使用交叉联接解决方案：

SELECT t1.*, t2.*
FROM table1 t1
INNER JOIN table2 t2 
ON 1=1;

SELECT
    a.name,
    b.title
FROM
    author AS a
LEFT JOIN
    book AS b
    ON a.id = b.author_id
UNION
SELECT
    a.name,
    b.title
FROM
    author AS a
RIGHT JOIN
    book AS b
    ON a.id = b.author_id

我修复了响应，工作包括所有行（基于Pavle Lekic的响应）：

    (
    SELECT a.* FROM tablea a
    LEFT JOIN tableb b ON a.`key` = b.key
    WHERE b.`key` is null
    )
    UNION ALL
    (
    SELECT a.* FROM tablea a
    LEFT JOIN tableb b ON a.`key` = b.key
    where  a.`key` = b.`key`
    )
    UNION ALL
    (
    SELECT b.* FROM tablea a
    right JOIN tableb b ON b.`key` = a.key
    WHERE a.`key` is null
    );

这也是可能的，但您必须在select中提到相同的字段名。

SELECT t1.name, t2.name FROM t1
LEFT JOIN t2 ON t1.id = t2.id
UNION
SELECT t1.name, t2.name FROM t2
LEFT JOIN t1 ON t1.id = t2.id

您可以只转换一个完整的外部联接，例如。

SELECT fields
FROM firsttable
FULL OUTER JOIN secondtable ON joincondition

进入：

SELECT fields
FROM firsttable
LEFT JOIN secondtable ON joincondition
UNION ALL
SELECT fields (replacing any fields from firsttable with NULL)
FROM secondtable
WHERE NOT EXISTS (SELECT 1 FROM firsttable WHERE joincondition)

或者，如果firsttable中至少有一列（比如foo）不为NULL，则可以执行以下操作：

SELECT fields
FROM firsttable
LEFT JOIN secondtable ON joincondition
UNION ALL
SELECT fields
FROM firsttable
RIGHT JOIN secondtable ON joincondition
WHERE firsttable.foo IS NULL

使用联合查询将删除重复项，这与从不删除任何重复项的完全外部联接的行为不同：

[Table: t1]        [Table: t2]
value              value
-----------        -------
1                  1
2                  2
4                  2
4                  5

这是完全外部联接的预期结果：

value | value
------+-------
1     | 1
2     | 2
2     | 2
Null  | 5
4     | Null
4     | Null

这是使用左联合和右联合的结果：

value | value
------+-------
Null  | 5
1     | 1
2     | 2
4     | Null

SQL Fiddle

我建议的查询是：

select
    t1.value, t2.value
from t1
left outer join t2
  on t1.value = t2.value
union all      -- Using `union all` instead of `union`
select
    t1.value, t2.value
from t2
left outer join t1
  on t1.value = t2.value
where
    t1.value IS NULL

上述查询的结果与预期结果相同：

value | value
------+-------
1     | 1
2     | 2
2     | 2
4     | NULL
4     | NULL
NULL  | 5

SQL Fiddle

@史蒂夫·钱伯斯：（来自评论，非常感谢！）

注意：这可能是最佳的解决方案，既可以提高效率，也可以生成与FULL OUTER JOIN相同的结果。这篇博客文章也很好地解释了这一点，引用了方法2的话：“这可以正确处理重复的行，并且不包含任何不应该包含的内容。有必要使用UNION ALL而不是普通UNION，这将消除我想要保留的重复项。这在大型结果集上可能会更有效，因为不需要排序和删除重复项。”

我决定添加另一个来自完全外部连接可视化和数学的解决方案。它并不比上述内容更好，但更具可读性：

完全外部连接意味着（t1ût2）：都在t1或t2中（t1ût2）=（t1ğt2）+t1_only+t2_only：t1和t2中的所有值加上t1中不在t2中的全部值，以及t2中不在t1中的全部：

-- (t1 ∩ t2): all in both t1 and t2
select t1.value, t2.value
from t1 join t2 on t1.value = t2.value
union all  -- And plus
-- all in t1 that not exists in t2
select t1.value, null
from t1
where not exists( select 1 from t2 where t2.value = t1.value)
union all  -- and plus
-- all in t2 that not exists in t1
select null, t2.value
from t2
where not exists( select 1 from t1 where t2.value = t1.value)

SQL Fiddle