select * 
from youtablename  
where commit_time between to_char(sysdate,'yyyymmdd')||'0900' and to_char(sysdate,'yyyymmdd')||'1000' 
UNION ALL
select * 
from youtablename  
where commit_time between to_char(sysdate,'yyyymmdd')||'1500' and to_char(sysdate,'yyyymmdd')||'1600'

------解决方案--------------------
select * from A where substr(commit_time, 9) between '090000' and '100000' or substr(commit_time, 9) between '150000' and '160000'  
如果不含后面的时间点则应该为：
select * from A where substr(commit_time, 9) between '090000' and '095959' or substr(commit_time, 9) between '150000' and '155959'

------解决方案--------------------
commit_time 字段如果是字符类型 可以先将转换成日期类型to_date  然后再 to_char转换成字符 
  
 select to_char(to_date( '20080301112536 ', 'yyyymmddhhmiss '), 'Hh24:mi:ss ') from dual 
  
 下面的代码 可以试试。。。。自己把表名和列名改下。 
 select 
       a.commit_id, 
       a.commit_name, 
       a.commit_time 
 from 
     (select commit_id, 
             commit_name, 
             commit_time, 
             case when to_char(to_date( '20080301112536 ', 'yyyymmddhhmiss '), 'Hh24:mi ')> = '09:00 ' 
                   and to_char(to_date( '20080301112536 ', 'yyyymmddhhmiss '), 'Hh24:mi ') <= '10:00 ' then 1 
                  when to_char(to_date( '20080301112536 ', 'yyyymmddhhmiss '), 'Hh24:mi ')> = '09:00 ' 
                   and to_char(to_date( '20080301112536 ', 'yyyymmddhhmiss '), 'Hh24:mi ') <= '10:00 ' then 1 
                  else 0 
              end temp_column 
       from table_name ) a 
 where a.temp_column = 1