如何按照顶部的个人资料图片进行分类？

Question

所以我现在正在为我们的开发人员填充（被警告我是初学者），但我试图简单地将我的搜索结果按照包含个人资料图片的个人资料排序（即， t要空白的个人资料图片显示在结果的顶部...他们应该都在结果的最后）...请注意，有几个用户类型，这就是为什么有这么多的代码...

我敢肯定，我要去哪里错的是2号线...

ORDER BY $为了u.picture ISNULL DESC“; （这涉及按个人资料图片排序）。真的很感激任何和所有的帮助...... thx！

的代码如下：

if ($user_type == 1) { 
     $sql = "SELECT a.*, u.*, 
       (SELECT COUNT(DISTINCT userId) FROM LF_usertype_A WHERE usertype_BId = u.userId AND status = 1) as i_cnt, 
       (SELECT COUNT(productId) FROM LF_products WHERE userId = u.userId AND status = 1) as product_cnt, 
       (SELECT COUNT(transactionId) 
        FROM LF_Transactions 
        WHERE usertypeBId = u.userId 
         AND (status = 1 OR status = 2) 
         AND type = 9 
         AND userId != usertypeBId 
         AND userId != usertypeAId) AS cnt 
       FROM LF_Users u 
       JOIN LF_products a ON a.userId = u.userId 
       LEFT JOIN LF_Transactions t ON t.productId = a.productId 
       WHERE a.status = 1 
        AND u.status = 1 
        AND u.userType = :ut $where 
       GROUP BY u.userID 
       ORDER BY $order u.name DESC LIMIT 200"; 
    } elseif ($filter != "recent" && $user_type == 2) { 
     $sql = "SELECT u.*, 
       (SELECT COUNT(a.productId) FROM LF_usertypeA a INNER JOIN LF_products ON a.productId = m.productId INNER JOIN LF_Users uu ON uu.userId = a.usertypeAId WHERE a.userId = u.userId AND uu.status = 1 AND a.status = 1 AND m.status = 1) as product_cnt, 
       (SELECT COUNT(transactionId) 
        FROM LF_Transactions 
        WHERE usertypeBId = u.userId 
         AND (status = 1 OR status = 2) 
         AND type = 9 
         AND userId != usertypeAId 
         AND userId != usertypeBId) AS cnt 
       FROM LF_Users u 
       LEFT JOIN LF_Transactions t ON t.usertypeBId = u.userId 
       WHERE u.status = 1 
        AND u.userId != 1 
        AND u.userType = :ut $where 
       GROUP BY u.userID 
       ORDER BY $order u.name DESC LIMIT 200 
       ORDER BY $order u.picture ISNULL DESC"; 
    } else { 
     $sql = "SELECT u.*, 
       (SELECT COUNT(a.productId) FROM LF_usertype_A a INNER JOIN LF_products m ON a.productId = m.productId INNER JOIN LF_Users uu ON uu.userId = a.usertypeAId WHERE a.userId = u.userId AND uu.status = 1 AND a.status = 1 AND m.status = 1) as product_cnt, 
       (SELECT COUNT(transactionId) 
        FROM LF_Transactions 
        WHERE usertypeBId = u.userId 
         AND (status = 1 OR status = 2) 
         AND type = 9 
         AND userId != usertypeAId 
         AND userId != usertypeBId) AS cnt 
       FROM LF_Users u 
       WHERE u.status = 1 
       AND u.userId != 1 
       AND u.userType = :ut $where 
       GROUP BY u.userID 
       ORDER BY $order u.name DESC LIMIT 200 
     ORDER BY $order u.picture ISNULL DESC"; 
    } 

Answer 1

你将有你的规则排序顺序之前把isnull条件，如果你希望它的优先级：

ORDER BY ISNULL(u.picture), $order