有没有最简单的方法从oracle中的电子邮件地址中仅选择域名？

Question

我有一个名为EMAIL_ID和列名作为EMAIL_ID表，我有这些电子邮件ID的列

name@yahoo.com
name@msn.com
name@gmail.com
name@xmsoftware.com 

而我只需要域名，如这些作为结果

yahoo
msn
gmail
xmsoftware 

是否有最简单的方法（或其他方法）？

select substr(email_id.email_id,(instr(email_id.email_id,'@',1)+1),(instr(email_id.email_id,'.',1) - (instr(email_id.email_id,'@',1)+1))) as domain 
from email_id;

Answer 1

我更喜欢REGEXP_SUBSTR

select REGEXP_SUBSTR(email_id,'^.*@(.*)\.(.*)$',1,1,null,1) as domain 
 FROM emails;--                     ^ first match        ^ within()

演示

Answer 2

老式的方法是使用SUBSTR + INSTR组合。 与正则表达式相比，它需要输入更多的字母，但是在大型数据集上可能会更快地工作。

SQL> WITH email_id (email_id)
  2       AS (SELECT 'name@yahoo.com' FROM DUAL
  3           UNION ALL
  4           SELECT 'name@msn.com' FROM DUAL
  5           UNION ALL
  6           SELECT 'name@gmail.com' FROM DUAL
  7           UNION ALL
  8           SELECT 'name@xmsoftware.com ' FROM DUAL)
  9  select email_id,
 10         substr(email_id,
 11                instr(email_id, '@') + 1,
 12                instr(email_id, '.', instr(email_id, '@')) - instr(email_id, '@') - 1) result
 13  from email_id;

EMAIL_ID             RESULT
-------------------- --------------------
name@yahoo.com       yahoo
name@msn.com         msn
name@gmail.com       gmail
name@xmsoftware.com  xmsoftware

SQL>