Question

我正在尝试在PHP（特别是Laravel）中创建一个MySQL视图，我遇到了一个奇怪的错误：

[PDOException]
SQLSTATE[42000]: SSyntax error or access violation: 1142 ANY command denied to user 'user'@'localhost' for table '/tmp/#sql_475_0'

在MySQL中直接运行create语句可以正常工作。如果我删除了视图的连接，那么一切正常。用户具有完全权限（GRANT ALL）。广泛的谷歌搜索没有返回任何类似的东西。

我的代码如下，略有简化，运行第4个语句创建jobs_view时会产生错误。

DB::statement("
    CREATE VIEW quote_response_count AS (
        SELECT job_id, COUNT(quotes.id) as total FROM quotes 
        INNER JOIN quote_requests on quote_requests.quote_id = quotes.id 
        INNER JOIN quote_responses on quote_responses.quote_request_id = quote_requests.id
        GROUP BY job_id
    );
");

DB::statement("
    CREATE VIEW customer_paid AS (
        SELECT job_id, SUM(amount) as total FROM transactions 
        WHERE category = 'customer payment' AND is_verified = 1
        GROUP BY job_id, category
    );
");

DB::statement("
    CREATE VIEW company_paid AS (
        SELECT job_id, SUM(amount) as total FROM transactions 
        WHERE category = 'company payment' AND is_verified = 1
        GROUP BY job_id, category
    );
");

DB::statement("
    CREATE VIEW jobs_view AS (
        SELECT 
            jobs.*,         
            IFNULL(customer_paid.total, 0)              AS customer_paid,
            IFNULL(company_paid.total, 0)               AS company_paid,
            IFNULL(quote_response_count.total, 0)       AS responses_received,                  
            price - IFNULL(customer_paid.total, 0)      AS customer_owes,
            cost  - IFNULL(customer_paid.total, 0)      AS owes_company,
            (
                deposit > 0 AND IFNULL(customer_paid.total, 0) >= deposit
            )                                           AS deposit_paid             

         FROM jobs

         LEFT OUTER JOIN quote_response_count   AS quote_response_count ON quote_response_count.job_id  = jobs.id
         LEFT OUTER JOIN customer_paid          AS customer_paid        ON customer_paid.job_id         = jobs.id
         LEFT OUTER JOIN company_paid           AS company_paid         ON company_paid.job_id          = jobs.id
    );
");

PHP应用程序中的SHOW GRANTS输出如下：

[Grants for user@localhost] => GRANT USAGE ON *.* TO 'user'@'localhost' IDENTIFIED BY PASSWORD '****************************'

[Grants for user@localhost] => GRANT ALL PRIVILEGES ON `dbname`.* TO 'user'@'localhost'

以下非常简化的示例也会产生相同的结果：

DB::statement("
    CREATE TABLE table1 (
        id int(11) NOT NULL AUTO_INCREMENT,
        foo varchar(45) DEFAULT NULL,
        PRIMARY KEY (id)
    );
");
DB::statement("
    CREATE VIEW view1 AS (
        SELECT id, foo FROM table1
    );
");
DB::statement("
    CREATE VIEW view2 AS (
        SELECT table1.id, view1.foo FROM table1
        INNER JOIN view1 ON view1.id = table1.id
    );
");

如果只是从view1中选择而不是加入，则会发生同样的错误。

我遇到此问题的系统是运行PHP 5.5.23和MySQL 5.5.41的Ubuntu 12.04服务器。

Answer 1

尤里卡！对于遇到此问题的任何其他人，由于Laravel设置以下PDO连接选项而出现问题：

PDO::ATTR_EMULATE_PREPARES => false

我的解决方案是克隆我的数据库配置，覆盖PDO选项，然后在创建我的视图时使用该连接，而不是为我的整个应用程序启用Emulate Prepares：

配置/ database.php中

'mysql' => array(
    'driver'    => 'mysql',
    'host'      => 'localhost',
    'database'  => 'database',
    'username'  => 'user',
    'password'  => 'password',
    'charset'   => 'utf8',
    'collation' => 'utf8_unicode_ci',
    'prefix'    => '',
),
'mysql-emulate-prepares' => array(
    'driver'    => 'mysql',
    'host'      => 'localhost',
    'database'  => 'database',
    'username'  => 'user',
    'password'  => 'password',
    'charset'   => 'utf8',
    'collation' => 'utf8_unicode_ci',
    'prefix'    => '',
    'options'   => array(
        PDO::ATTR_EMULATE_PREPARES => true,
    ),
),

移植

$rand = rand(10000, 99999);

DB::statement("
    CREATE TABLE table".$rand." (
        id int(11) NOT NULL AUTO_INCREMENT,
        foo varchar(45) DEFAULT NULL,
        PRIMARY KEY (id)
    );
");
DB::statement("
    CREATE VIEW view".$rand." AS (
        SELECT id, foo FROM table1
    );
");
DB::connection('mysql-emulate-prepares')->statement("
    CREATE VIEW view".($rand+2)." AS (
        SELECT table".$rand.".id, view".$rand.".foo FROM table".$rand."
        INNER JOIN view".$rand." ON view".$rand.".id = table".$rand.".id
    );
");

Ryan Vincent帮助我调试这个问题。{/ p>

Answer 2

这是另一个不涉及创建和指定单独配置的解决方案：

$dconn = DB::connection()->getDoctrineConnection();
$dconn->exec("CREATE OR REPLACE VIEW ...");

请注意：

调用getDoctrineConnection()方法会创建一个新的PDO对象，而DatabaseManager不会跟踪该

。

这意味着DB::reconnect()，DB::disconnect()或DB::purge()不会影响这个新的PDO对象。因此，它将作为请求其余部分的开放连接。

这通常不是Web请求或部署操作中的问题，因为脚本最终会结束。但是，如果一个人正在使用长期运行的妖魔化PHP进程与Gearman或Resque之类的东西，那么最终会抛出一条PDO异常并显示消息MySQL server has gone away (error 2006)

此问题的解决方案是显式关闭Doctrine连接，如下所示：

$dconn = DB::connection()->getDoctrineConnection();
$dconn->exec("CREATE OR REPLACE VIEW ...");
$dconn->close();

PDOException语法错误或访问冲突1142，在创建引用其他视图的视图时

2 个答案:

配置/ database.php中

移植