Mysql錯誤顯示

Question

我有一個mysql錯誤，因爲mysql_fetch_array():提供的參數不是有效的MySQL結果。 如何解決「Mysql_fetch_array提供的參數不是有效的MYSQL結果」錯誤？

function rating($did) 
{ 
    $sqls = "SELECT AVG(driver_rating.rate) 
      FROM `driver_rating` 
      JOIN `order` ON order.id = driver_rating.order_id 
      JOIN `driver` ON order.did = $did 
      GROUP BY $did"; 

    $results = mysql_query($sqls); 

    $num_of_result = mysql_affected_rows();    

    while($rows = mysql_fetch_array($results)) 
    { 
     $info_result=$rows[0]; 
    } 

    if($num_of_result == 0) 
     $infor = 'ok'; 
    else 
     $infor = $info_result; 

    return $infor; 
} 

Answer 1

您使用GROUP BY子句不選擇該字段列，在您選擇字段列添加$did。

SELECT $did, AVG(driver_rating.rate) 
FROM `driver_rating` 
JOIN `order` ON order.id = driver_rating.order_id 
JOIN `driver` ON order.did = $did 
GROUP BY $did 

華林：Please, don't use mysql_* functions in new code。他們不再維護and are officially deprecated。請參閱red box？請改爲了解prepared statements，並使用PDO或MySQLi - this article將幫助您決定哪個。如果您選擇PDO，here is a good tutorial。