如何使用空值自加入 postgresql

Question

我想显示客户的姓名以及推荐他们的人的姓名。 我以为我走在正确的轨道上，但结果搞砸了。 我在WHERE子句中尝试了条件ReferredBy NOT NULL ReferredBy NOT NULL在ON子句中尝试了 ReferredBy NOT NULL - 不走运。 我将不胜感激任何帮助！ sqlfiddle

CREATE TABLE Customers(
          Id int NOT NULL,
          Name varchar(50) NOT NULL,
          ReferredBy int  REFERENCES Customers(Id),
          PRIMARY KEY (Id)
);

INSERT INTO Customers VALUES
(11, 'Peter', 22),
(22, 'Ariel', NULL),
(33, 'Tom', 11);

我的做法：

 SELECT c.Id, c.Name, c.ReferredBy, n.Name as ReferredBy_name
 FROM Customers c
 LEFT JOIN Customers n
 ON c.Id = n.ReferredBy 

所需的 Output：

Answer 1

我认为JOIN条件使表格倒置：

SELECT c.Id, c.Name, c.ReferredBy, n.Name as ReferredBy_name
FROM Customers c LEFT JOIN
     Customers n
     ON n.Id = c.ReferredBy ;