Question

我刚刚在Stackoverflow上得到了一些帮助并获得了正确的查询。但是他不知道任何PHP，所以现在我也需要帮助。

这是查询和PHP：

if($do_resync) {
    $sql = "
        with MY_VIEW as
        (
        SELECT
            room_user.*,
            user.username AS user_name,
            user.group_id AS group_id, USER.USER_ID
        FROM {$x7->dbprefix}room_users room_user
        INNER JOIN {$x7->dbprefix}users user ON
            user.id = room_user.user_id
        WHERE
            room_id IN ({$rooms})
        )
        SELECT COUNT(id) AS block FROM
            blocks AS B, MY_VIEW AS M
        WHERE
            B.user_id = M.user_id
        AND B.blocked_id = B.user_id
    ";
    $st = $db->prepare($sql);
    $st->execute(); (THIS IS LINE 230 that the error points at)
    $users = $st->fetchAll();
    $output['users'] = $users;
}

echo json_encode($output);

这是我得到的错误：

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 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 'MY_VIEW as ( SELECT room_user.*, user.username AS user_name, u' at line 6' in pages/sync.php:230 Stack trace: #0 pages/sync.php(230): PDOStatement->execute() #1 index.php(50): require('/home/vol5_3/by...') #2 {main} thrown in pages/sync.php on line 230

编辑：

SELECT COUNT(id) AS block FROM
    blocks AS B, 
(
    SELECT
        room_user.*,
        user.username AS user_name,
        user.group_id AS group_id, USER.USER_ID
    FROM {$x7->dbprefix}room_users room_user
    INNER JOIN {$x7->dbprefix}users user ON
        user.id = room_user.user_id
    WHERE
        room_id IN ({$rooms})
) AS M
WHERE
    B.user_id = M.user_id
AND B.blocked_id = B.user_id

Answer 1

Mysql不支持WITH子句如何执行此操作，如下所示，从查询中加入数据集

SELECT * FROM 
        (
        SELECT
            room_user.*,
            user.username AS user_name,
            user.group_id AS group_id, USER.USER_ID
        FROM {$x7->dbprefix}room_users room_user
        INNER JOIN {$x7->dbprefix}users user ON
            user.id = room_user.user_id
        WHERE
            room_id IN ({$rooms})
        ) AS M
JOIN (SELECT * FROM blocks ) AS B
ON (B.user_id = M.user_id)
WHERE B.blocked_id = B.user_id

修改

SELECT * FROM ( SELECT room_user.*, user.username AS user_name, user.group_id AS group_id, USER.USER_ID FROM {$x7->dbprefix}room_users room_user INNER JOIN {$x7->dbprefix}users user ON user.id = room_user.user_id WHERE room_id IN ({$rooms}) ) AS M JOIN blocks B ON (B.user_id = M.user_id) WHERE B.blocked_id = B.user_id

或移动你在哪里的条件

JOIN blocks B ON (B.user_id = M.user_id AND B.blocked_id = B.user_id)

Answer 2

我可以想出两种解决问题的方法：

1创建一个视图：

    $sql="CREATE VIEW MY_VIEW as
    SELECT
        room_user.*,
        user.username AS user_name,
        user.group_id AS group_id, USER.USER_ID
    FROM {$x7->dbprefix}room_users room_user
    INNER JOIN {$x7->dbprefix}users user ON
        user.id = room_user.user_id
    WHERE
        room_id IN ({$rooms})";

然后执行SELECT

2创建一个临时表：（http://dev.mysql.com/doc/refman/5.1/en/create-table.html）

    $sql1 = "CREATE TEMPORARY TABLE tmp_room_users_x0001 (
     -- the list of fields here ...
    )";


    // populate the table ...
    $sql2 = "INSERT INTO tmp_room_users_x0001 (SELECT
        room_user.*,
        user.username AS user_name,
        user.group_id AS group_id, USER.USER_ID
    FROM {$x7->dbprefix}room_users room_user
    INNER JOIN {$x7->dbprefix}users user ON
        user.id = room_user.user_id
    WHERE
        room_id IN ({$rooms})";

为什么我的MySQL查询不能在PHP中工作，我收到错误？

2 个答案: