PostgreSQL条件连接

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\'),\'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，如果缺少某人的地址，我该如何获取以下信息，即显示其伴侣的地址？ -

    name    |  address  
------------+-----------
 Clark Kent | Manhattan
 Loise Lane | Manhattan
(2 rows)

谢谢。     

解决方法

        

select c.name,coalesce(a.address,a1. address) from contacts c
left outer join addresses a on c.address_id = a.address_id 
left outer join partners on c.name=partners.name1 
left outer join contacts c1 on c1.name=partners.name2 
left outer join addresses a1 on c1.address_id = a1.address_id;