Question

我有以下PHP代码：

$result = mysql_query('
            set @num := 0, @type := "";
            UPDATE orders INNER JOIN (
                SELECT id, user_id, created, row_number 
                FROM 
                (
                    SELECT id, user_id, created,
                    @num := if(@type = user_id, @num + 1, 1) AS row_number,
                    @type := user_id AS dummy
                    FROM orders
                    WHERE status = "queue" AND type="order"
                    ORDER BY user_id, created asc 
                ) AS grouped_orders 
                WHERE grouped_orders.row_number <= 2
            ) m ON orders.id = m.id 
            SET orders.status = "process", orders.lock_id = "hash";
        ');

当我跳过$result = mysql_query('和');并将此查询复制粘贴到我的PHP MyAdmin面板时，它可以工作 - 但是当我使用PHP的mysql_query（...）执行它时，我收到以下错误：

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 'UPDATE orders INNER JOIN ( SELECT id, user_id, created, row_number FROM' at line 2

我尝试用括号做一些事情，向上/向下移动并折叠到一行，但它不起作用......

不幸的是，在服务器上，我正在努力，没有mysqli。

Answer 1

试试这个：

$result = mysql_query('
            set @num := 0, @type := "";
            UPDATE orders INNER JOIN (
                SELECT id, user_id, created, row_number 
                FROM 
                (
                    SELECT id, user_id, created,
                    @num := if(@type = user_id, @num + 1, 1) AS row_number,
                    @type := user_id AS dummy
                    FROM orders
                    WHERE status = "queue" AND type="order"
                    ORDER BY user_id, created asc 
                ) AS grouped_orders 
                WHERE grouped_orders.row_number <= 2
            ) m ON orders.id = grouped_orders.id 
            SET orders.status = "process", orders.lock_id = "hash";
        ');

组顺序是要加入的第二个表的名称

编辑：

你的查询看起来是O.K（至少没有错误），但你的问题依赖于PHP你正在使用mysql_query。

“mysql_query（）向与指定link_identifier关联的服务器上的当前活动数据库发送唯一查询（不支持多个查询）。”（http://php.net/manual/en/function.mysql -query.php）

所以你需要另一种方法来执行多个查询，或者找到另一个解决方案，尝试强制使用mysqli，因为你也可以防止攻击，并且能够进行多次查询....

Answer 2

尝试multi-query而不是mysql_query，因为您正在执行更多的一个查询

修改

尝试在select due subselects中定义变量：

UPDATE orders INNER JOIN (
      SELECT id, user_id, created, row_number 
      FROM 
      (
           SELECT id, user_id, created,
           @num := if(@type = user_id, @num + 1, 1) AS row_number,
           @type := user_id AS dummy
           FROM orders
           WHERE status = "queue" AND type="order"
           ORDER BY user_id, created asc 
            ) AS grouped_orders, (SELECT @num:=0) r, (SELECT @type:="") t
            WHERE grouped_orders.row_number <= 2
        ) m ON orders.id = grouped_orders.id 
        SET orders.status = "process", orders.lock_id = "hash";

通过mysql_query（...）执行时，有效的mysql查询返回错误

2 个答案:

修改