您提供的数据样本表明这是单行,但已将其显示为两个不同的行。所以这个例子基于你的话。
-- Sample of data from your question + extra row for the sake of demonstration
-- id column is added to distinguish the rows(I assume you have one)
with t1(id, col) as(
select 1, '^FO360,065^AEN,25,10^FD{CUSTOMERS1.CUST_NAME}^FS^FO360,095^AAN,15,12^FD{CUSTOMERS1.CUST_ADDR1}^FS' from dual union all
select 2, '^FO360,065^AEN,25,10^FD{CUSTOMERS2.CUST_NAME}^FS^FO360,095^AAN,15,12^FD{CUSTOMERS2.CUST_ADDR2}^FS' from dual
),
cnt(c) as(
select level
from (select max(regexp_count(col, '{\w+.\w+}')) as o_c
from t1
) z
connect by level <= z.o_c
)
select t1.id, listagg(regexp_substr(t1.col, '{\w+.\w+}', 1, cnt.c)) within group(order by t1.id) res
from t1
cross join cnt
group by t1.id
结果:
ID RES
---------------------------------------------------------
1 {CUSTOMERS1.CUST_ADDR1}{CUSTOMERS1.CUST_NAME}
2 {CUSTOMERS2.CUST_ADDR2}{CUSTOMERS2.CUST_NAME}
根据@a_horse_with_no_name对该问题的评论,实际上,只需替换与该模式不匹配的所有其他内容要简单得多。这是一个例子:
with t1(col) as(
select '^FO360,065^AEN,25,10^FD{CUSTOMERS.CUST_NAME}^FS^FO360,095^AAN,15,12^FD{CUSTOMERS.CUST_ADDR1}^FS' from dual
)
select regexp_replace(t1.col, '({\w+.\w+})|.', '\1') res
from t1
结果:
RES
-------------------------------------------
{CUSTOMERS.CUST_NAME}{CUSTOMERS.CUST_ADDR1}