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ÓÐÁ½ÕÅ±í£º²¿Ãűídepartment ²¿ÃűàºÅdept_id ²¿ÃÅÃû³Ædept_name
Ô±¹¤±íemployee Ô±¹¤±àºÅemp_id Ô±¹¤ÐÕÃûemp_name ²¿ÃűàºÅdept_id ¹¤×Êemp_wage
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answer:
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select emp_id,emp_wage,dept_name from employee as e inner join department as d on e.dept_id=d.dept_id where e.emp_wage>5000 group by e.emp_id;
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select dept_name,emp_id from department d left join employee e on e.dept_id=d.dept_id group by e.emp_id;
+------------+--------+
| dept_name | emp_id |
+------------+--------+
| ×Éѯ²¿ | NULL |
| Èí¼þ¿ª·¢²¿ | 1001 |
| Êг¡²ß»®²¿ | 1002 |
| ÏúÊÛ²¿ | 1003 |
| HR | 1004 |
| HR | 1005 |
| HR | 1006 |
| Èí¼þ¿ª·¢²¿ | 1007 |
+------------+--------+
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select dept_name from department d [inner] join employee e on d.dept_id=e.dept_id group by dept_name
having count(e.dept_id) >=2;
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GO
SET QUOTED_IDENTIFIER ON
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