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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;
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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 ÖÐÆ½¾ù¹¤×ÊÊǶàÉÙ
Ê×ÏÈҪȷ¶¨ÄÄЩÈËÊÇmanager
¹«Ë¾ÓÐÄÄЩÈËÊÇ manager :
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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
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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
*/
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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
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