SELECT NAME, kemu, shunxu
  FROM xueke a
 WHERE EXISTS (SELECT '*'
                 FROM persons b
                WHERE b.NAME = a.NAME)

------解决方案--------------------
WITH persons AS
  (SELECT 'yiyang' NAME,'01' banji,1 shunxu FROM dual
   UNION ALL
   SELECT 'yiyang' NAME,'02' banji,1 shunxu FROM dual
   UNION ALL
   SELECT 'lili' NAME,'01' banji,2 shunxu FROM dual),
   xueke AS
   (SELECT 'yiyang' name,'01-yingyu' kemu FROM dual
    UNION ALL
    SELECT 'yiyang' name,'01-shuxue' kemu FROM dual
    UNION ALL
    SELECT 'yiyang' name,'02-yingyu' kemu FROM dual
    UNION ALL
    SELECT 'lili' name,'01-yuwen' kemu FROM dual)
SELECT a.NAME,b.kemu FROM  
  (SELECT DISTINCT NAME,shunxu FROM persons ORDER BY shunxu) a,
  xueke b
where a.name = b.NAME
ORDER BY a.shunxu;

------解决方案--------------------
不同班级？distinct?不太明白
select a.name,b.kemu from persons a,xueke b where a.name = b.name
 and b.kemu like a.banji||'%' order by shunxu

------解决方案--------------------
10g以上用正则
regexp_like(Colname,'^[A|B]')

------解决方案--------------------
10g 正则 regexp_like(table_name,'^A|^B')


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

SQL> edi
已写入 file afiedt.buf

  1* select a.name,kemu from xueke a where exists(select 1 from Persons b where a.name=b.name)
SQL> /

NAME     KEMU
-------- --------------------
yiyang   02-yingyu
yiyang   01-shuxue
yiyang   01-yingyu
lili     01-yuwen

SQL>

------解决方案--------------------
为什么没有引入学号字段呢？用名字重复肯定很多，而且现有模式sql执行效率不会很高

------解决方案--------------------
select b.* from xueke b inner join (select distinct name from persons) a
on a.name=b.name

------解决方案--------------------
探讨

两个表

请看 "Persons" 表：
name    banji  shunxu
yiyang   01     1
yiyang   02     1
lili     01     2
  
接下来请看 "xueke" 表：

name    kemu       shunxu
yiyang  01-yingyu    1
yiyang  01-shuxue    ……