与 NULL 值的表比较

Question

我正在寻找比较两个表并返回包含两个表之间差异的行的最佳方法。

//表A

CREATE TABLE `mydb`.`a` (
  `id` int(10) unsigned NOT NULL,
  `color` varchar(45) DEFAULT NULL,
  `animal` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `mydb`.`a` 
    (id,color,animal) 
VALUES 
    (1,"red",""),
    (2,"blue","cat"),
    (3,"green","dog");
    (4,"orange","fish");

//表B

CREATE TABLE `mydb`.`b` (
  `id` int(10) unsigned NOT NULL,
  `color` varchar(45) DEFAULT NULL,
  `animal` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `mydb`.`b` 
    (id,color,animal) 
VALUES 
    (1,"red",""),
    (2,"blue","cat"),
    (3,"green","bird");
    (4,"brown","fish");

该语句将返回表 B 的第 3,4 行，因为 bird 和 brown 与 dog 和 orange 不同。

Answer 1

NULL 表示缺少信息。它永远不会匹配其他值（包括其他 NULL）。因此，例如，虽然零等于零：

SQL> select case when 0 = 0 then 'true' else 'false' end ;
case when 0 = 0 then 'true' else 'false' end
--------------------------------------------
true                                    

NULL 不等于 NULL：

SQL> select case when null = null then 'true' else 'false' end ;
case when null = null then 'true' else 'false' end
--------------------------------------------------
false                             

如果你想匹配它们，你必须将它们转换 NULL 到其他值（例如零）：

SQL> select case when ifnull(null, 0) = ifnull(null, 0) then 'true' else 'false' end ;
case when ifnull(null, 0) = ifnull(null, 0) then 'true' else 'false' end
------------------------------------------------------------------------
true   

编辑

假设你有：

mysql> select * from a;
+----+--------+--------+
| id | color  | animal |
+----+--------+--------+
|  1 | red    |        |
|  2 | blue   | cat    |
|  3 | green  | dog    |
|  4 | orange | fish   |
+----+--------+--------+
4 rows in set (0.00 sec)

mysql> select * from b;
+----+-------+--------+
| id | color | animal |
+----+-------+--------+
|  1 | red   |        |
|  2 | blue  | cat    |
|  3 | green | bird   |
|  4 | brown | fish   |
+----+-------+--------+
4 rows in set (0.00 sec)

我会使用：

mysql> select b.* from b left outer join a on a.id = b.id and 
       a.animal = b.animal and a.color = b.color 
       where a.id is null and a.animal is null and a.color is null ;
+----+-------+--------+
| id | color | animal |
+----+-------+--------+
|  3 | green | bird   |
|  4 | brown | fish   |
+----+-------+--------+
2 rows in set (0.00 sec)

查找 B 中与 A 中对应行不同的行。 并且：

mysql> select a.* from a left outer join b on a.id = b.id and 
       a.animal = b.animal and a.color = b.color 
       where b.id is null and b.animal is null and b.color is null ;
+----+--------+--------+
| id | color  | animal |
+----+--------+--------+
|  3 | green  | dog    |
|  4 | orange | fish   |
+----+--------+--------+
2 rows in set (0.00 sec)

在 A 中查找与 B 中相应行不同的行。