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我有一个查询，它根据 $oneid 从等待表中获取所有用户。这工作得很好，但我想计算一个用户留下了多少条消息。这些在另一个表 one_msg 中。

function waiting_users($oneid)
    {

    $query_str ="SELECT a.user_id, b.email, b.username, b.fbook, c.user_id as saved_user, d.type as user_type,
                    FROM wait a 
                    LEFT JOIN users b ON a.user_id=b.id
                    JOIN user_profiles d ON a.user_id=d.user_id
                    LEFT JOIN save_one c ON a.user_id=c.user_id AND c.one_id=?
                    WHERE a.post_id = ? 
                    ORDER BY a.date ASC";
}
            $query = $this->db->query($query_str, array( $oneid, $oneid ) );

one_msg 表

+----+--------+---------+---------+---------------------+
| id | one_id | host_id | user_id | date                |
+----+--------+---------+---------+---------------------+
|  3 |    127 |     268 |     270 | 2012-06-11 18:57:58 | 
|  4 |    127 |     268 |     270 | 2012-06-11 21:45:11 | 
|  5 |    127 |     268 |     270 | 2012-06-12 09:10:01 | 
+----+--------+---------+---------+---------------------+

所以我试图像这样计算来自 one_msg 的消息，但它为所有用户返回相同的值。

function waiting_users($oneid)
    {

    $query_str ="SELECT a.user_id, b.email, b.username, b.fbook, c.user_id as saved_user, d.type as user_type,
                    (SELECT COUNT(id) FROM one_msg WHERE post_id = ?) AS count
                    FROM wait a 
                    LEFT JOIN users b ON a.user_id=b.id
                    JOIN user_profiles d ON a.user_id=d.user_id
                    LEFT JOIN save_one c ON a.user_id=c.user_id AND c.one_id=?
                    WHERE a.post_id = ? 
                    ORDER BY a.date ASC";

}
            $query = $this->db->query($query_str, array( $oneid, $oneid, $oneid ) );
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1 回答 1

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Your sub-query should check for the user_id instead of post_id (which does not exist in one_msg, so it is taken from wait:

(SELECT COUNT(id) FROM one_msg m WHERE m.user_id = a.user_id) AS count

instead of

(SELECT COUNT(id) FROM one_msg WHERE post_id = ?) AS count
于 2012-06-12T07:27:47.957 回答