Question

我不确定如何正确地提出这个问题，但现在是：

表1

id   |    email
---------------------------
1         test@test.com
---------------------------
2         random@test.com
---------------------------
3         magic@test.com
---------------------------
4         example@test.com
---------------------------
5         noreply@test.com
---------------------------
6         admin@test.com
---------------------------
7         editor@test.com

表2

id   |    email_two
---------------------------
7         test@test.com
---------------------------
10        random@test.com
---------------------------
33        magic@test.com
---------------------------
99        example@test.com
---------------------------
109       master@test.com
---------------------------
299       blaster@test.com

问题：

如何正确连接两个表，并通过电子邮件获取不匹配的结果？例如，我需要从两个表中得到的是：

noreply@test.com
admin@test.com
editor@test.com
master@test.com
blaster@test.com

因为其他电子邮件相互匹配。

代码

SELECT email_two FROM table2 b WHERE NOT EXISTS (SELECT * FROM table1 a WHERE a.email = b.email_two

此代码仅返回table2中缺少的代码，但我无法找到在一个查询中从两个表返回缺失结果的正确方法。

感谢您的回答。

Answer 1

你想要的最合适的操作是FULL OUTER JOIN，遗憾的是MySQL不支持它。

您可以改为使用UNION ALL：

SELECT email_two AS email
FROM table2 b 
WHERE NOT EXISTS (SELECT * FROM table1 a 
                  WHERE a.email = b.email_two)

UNION ALL

SELECT email 
FROM table1 a 
WHERE NOT EXISTS (SELECT * FROM table2 b 
                  WHERE a.email = b.email_two)

Answer 2

您可以使用LEFT JOIN和UNION ALL来获取此信息，下面是可以帮助您获取这些记录的示例查询。

select t1.email as 'email'  from  t1 LEFT JOIN t2 on t1.email = t2.email
where t2.email is null
UNION ALL
select t2.email as 'email'  from  t2 LEFT JOIN t1 on t2.email = t1.email 
where t1.email is null;

连接两个表后返回不匹配的行

2 个答案: