如何按照頂部的個人資料圖片進行分類？

Question

所以我現在正在爲我們的開發人員填充（被警告我是初學者），但我試圖簡單地將我的搜索結果按照包含個人資料圖片的個人資料排序（即， t要空白的個人資料圖片顯示在結果的頂部...他們應該都在結果的最後）...請注意，有幾個用戶類型，這就是爲什麼有這麼多的代碼...

我敢肯定，我要去哪裏錯的是2號線...

ORDER BY $爲了u.picture ISNULL DESC「; （這涉及按個人資料圖片排序）。真的很感激任何和所有的幫助...... thx！

的代碼如下：

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

你將有你的規則排序順序之前把isnull條件，如果你希望它的優先級：

ORDER BY ISNULL(u.picture), $order