【发布时间】:2011-06-11 09:46:42
【问题描述】:
最近 Django 社区出现了一个关于测试 MySQL(使用 MyISAM)的问题。
这是 django 票:http://code.djangoproject.com/ticket/14661
其中一位 Django 核心开发人员提出了这个测试,我们中的许多人已经能够复制它。有人猜到我们在这里遇到了什么吗?它只是 MySQL 中的错误还是我遗漏了什么?
这是测试代码和查询:
DROP TABLE IF EXISTS `testapp_tag`;
CREATE TABLE `testapp_tag` (
`id` integer AUTO_INCREMENT NOT NULL PRIMARY KEY,
`name` varchar(10) NOT NULL,
`parent_id` integer
);
INSERT INTO `testapp_tag` (`name`, `parent_id`) VALUES ("t1", NULL);
INSERT INTO `testapp_tag` (`name`, `parent_id`) VALUES ("t2", 1);
INSERT INTO `testapp_tag` (`name`, `parent_id`) VALUES ("t3", 1);
INSERT INTO `testapp_tag` (`name`, `parent_id`) VALUES ("t4", 3);
INSERT INTO `testapp_tag` (`name`, `parent_id`) VALUES ("t5", 3);
SELECT `testapp_tag`.`id`, `testapp_tag`.`name`, `testapp_tag`.`parent_id` FROM `testapp_tag` WHERE NOT ((`testapp_tag`.`id` IN (SELECT U0.`id` FROM `testapp_tag` U0 LEFT OUTER JOIN `testapp_tag` U1 ON (U0.`id` = U1.`parent_id`) WHERE U1.`id` IS NULL) AND `testapp_tag`.`id` IS NOT NULL)) ORDER BY `testapp_tag`.`name` ASC;
SELECT `testapp_tag`.`id`, `testapp_tag`.`name`, `testapp_tag`.`parent_id` FROM `testapp_tag` WHERE NOT ((`testapp_tag`.`id` IN (SELECT U0.`id` FROM `testapp_tag` U0 LEFT OUTER JOIN `testapp_tag` U1 ON (U0.`id` = U1.`parent_id`) WHERE U1.`id` IS NULL) AND `testapp_tag`.`id` IS NOT NULL)) ORDER BY `testapp_tag`.`name` ASC;
这是输出:
mysql> SELECT `testapp_tag`.`id`, `testapp_tag`.`name`, `testapp_tag`.`parent_id` FROM `testapp_tag` WHERE NOT ((`testapp_tag` .`id` IN (SELECT U0.`id` FROM `testapp_tag` U0 LEFT OUTER JOIN `testapp_tag` U1 ON (U0.`id` = U1.`parent_id`) WHERE U1.`id` IS NULL) AND `testapp_tag`.`id` IS NOT NULL)) ORDER BY `testapp_tag`.`name` ASC;
+----+------+-----------+
| id | name | parent_id |
+----+------+-----------+
| 1 | t1 | NULL |
| 3 | t3 | 1 |
| 5 | t5 | 3 |
+----+------+-----------+
3 rows in set (0.00 sec)
mysql> SELECT `testapp_tag`.`id`, `testapp_tag`.`name`, `testapp_tag`.`parent_id` FROM `testapp_tag` WHERE NOT ((`testapp_tag` .`id` IN (SELECT U0.`id` FROM `testapp_tag` U0 LEFT OUTER JOIN `testapp_tag` U1 ON (U0.`id` = U1.`parent_id`) WHERE U1.`id` IS NULL) AND `testapp_tag`.`id` IS NOT NULL)) ORDER BY `testapp_tag`.`name` ASC;
+----+------+-----------+
| id | name | parent_id |
+----+------+-----------+
| 1 | t1 | NULL |
| 3 | t3 | 1 |
+----+------+-----------+
2 rows in set (0.01 sec)
【问题讨论】:
-
省略提示信息 (mysql>) 使我们更容易复制/粘贴,因此我们可以在可能的情况下在我们的系统上进行测试。
-
哪一部分错了?只是最后一个查询?我明白了.. 相同的查询在第二次运行时丢失了 1 条记录