oracle sql 将字符串转换成多行

    今天遇到客户要求将字符串转换成多行的形式，自己摸索了一些；大家直接copy去用吧；

     with tt as(select zhbh as b,replace(yhbh,'，',',') as a from ptsx_tmpjyzhxx where zhbh in(6568,13810,6529))
select a, b from
(
select level,b,substr(a,instr(','||a,',',1,level),instr(a||',',',',1,level)-instr(','||a,',',1,level))a
from tt
connect by level<=length(a)-length(replace(a,','))+1 ) group by b,a

   在网上也搜索了几种，大家可以试试不错的

    with tmp_t as
(select '1001' as userId, '10,12,15' as workgroups
from dual
union
select '1002' as userId, '2,4,5' as workgroups from dual)
select userid,
substr(tempgroups,
instr(tempgroups, ',', 1, lv) + 1,
instr(tempgroups, ',', 1, lv + 1) -
(instr(tempgroups, ',', 1, lv) + 1))
from (select userid,
',' || workgroups || ',' AS tempgroups,
length(workgroups || ',') -
nvl(length(replace(workgroups, ',')) ， 0) AS groupcount
FROM tmp_t) a,
(select LEVEL lv from dual CONNECT BY LEVEL <= 5) b
where b.lv <= a.groupcount
order by userid, lv
   最简单的这种：

with t as
(select 'a;b;c;d;e' as str from dual)
select level,
t.str,
substr(t.str, 2 * (level - 1) + 1, 1) as str_signle
from t
connect by level <= length(t.str) - length(replace(t.str, ';', '')) + 1;

with t as

 (select 'i;am;a;test;mine' as str from dual)

select level,
  str,
  regexp_substr(t.str, '[^;]+', 1, level) str_single

  from t

connect by level <= length(t.str) - length(replace(t.str, ';', '')) + 1;