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我们的数据库已设置好，因此我们有一个credentials包含多种不同类型凭据（登录名等）的表。还有一个credential_pairs表将其中一些类型关联在一起（例如，用户可能有密码和安全令牌）。

在尝试查看一对是否匹配时，有以下查询：

SELECT DISTINCT cp.credential_id FROM credential_pairs AS cp
INNER JOIN credentials AS c1 ON (cp.primary_credential_id   = c1.credential_id)
INNER JOIN credentials AS c2 ON (cp.secondary_credential_id = c2.credential_id)
WHERE c1.data = AES_ENCRYPT('Some Value 1', 'encryption key')
AND   c2.data = AES_ENCRYPT('Some Value 2', 'encryption key');

此查询工作正常，并为我们提供了我们所需要的。但是，它不断显示在慢查询日志中（可能是由于缺少索引？）。当我要求 MySQL“解释”它给我的查询时：

+----+-------------+-------+------+--------------------------------------------------------+---------------------+---------+-------+-------+--------------------------------+
| id | select_type | table | type | possible_keys                                          | key                 | key_len | ref   | rows  | Extra                          |
+----+-------------+-------+------+--------------------------------------------------------+---------------------+---------+-------+-------+--------------------------------+
|  1 | SIMPLE      | c1    | ref  | credential_id_UNIQUE,credential_id,ix_credentials_data | ix_credentials_data | 22      | const |     1 | Using where; Using temporary   |
|  1 | SIMPLE      | c2    | ref  | credential_id_UNIQUE,credential_id,ix_credentials_data | ix_credentials_data | 22      | const |     1 | Using where                    |
|  1 | SIMPLE      | cp    | ALL  | NULL                                                   | NULL                | NULL    | NULL  | 69197 | Using where; Using join buffer |
+----+-------------+-------+------+--------------------------------------------------------+---------------------+---------+-------+-------+--------------------------------+

我有一种感觉，最后一个条目（它显示 69197 行）可能是问题所在，但我与 DBA 相差甚远......帮助？

凭证表：

CREATE TABLE  `credentials` (
  `hidden_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `credential_id` varchar(255) NOT NULL,
  `data` blob NOT NULL,
  `credential_status` varchar(100) NOT NULL,
  `insert_date` datetime NOT NULL,
  `insert_user` int(10) unsigned NOT NULL,
  `update_date` datetime DEFAULT NULL,
  `update_user` int(10) unsigned DEFAULT NULL,
  `delete_date` datetime DEFAULT NULL,
  `delete_user` int(10) unsigned DEFAULT NULL,
  `is_deleted` tinyint(1) NOT NULL DEFAULT '0',
  PRIMARY KEY (`hidden_id`,`credential_id`),
  UNIQUE KEY `credential_id_UNIQUE` (`credential_id`),
  KEY `credential_id` (`credential_id`),
  KEY `data` (`data`(10)),
  KEY `credential_status` (`credential_status`(10))
) ENGINE=InnoDB AUTO_INCREMENT=1572 DEFAULT CHARSET=utf8;

credential_pairs 表：

CREATE TABLE  `credential_pairs` (
  `hidden_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `credential_id` varchar(255) NOT NULL,
  `primary_credential_id` varchar(255) NOT NULL,
  `secondary_credential_id` varchar(255) NOT NULL,
  `is_deleted` tinyint(1) DEFAULT NULL,
  PRIMARY KEY (`hidden_id`,`credential_id`),
  KEY `primary_credential_id` (`primary_credential_id`(10)),
  KEY `secondary_credential_id` (`secondary_credential_id`(10))
) ENGINE=InnoDB AUTO_INCREMENT=500 DEFAULT CHARSET=latin1;

凭证索引：

+-------------+------------+----------------------+--------------+---------------+-----------+-------------+----------+--------+------+------------+---------+
| Table       | Non_unique | Key_name             | Seq_in_index | Column_name   | Collation | Cardinality | Sub_part | Packed | Null | Index_type     | Comment |
+-------------+------------+----------------------+--------------+---------------+-----------+-------------+----------+--------+------+------------+---------+
| credentials |          0 | PRIMARY              |            1 | hidden_id     | A         |      186235 |     NULL | NULL   |      | BTREE      |         |
| credentials |          0 | PRIMARY              |            2 | credential_id | A         |      186235 |     NULL | NULL   |      | BTREE      |         |
| credentials |          0 | credential_id_UNIQUE |            1 | credential_id | A         |      186235 |     NULL | NULL   |      | BTREE      |         |
| credentials |          1 | credential_id        |            1 | credential_id | A         |      186235 |     NULL | NULL   |      | BTREE      |         |
| credentials |          1 | ix_credentials_data  |            1 | data          | A         |      186235 |       20 | NULL   |      | BTREE      |         |
+-------------+------------+----------------------+--------------+---------------+-----------+-------------+----------+--------+------+------------+---------+

