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
    );

在SQLite中，您应该执行以下操作：

SELECT * 
FROM leftTable lt 
LEFT JOIN rightTable rt ON lt.id = rt.lrid 
UNION
SELECT lt.*, rl.*  -- To match column set
FROM rightTable rt 
LEFT JOIN  leftTable lt ON lt.id = rt.lrid

您可以执行以下操作：

(SELECT 
    *
FROM
    table1 t1
        LEFT JOIN
    table2 t2 ON t1.id = t2.id
WHERE
    t2.id IS NULL)
UNION ALL
 (SELECT 
    *
FROM
    table1 t1
        RIGHT JOIN
    table2 t2 ON t1.id = t2.id
WHERE
    t1.id IS NULL);

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

前面的答案实际上都不正确，因为当存在重复值时，它们不遵循语义。

对于查询，例如（来自此副本）：

SELECT * FROM t1 FULL OUTER JOIN t2 ON t1.Name = t2.Name;

正确的等效值为：

SELECT t1.*, t2.*
FROM (SELECT name FROM t1 UNION  -- This is intentionally UNION to remove duplicates
      SELECT name FROM t2
     ) n LEFT JOIN
     t1
     ON t1.name = n.name LEFT JOIN
     t2
     ON t2.name = n.name;

如果需要使用NULL值（这可能也是必要的），则使用NULL安全比较运算符<=>而不是=。

使用交叉联接解决方案：

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