问题描述
我有很多地址需要相互匹配(母亲/孩子),某些地址在行尾有apt,suite等,所以-试图找到使最后一行不带整行的方法-
with addy as (select '22 W JAMESTOWN ST APT 22' as addy from dual union
select '22 W JAMESTOWN ST 22' as addy from dual)
select addy.*,regexp_substr(addy,'(\d*)(\D*)(\s)',1,'i') as no_ from addy;
最终结果应为:22 W JAMESTOWN ST 这在oracle中-向前功能似乎不起作用- '/.+?(?=APT)/'无效
第一行有效,第二行无效。任何输入表示赞赏-TIA 劳伦斯
解决方法
您可以尝试使用REGEX_SUBSTR
来删除地址末尾的多余信息,而不是REGEX_REPLACE
。
查询
WITH
addy
AS
(SELECT '22 W JAMESTOWN ST APT 22' AS addy FROM DUAL
UNION
SELECT '22 W JAMESTOWN ST 22' AS addy FROM DUAL)
SELECT addy.*,REGEXP_REPLACE (addy,'\s?(APT)?\s?\d+$') AS no_
FROM addy;
结果
ADDY NO_
___________________________ ____________________
22 W JAMESTOWN ST 22 22 W JAMESTOWN ST
22 W JAMESTOWN ST APT 22 22 W JAMESTOWN ST