自我加入一张桌子

Question

 CustID       Name         ReferredBy
 1         Neeta Sayam  
 2         Dolly Dilly         1
 3         Meena Kimi          2

找到其他人推荐的所有客户的名称.

输出应该是dolly dilly和meena kimi.

我已经成功地找到了通过查询重新评估其他人的客户

 SELECT c1.name FROM Customer c1 JOIN Customer c2 ON c1.custid=c2. ReferredBy

Answer 1

除非我遗漏了什么:

SELECT *
FROM Customer
WHERE ReferredBy IS NOT NULL