oracle 将多选字典列拆分为多行

oracle 将多值字典列拆分为多行

需求描述：

一般情况下，业务数据中存储的字典值为单值，但有时业务需要存储多值，需要通过sql解析多值字典项的内容；

字典表：

GUPID GUPNAME
1 分组1
2 分组2
3 分组3
4 分组4
5 分组5
11 分组11
12 分组12

数据表：

ID GUPIDS
1 1,2,3
2 1,11,12
3 4,5

最终需要解析成如下结果：

ID GUPID GUPNAME
1 1 分组1
1 2 分组2
1 3 分组3
2 1 分组1
2 11 分组11
2 12 分组12
3 4 分组4
3 5 分组5

解决办法：

通过组合使用 CONNECT BY LEVEL，regexp_substr 语法可以解析出字典内容

/* 业务字典表 */
create table tb_dic as 
  select 1 as gupId, '分组1' as gupName from dual union all 
  select 2 as gupId, '分组2' as gupName from dual union all 
  select 3 as gupId, '分组3' as gupName from dual union all 
  select 4 as gupId, '分组4' as gupName from dual union all 
  select 5 as gupId, '分组5' as gupName from dual union all 
  select 11 as gupId, '分组11' as gupName from dual union all 
  select 12 as gupId, '分组12' as gupName from dual 
;

/* 业务数据表 */
create table tb_data as 
  select 1 as id, '1,2,3' as gupIds from dual union all 
  select 2 as id, '1,11,12' as gupIds from dual union all 
  select 3 as id, '4,5' as gupIds from dual 
;

/* 分析业务数据中，存在的字典值数 */
select x1.id, x1.gupIds, x0.gupId, x0.gupName
from tb_dic x0,
  (select id, gupIds,length(gupIds) - length(replace(gupIds, ',')) + 1 as dics
  from tb_data
  ) x1 ,
  (/* 用于生成连续数值 */
  select LEVEL as rid from dual CONNECT BY LEVEL <= (select max(gupId) from tb_dic x0)
  ) x2 
where regexp_substr(x1.gupIds, '[^,]+',1,x2.rid)=x0.gupId
and x2.rid <=x1.dics
order by 1,3
;

或者，整合上面的内容为一条sql，输出结果一样，如下：

with  x0 as (/* 业务字典表 */
  select 1 as gupId, '分组1' as gupName from dual union all 
  select 2 as gupId, '分组2' as gupName from dual union all 
  select 3 as gupId, '分组3' as gupName from dual union all 
  select 4 as gupId, '分组4' as gupName from dual union all 
  select 5 as gupId, '分组5' as gupName from dual union all 
  select 11 as gupId, '分组11' as gupName from dual union all 
  select 12 as gupId, '分组12' as gupName from dual 
), x1 as (/* 业务数据表 */
  select 1 as id, '1,2,3' as gupIds from dual union all 
  select 2 as id, '1,11,12' as gupIds from dual union all 
  select 3 as id, '4,5' as gupIds from dual 
), x2 as (/* 分析业务数据中，存在的字典值数 */
select id, gupIds,length(gupIds) - length(replace(gupIds, ',')) + 1 as dics
from x1
)
select x2.id, x2.gupIds, x0.gupId, x0.gupName
  from x2,x0,
    (/* 用于生成连续数值 */
    select LEVEL as rid from dual CONNECT BY LEVEL <= (select max(gupId) from x0)
    ) t 
where regexp_substr(x2.gupIds, '[^,]+',1,t.rid)=x0.gupId
and t.rid <=x2.dics
order by 1,3