credential_pair 索引：

+------------------+------------+---------------------------------------------+--------------+-------------------------+-----------+-------------+----------+--------+------+------------+---------+
| Table            | Non_unique | Key_name                                    | Seq_in_index | Column_name             | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment |
+------------------+------------+---------------------------------------------+--------------+-------------------------+-----------+-------------+----------+--------+------+------------+---------+
| credential_pairs |          0 | PRIMARY                                     |            1 | hidden_id               | A         |       69224 |     NULL | NULL   |      | BTREE      |         |
| credential_pairs |          0 | PRIMARY                                     |            2 | credential_id           | A         |       69224 |     NULL | NULL   |      | BTREE      |         |
| credential_pairs |          1 | ix_credential_pairs_credential_id           |            1 | credential_id           | A         |       69224 |       36 | NULL   |      | BTREE      |         |
| credential_pairs |          1 | ix_credential_pairs_primary_credential_id   |            1 | primary_credential_id   | A         |       69224 |       36 | NULL   |      | BTREE      |         |
| credential_pairs |          1 | ix_credential_pairs_secondary_credential_id |            1 | secondary_credential_id | A         |       69224 |       36 | NULL   |      | BTREE      |         |
+------------------+------------+---------------------------------------------+--------------+-------------------------+-----------+-------------+----------+--------+------+------------+---------+

更新说明：

AFAICT：DISTINCT 是多余的……没有什么真正需要它，所以我放弃了它。为了遵循 Fabrizio 的建议，在 credential_pairs 查找中找到位置，我随后将语句更改为：

SELECT credential_id 
FROM credential_pairs cp
WHERE cp.primary_credential_id = (SELECT credential_id FROM credentials WHERE data = AES_ENCRYPT('value 1','enc_key')) AND
    cp.secondary_credential_id = (SELECT credential_id FROM credentials WHERE data = AES_ENCRYPT('value 2','enc_key'))

没事了。该语句需要同样长的时间，并且解释看起来几乎相同。因此，我在主列和辅助列中添加了一个索引：

ALTER TABLE credential_pairs ADD INDEX `idx_credential_pairs__primary_and_secondary`(`primary_credential_id`, `secondary_credential_id`);

没事了。

+----+-------------+-------------+-------+---------------------+---------------------------------------------+---------+------+-------+--------------------------+
| id | select_type | table       | type  | possible_keys       | key                                         | key_len | ref  | rows  | Extra                    |
+----+-------------+-------------+-------+---------------------+---------------------------------------------+---------+------+-------+--------------------------+
|  1 | PRIMARY     | cp          | index | NULL                | idx_credential_pairs__primary_and_secondary | 514     | NULL | 69217 | Using where; Using index |
|  3 | SUBQUERY    | credentials | ref   | ix_credentials_data | ix_credentials_data                         | 22      |      |     1 | Using where              |
|  2 | SUBQUERY    | credentials | ref   | ix_credentials_data | ix_credentials_data                         | 22      |      |     1 | Using where              |
+----+-------------+-------------+-------+---------------------+---------------------------------------------+---------+------+-------+--------------------------+

它说它正在使用索引，但它看起来仍然像是在扫描表。因此，我添加了一个联合密钥（根据下面的 a'r 评论）：

ALTER TABLE credential_pairs ADD KEY (primary_credential_id, secondary_credential_id);

而且...与索引相同的结果（这些功能是否相同？）。

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1 回答 1

0

DISTINCT 是生成“临时使用”的原因，您通常希望尽可能避免使用这些

另外，您正在扫描整个credential_pair表，因为您没有任何条件反对它，因此不使用索引，并且在应用 WHERE 之前返回整个表

希望这是有道理的

编辑/添加

尝试从另一个表开始，如果我理解正确，您有表 A、表 B 和表 AB，并且您从 AB 开始选择，尝试从 A 开始

我没有对此进行测试，但您可以尝试：

SELECT cp.credential_id
FROM credentials AS c1
    LEFT JOIN credential_pairs AS cp ON (c1.credential_id = cp.primary_credential_id)
    LEFT JOIN credentials AS c2 ON (cp.secondary_credential_id = c2.credential_id)
WHERE
    c1.data = AES_ENCRYPT('Some Value 1', 'encryption key')
    AND c2.data = AES_ENCRYPT('Some Value 2', 'encryption key');

过去我通过移动选择表来获得好运

于 2011-11-07T17:19:42.147 回答