使用双引号字符串进行选择 mysql 上的错误 1064 <42000>

Question

Zend Framework 2产生的查询:

SELECT "uc".*, "c".* FROM "user_contacts" AS "uc" INNER JOIN "contacts" AS "c" ON "uc"."contact_id" = "c"."contact_id" WHERE "uc"."user_id" = '2' AND "c"."user_id" = '1';

导致以下错误(在命令行上运行时):

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.*, "c".* FROM "user_contacts" AS "uc" INNER JOIN "contacts" AS "c" ON "uc"."con' at line 1

这个查询(完全相同的查询减去双引号)运行良好:

SELECT uc.*, c.* FROM user_contacts AS uc INNER JOIN contacts AS c ON uc.contact_id = c.contact_id WHERE c.user_id = 2 AND uc.user_id = 1;
+---------+------------+------------+---------+
| user_id | contact_id | contact_id | user_id |
+---------+------------+------------+---------+
|       1 |          7 |          7 |       2 |
+---------+------------+------------+---------+
1 row in set (0.00 sec)

为什么是这样，我怎么能解决这个问题?

在Ubuntu 12.10上使用AMP堆栈

表看起来像这样:

CREATE TABLE `contacts` (
    `contact_id` int(11) NOT NULL AUTO_INCREMENT,
    `user_id` int(11) NOT NULL,
    PRIMARY KEY (`contact_id`),
    UNIQUE KEY `contact_id_UNIQUE` (`contact_id`)
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8
CREATE TABLE `user_contacts` (
    `user_id` int(11) NOT NULL,
    `contact_id` int(11) NOT NULL,
    PRIMARY KEY (`user_id`,`contact_id`),
    KEY `user_contacts_user_id_fkey_idx` (`user_id`),
    KEY `user_contacts_contact_id_idx` (`contact_id`),
    CONSTRAINT `user_contacts_contact_id_fkey` FOREIGN KEY (`contact_id`) REFERENCES    `contacts` (`contact_id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
    CONSTRAINT `user_contacts_user_id_fkey` FOREIGN KEY (`user_id`) REFERENCES `user` (`user_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf8

Zend db适配器代码:

return array(
    'db' => array(
        'driver'         => 'Pdo',
        'dsn'            => 'mysql:dbname=' . $dbName . ';host=' . $host,
        'driver_options' => array(
            PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES 'UTF8''
        ),
    ),
    'service_manager' => array(
        'factories' => array(
            'ZendDbAdapterAdapter'
                => 'ZendDbAdapterAdapterServiceFactory',
        ),
    ),
);

选择代码:

public function checkIfFriends($currentUserId,$requestedUserId) {
    $currentUserId      = (int) $currentUserId;
    $requestedUserId    = (int) $requestedUserId;
    $sql = new Sql($this->tableGateway->getAdapter());
    $select = $sql->select();
    $select->from(array('uc' => $this->tableGateway->getTable()))
        ->join(array('c' => 'contacts'), 'uc.contact_id = c.contact_id');
    $where = new Where();
    $where
        ->equalTo('uc.user_id', $currentUserId)
        ->equalTo('c.user_id', $requestedUserId);
    $select->where($where);

    //echo $select->getSqlString();
    $rowSet = $this->tableGateway->selectWith($select);
    $row = $rowSet->current();
    return ($row) ? true: false;  
}

为什么这个被关闭为完全相同的副本?它不是。我理解的问题可能是相同的，但ZF2正在产生一个查询不运行由于引用。

Answer 1

在Zend Framework论坛帖子中找到的解决方案:

工作，但不是一个真正可接受的解决方案，一个更好的解决方案将不胜感激。

public function checkIfFriends($currentUserId,$requestedUserId) {
    $currentUserId      = (int) $currentUserId;
    $requestedUserId    = (int) $requestedUserId;
    $sql = new Sql($this->tableGateway->getAdapter());
    $select = $sql->select();
    $select->from(array('uc' => $this->tableGateway->getTable()))
        ->join(array('c' => 'contacts'), 'uc.contact_id = c.contact_id');
    $where = new Where();
    $where
        ->equalTo('uc.user_id', $currentUserId)
        ->equalTo('c.user_id', $requestedUserId);
    $select->where($where);
    $dbAdapter = $this->tableGateway->getAdapter();
    $string = $sql->getSqlStringForSqlObject($select);
    $rowSet = $dbAdapter->query($string, $dbAdapter::QUERY_MODE_EXECUTE);
    $row = $rowSet->current();
    return ($row) ? true: false;  
}

Answer 2

我通过关闭操作系统中的死键设置来解决这个问题。如何在Linux Mint中做到这一点:http://forums.linuxmint.com/viewtopic.php?f=55&t=97812