Question

我正在尝试这个陈述

with value_table as
(select 1 id, '1/2/3' objnr from dual
union all 
select 2, '4/5/6' from dual),
 test as
(select id, objnr col from value_table  where id in (1, 2))
    select id, regexp_substr(col, '[^/]+', 1, level) result
    from test 
    connect by level <= length(regexp_replace(col, '[^/]+')) + 1
    order by 1

我想获得6行

但我多次获取这些行。当我只用一个id尝试它时，它没有问题。作为一种解决方法，我只为每个id创建一个循环，另一个解决方案是使用distinct，但是当我尝试使用真实数据（超过1000个条目）时需要花费很长时间。有人能提供更复杂的解决方案吗？

Answer 1

尝试使用下面的

with value_table as
(select 1 id, '1/2/3' objnr from dual
union all 
select 2, '4/5/6' from dual),
 test as
(select id, objnr col from value_table  where id in (1, 2))
    select distinct id, regexp_substr(col, '[^/]+', 1, level) result
    from test 
    connect by level <= regexp_count(col, '[^/]+')
order by 1

尝试使用sqlfiddle http://sqlfiddle.com/#!4/03d80/14

修改

如果你不想要分别尝试下面的

with value_table as (select 1 id, '1/2/3' objnr from dual union all select 2, '4/5/6' from dual), test as (select id, objnr col from value_table where id in (1, 2)) select id, regexp_substr(col, '[^/]+', 1, level) result from test connect by level <= regexp_count(col, '[^/]+') and id = prior id and prior dbms_random.value is not null order by 1

使用模型子句PRIOR sqlfiddle链接为http://sqlfiddle.com/#!4/03d80/31

Answer 2

执行此查询

with test as
(
select 1 id, '1/2/3' objnr from dual union all
select 2 id, '4/5/6' objnr from dual 
)
select id, regexp_substr (objnr, '[^/]+', 1, rn) result
from test
cross
join (select rownum rn
       from (select max (length (regexp_replace (objnr, '[^/]+'))) + 1 mx
               from test
            )
     connect by level <= mx
     )
where regexp_substr (objnr, '[^/]+', 1, rn) is not null
and id in (1, 2)
order by id,result ;

ID结果

1 1
1 2
1 3
2 4
2 5
2 6

如果您使用的是Oracle 11g，您还可以使用REGEXP_COUNT而不是REGEXP_REPLACE和LENGTH的组合，如下所示：

cross
join (select rownum rn
    from (select max (regexp_count (objnr, '/') + 1) mx
            from test
         )
 connect by level <= mx
 )

具有外连接行为，因此如下所示略有变化：

with test as
  (
  select 1 id, '1/2/3' objnr from dual union all
  select 2 id, '4/5/6' objnr from dual 
  )
  select id, regexp_substr (objnr, '[^/]+', 1, rn) result
    from test
    left outer join (select rownum rn
           from (select max (regexp_count (objnr, '/') + 1) mx
                   from test
                )
         connect by level <= mx
         ) splits 
    ON  splits.rn <= length(regexp_replace (objnr, '[^/]+'))+1    
   and id in (1, 2)
   order by id,result ;

在许多行中拆分列会产生太多行

2 个答案: