替换字符串与另一个字符串加上各自的位置

Question

我想知道是否有可能用另一个字符串加位置替换特定字符串而不使用PL/SQL块（循环，用户定义函数/存储过程，带函数构造...）。

这里：st - >pos_num

输入：

"aa bbb st cccc dddd st eeeeeeeeeee ffff g st g h i st j k l m st" 

输出：

"aa bbb pos_1 cccc dddd pos_2 eeeeeeeeeee ffff g pos_3 g h i pos_4 j k l m pos_5" 

DBFiddle

我觉得这是有可能实现它的Wi第一行操作（也许是正则表达式）。

Answer 1

使用MODEL条款：

select m_1 
from dual 
model dimension by (0 as key) 
measures (cast('st post aa bbb st cccc dddd st ee ffff g st g h i st j k l m st' 
       as varchar2(500)) as m_1) 
rules iterate (100) until(not regexp_like(m_1[0], '(|^)(st)(|$)')) 
(m_1[0] = regexp_replace(m_1[0], 
      '(|^)st(|$)','\1pos_'||to_char(ITERATION_NUMBER+1)||'\2',1,1)); 

DBFiddle Demo

输出：

pos_1 post aa bbb pos_2 cccc dddd pos_3 ee ffff g pos_4 g h i pos_5 j k l m pos_6 

Answer 2

递归cte方法。

with cte(string,col,cnt,repl) as 
(select string,1,regexp_count(string,'st'),regexp_replace(string,'st','pos_'||to_char(1),1,1) as repl 
from test 
union all 
select string,col+1,cnt,regexp_replace(repl,'st','pos_'||to_char(col+1),1,1) as repl 
from cte 
--join it to the original table if there are multiple rows, on string column. 
where col<cnt 
) 
cycle col set cycle to 1 default 0 
select string,repl 
from cte 
where cnt=col 

Answer 3

这是一个稍微不同的使用递归CTE的解决方案。仅当它被空格（或字符串的开头或结尾）包围时，它才会查找st。

with 
    inputs (str) as (
    select 'aa bbb st sccc dddd st eee fff g st g h i st j k l m st' from dual 
    union all 
    select 'st abc st st st where st is not st'      from dual 
    union all 
    select 'post st stop postal'          from dual 
), 
    r (lvl, str, new_str) as (
    select 1, str, str 
     from inputs 
    union all 
    select lvl + 1, str, 
      regexp_replace(new_str, '(|^)st(|$)', '\1pos_' || lvl || '\2', 1, 1) 
     from r 
     where regexp_like(new_str, '(|^)(st)(|$)') 
) 
select str, new_str 
from r 
where not regexp_like(new_str, '(|^)(st)(|$)') 
; 

STR              NEW_STR 
------------------------------------------------------- ---------------------------------------------------------------------- 
post st stop postal          post pos_1 stop postal 
aa bbb st sccc dddd st eee fff g st g h i st j k l m st aa bbb pos_1 sccc dddd pos_2 eee fff g pos_3 g h i pos_4 j k l m pos_5 
st abc st st st where st is not st      pos_1 abc pos_2 pos_3 pos_4 where pos_5 is not pos_6