Mysql 当SQL中存在变量时无法命中索引？

下面的SQL无法命中索引idx\u user\u userid，我不知道如何解决它

SET @q = 'abcd';
EXPLAIN SELECT user_id, mobile_num FROM user WHERE user_id = @q;

用户说明：

CREATE user(
  row_id INT AUTO_INCREMENT NOT NULL PRIMARY KEY,
  user_id CHAR(20) NOT NULL,
  mobile_num CHAR(15) NOT NULL DEFAULT ''
) ENGINE = InnoDB
CREATE UNIQUE INDEX idx_user_userid ON user(user_id);

---

MySQL版本是5.1.36

这与我在MySQL 5.0.51a-24中预期的一样有效。“user_id”列中的任何行是否实际包含@q变量中的值？以下是我的系统的输出：

CREATE TABLE `user` (
    `row_id` int(11) NOT NULL auto_increment,
    `user_id` char(20) NOT NULL,
    `mobile_num` char(15) NOT NULL default '',
    PRIMARY KEY  (`row_id`),
    UNIQUE KEY `idx_user_userid` (`user_id`)
) ENGINE=InnoDB;

INSERT INTO `user` VALUES
    (1, 'user1', '1234567890'),
    (2, 'user2', '1234567890');

SELECT * FROM `user`;

+--------+---------+------------+
| row_id | user_id | mobile_num |
+--------+---------+------------+
|      1 | user1   | 1234567890 |
|      2 | user2   | 1234567890 |
+--------+---------+------------+

SET @q = 'user1';

EXPLAIN SELECT user_id, mobile_num FROM user WHERE user_id = @q;

+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+
| id | select_type | table | type  | possible_keys   | key             | key_len | ref   | rows | Extra |
+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+
|  1 | SIMPLE      | user  | const | idx_user_userid | idx_user_userid | 20      | const |    1 |       |
+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+

SET @q = 'abcd';

EXPLAIN SELECT user_id, mobile_num FROM user WHERE user_id = @q;
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows | Extra                                               |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
|  1 | SIMPLE      | NULL  | NULL | NULL          | NULL | NULL    | NULL | NULL | Impossible WHERE noticed after reading const tables |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+

请注意，当使用值“abcd”时，EXPLAIN SELECT语句如何返回“不可能在哪里”消息。这是因为MySQL无法匹配WHERE语句，因为表中不存在提供的值。但是，如果提供了有效值，则会选择正确的索引

如果使用文字值而不是传递用户定义的变量，则返回相同的结果：

EXPLAIN SELECT user_id, mobile_num FROM user WHERE user_id = 'abcd';
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows | Extra                                               |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
|  1 | SIMPLE      | NULL  | NULL | NULL          | NULL | NULL    | NULL | NULL | Impossible WHERE noticed after reading const tables |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+

EXPLAIN SELECT user_id, mobile_num FROM user WHERE user_id = 'user1';
+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+
| id | select_type | table | type  | possible_keys   | key             | key_len | ref   | rows | Extra |
+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+
|  1 | SIMPLE      | user  | const | idx_user_userid | idx_user_userid | 20      | const |    1 |       |
+----+-------------+-------+-------+-----------------+-----------------+---------+-------+------+-------+

---

但是在mysql 5.1.36中，下面的SQL显示了类型=ALL。解释从用户中选择用户id，mobild_num，其中用户id=@qYes，这很奇怪。您可以发布SHOW CREATE TABLE user和EXPLAIN SELECT的完整输出吗？另外，SELECT查询返回多少行？我发现了一个问题：1。在bash中的mysql客户端中，以下SQL无法命中索引。设置@q='abcd'；选择user_id，mobile_num FROM user WHERE user_id=@q；但是下面的内容将被点击，从用户中选择用户id，mobile_num，其中用户id='abcd'；2.如果在一个过程中，所有这些SQL都将同时命中索引。