
         1          1
         1          2
         1          3
         1          4
         1          5
         2          9
         2         10
         2         11
         2         12

具体而言，如果不将新行与同一id的新行链接，则会产生疯狂数量的不必要的行。这就是您需要prior id = id的原因。需要附加条件prior sys_guid() is not null，以便Oracle不会看到它不应该看到它们的周期（这些周期是由＆＃34; prior id = id＆＃34;）引起的。< / p>

在Oracle 12.1或更高版本中，您可以使用lateral子句：

select   n.id, l.nmbr
from     num_ranges n,
         lateral ( select start_number + level - 1 as nmbr from dual
                   connect by level <= end_number - start_number + 1 ) l
order by id, nmbr  --  If needed
;

Answer 3

这应该有效。也很快。

with cte as (select 0 as c from dual)
, cte4 as (select c from cte union all select c from cte union all select c from cte union all select c from cte)
, cte256 as (select t0.c from cte4 t0, cte4 t1, cte4 t2, cte4 t3)
, nums as (select row_number() over(order by null) as n from cte256 t0, cte256 t1, cte256 t2)
select NR.id, nums.n as NUMBER_
from nums
join NUM_RANGES NR on nums.n between NR.START_NUMBER and NR.END_NUMBER
;

生成给定范围的数字（sql查询）

3 个答案: