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1  (ÓÐÔ±¹¤µÄ³ÇÊÐ)¸÷¸ö³ÇÊеÄƽ¾ù¹¤×Ê
select city,avg(salary) from employees e,departments d locations l
where e.department_id = d.department_id and d.location_id = l.location_id group by city;
2  ²éѯÄÄЩ³ÇÊÐÓÐÔ±¹¤£¬ÄÄЩ³ÇÊÐûÓÐÔ±¹¤
select city from employees e,departments d,locations l
where e.department_id = d.department_id and d.location_id = l.location_id
group by city having count(employee_id) > 0
1)×麯Êý²»ÄÜ·ÅÔÚ where×Ó¾äÖУ¬ ÐèÒª·ÅÔÚ having ×Ó¾äÖÐ
2) Ðè°´ city ½øÐзÖ×é
3 manager ÖÐ×î¸ß¹¤×ÊÊÇË­   manager ÖÐ×îµÍ¹¤×ÊÊǶàÉÙ  manager ÖÐƽ¾ù¹¤×ÊÊǶàÉÙ
  Ê×ÏÈҪȷ¶¨ÄÄЩÈËÊÇmanager
     ¹«Ë¾ÓÐÄÄЩÈËÊÇ manager :
    ·½°¸1 
      select distinct m.employee_id,m.last_name from employees e,employees m
      where e.manager_id = m_employee_id  order by m.employee_id
    ·½°¸2
 select distinct manager_id from employees where manager_id is not null  order by manager_id
   ƽ¾ù¹¤×Ê:
 select avg(m.salary) from employees e,employees m
 where e.manager_id = m.employee_id  order by m.employee_id
  ×î¸ß¹¤×Ê:
/*
   select max(m.salary)  from employees e,employees m
 where e.manager_id= m.employee_id order by m.employee_id
*/
   ¿ÉÒÔÓÃ×Ó²éѯ
 select last_name,salary from employees  where salary =
 (select max(m.salary)  from employees e,employees m  where e.manager_id = m.employee_id )
 order by employee_id
5. Ñ¡ÔñËùÓÐÔ±¹¤µÄÐÕÃû£¬Ô±¹¤ºÅ£¬ÒÔ¼°ËûµÄ¹ÜÀíÕßµÄÐÕÃûºÍÔ±¹¤ºÅ£¬½á¹ûÀàËÆÓÚÏÂÃæµÄ¸ñʽ
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