【发布时间】:2011-09-04 07:57:19
【问题描述】:
假设我有以下表格--
smallville=# create table contacts (name varchar(16), address_id int);
CREATE TABLE
smallville=# create table addresses (address_id int, address varchar(16));
CREATE TABLE
smallville=# create table partners (name1 varchar(16), name2 varchar(16));
CREATE TABLE
smallville=# insert into contacts values ('Clark Kent', NULL), ('Loise Lane', 1);
INSERT 0 2
smallville=# insert into addresses values (1, 'Manhattan'), (2, 'North Pole');
INSERT 0 2
smallville=# insert into partners values ('Clark Kent', 'Loise Lane'),
('Loise Lane', 'Clark Kent') ;
INSERT 0 2
我可以得到姓名和地址 --
smallville=# select c.name, a.address from contacts c
left outer join addresses a
on c.address_id = a.address_id ;
name | address
------------+-----------
Clark Kent | (NULL)
Loise Lane | Manhattan
(2 rows)
但是我如何获得以下信息,即,如果一个人的地址丢失,则显示他/她的合作伙伴的地址? --
name | address
------------+-----------
Clark Kent | Manhattan
Loise Lane | Manhattan
(2 rows)
谢谢。
【问题讨论】:
标签: sql database postgresql stored-procedures join