sql - 自加入表格

Question

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

查找其他人推荐的所有客户的姓名。

输出应该是 dolly dilly 和 meena kimi。

我已成功找到通过查询转发他人的客户

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

Answer 1

除非我错过了什么：

SELECT *
FROM Customer
WHERE ReferredBy IS NOT NULL

Answer 2

有很多方法可以实现您想要的，但一种有趣的是使用 CTE，因为它允许您按级别获取参考，例如，在您的情况下，零级将是Neeta syam没有参考，一级是dolly dilly和meena kimi。以下查询将返回dolly dilly，meena kimi并且在指定的级别中，where reference = 1如下所示：

WITH CTEs (Id, CustomerName, Reference, RefCustomer)
AS
(
  SELECT 
    Id,
    Name,
    0 As Reference, 
    CONVERT(VARCHAR(length), 'No Reference') AS RefCustomer 
  FROM Customers 
  WHERE ReferredBy IS NULL
  UNION ALL
  SELECT
    c.CustId,
    c.Name, 
    cs.Reference + 1, 
    cs.CustomerName 
  FROM Customers c 
  INNER JOIN Ctes cs ON c.ReferedBy = cs.Id
)
SELECT CustomerName, RefCustomer 
FROM Ctes 
WHERE Reference = 1;