-3
ID EMP_ID EMP_NAE MANAGER_ID
1 1 KARAN 2
2 2 K VEDI 3
3 3 SOMYA 2
4 4 KAVITA 3
5 5 AMAN 2
ID EMP_ID EMP_NAE MANAGER_ID
1 1 KARAN 2
2 2 K VEDI 3
3 3 SOMYA 2
4 4 KAVITA 3
5 5 AMAN 2
所有經理對於員工的名單和他們的經理用這樣的:
select emp_nae as [Employee name], man.emp_nae as [Manager name]
from tablename as emp
left join tablename as man on man.emp_id = emp.manager_id
管理者的列表,並thier員工使用:
select emp_nae as [Employee name], man.emp_nae as [Manager name]
from tablename as man
join tablename as emp on man.emp_id = emp.manager_id
是你的輸出預期,還是你的結構表? MySql或Sql服務器? – 2012-07-27 17:41:20
我們應該重新開放,因爲它明確要問什麼,這實際上是一個常見的混淆/問題 – Hogan 2012-12-11 15:45:26