在MySQL慢查詢日誌中優化查詢

Question

我們的數據庫已設置好，因此我們有一個包含多種不同類型憑證（登錄等）的credentials表。還有一個將某些類型關聯在一起的credential_pairs表（例如，用戶可能有密碼和安全令牌）。

在嘗試，看是否有對比賽，有以下查詢：

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'); 

此查詢工作正常，給我們正是我們需要的。但是，它一直在緩慢的查詢日誌中顯示（可能是由於缺乏索引？）。當我問MySQL來「解釋」查詢它給了我：

+----+-------------+-------+------+--------------------------------------------------------+---------------------+---------+-------+-------+--------------------------------+ 
| 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 | 
+----+-------------+-------+------+--------------------------------------------------------+---------------------+---------+-------+-------+--------------------------------+ 

我有一種感覺，最後一個條目（如果它顯示69197行）可能是問題，但我從一個DBA很遠......幫幫我？

---

憑證表：

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; 

憑證索引：

+-------------+------------+----------------------+--------------+---------------+-----------+-------------+----------+--------+------+------------+---------+ 
| 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指標：

+------------------+------------+---------------------------------------------+--------------+-------------------------+-----------+-------------+----------+--------+------+------------+---------+ 
| 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  |   | 
+------------------+------------+---------------------------------------------+--------------+-------------------------+-----------+-------------+----------+--------+------+------------+---------+ 

---

更新備註：

AFAICT：本DISTINCT是多餘的......真的沒什麼需要它，所以我放棄了它。在試圖遵循法布里奇奧的建議得到對credential_pairs查找一個，我則改變了語句來爲已讀：

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')) 

與....無關。該聲明只需要很長時間，解釋看起來幾乎相同。所以，我給主要和次要列添加了一個索引：

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

而且......沒什麼。

+----+-------------+-------------+-------+---------------------+---------------------------------------------+---------+------+-------+--------------------------+ 
| 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    | 
+----+-------------+-------------+-------+---------------------+---------------------------------------------+---------+------+-------+--------------------------+ 

它說它使用索引，但它仍然看起來像它的表掃描。所以，我添加了一個聯合鍵（按下面a'r的評論）有：

ALTER TABLE credential_pairs ADD KEY (primary_credential_id, secondary_credential_id); 

而且......相同的結果與指數（是這些功能一樣嗎？）。

Answer 1

的DISTINCT是什麼產生了「使用臨時」，你平時要避免那些可能時

另外，你要掃描整個credential_pair表作爲你沒有反對任何條件，使不使用索引和WHERE

希望這是有道理的

編輯整個表應用之前返回/由加載

嘗試從不同的表開始，如果我理解正確的話，你有表A，表B和表AB，並且你從AB開始選擇，嘗試從A開始它A

我還沒有測試過這個，但是你可以嘗試：

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'); 

我已經走動