正则表达式解析地址不包括最后的 apt 或 suite

Question

我有许多地址需要相互匹配 (mother/children) 有些地址在行尾有 apt、suite 等 - 一直在寻找方法来获取整行 w/o 最后一部分-

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)',1,1,'i') as no_ from addy;

最终结果应为：22 W JAMESTOWN ST 这是在oracle中-前瞻功能似乎不起作用- '/.+?(?=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

Answer 1

代替REGEX_SUBSTR，您可以尝试REGEX_REPLACE删除地址末尾的额外信息。

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

正则表达式解析地址不包括最后的 apt 或 suite

regex parsing address to not include apt or suite at end

oracle

regexp-substr

查询

结果