日期:2014-05-17  浏览次数:20850 次

regexp_like的问题
表一:
anz_asset_product
oid asset_product_code
01 CA000
02 EQU111
03 UTX222

表二:
product_subtype
oid product_subtype_code
11 CA
12 EQU

题目是找出anz_asset_product中下面的记录
03 UTX222

因为UTX不存在于product_subtype表中,该如何实现呢,我知道有
SELECT *
FROM anz_asset_product anz
WHERE NOT EXISTS
 (SELECT 1
  FROM product_subtype p
  WHERE p.product_subtype_code =
  substr(anz.asset_product_code, 1, length(regexp_replace(anz.asset_product_code, '([[:alpha:]]*)([[:digit:]]*)', '\1'))));

可实现,可有没有更简单的方法,用regexp_like自动去匹配呢?

------解决方案--------------------
instr 函数
SQL code
with anz_asset_product as
(select  '01' oid, 'CA000' asset_product_code from dual union all 
select '02', 'EQU111' from dual union all
select '03', 'UTX222' from dual),
product_subtype as
(select '11' oid, 'CA' product_subtype_code from dual union all
select '12' , 'EQU' from dual)
select * from anz_asset_product t1 
where not exists 
(select 1 from product_subtype t2 where instr(t1.asset_product_code,t2.product_subtype_code,1,1)<>0)

------解决方案--------------------
SQL code

with anz_asset_product as(
select '01' oid,'CA000' asset_product_code from dual
union all
select '02','EQU111' from dual
union all
select '03','UTX222' from dual
),
product_subtype as(
select '11' oid,'CA' product_subtype_code from dual
union all
select '12','EQU' from dual
)SELECT *
FROM anz_asset_product anz
WHERE NOT EXISTS
 (SELECT 1
  FROM product_subtype p
  WHERE anz.asset_product_code like p.product_subtype_code||'%' );   --根据你的要求,用like就可以了
 
OID ASSET_PRODUCT_CODE
--- ------------------
03  UTX222

------解决方案--------------------
这样呢:
SQL code
select a.asset_product_code from anz_asset_product a
minus
select a.asset_product_code from anz_asset_product a , product_subtype b
where regexp_like(a.asset_product_code, b.product_subtype_code, 'i');