PHP (MySQL) 错误 : "Warning: mysql_num_rows() expects parameter 1 to be resource"

Question

这个问题在这里已经有了答案:





mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource

(31 个回答)


去年关闭。

if (!empty($_POST)){

    $email_to=$_POST['email_to'];
    $email_to=mysql_real_escape_string($_POST['email_to']);
    
    $sql = "UPDATE `cosmos`.`members` SET `conf` = '2' WHERE `members`.`email` = '$email_to';";
    $result=mysql_query($sql) or trigger_error(mysql_error().$sql);
    
    $count=mysql_affected_rows($result);                  // line 20
    if($count==1){
    
    $rows=mysql_fetch_array($result);
    $unique=$rows['u_code'];
    $name=$rows['username'];
    // ---------------- SEND MAIL FORM ---------------- 
    $to=$email_to; 
    $subject="Your Account Password Request! - Cosmos"; 
    $header="from: Tayal's/Cosmos <cosmos@gmail.com>"; 
    $messages= "Hey $name ,\r\n";
    $messages.="You recently requested a new password";
    $messages.="<br /><a href='confirm.php?uid" . $unique . "'>Confirmation Link</a> \r\n";
    $sentmail = mail($to,$subject,$messages,$header); 
    echo $messages; 
    }   else {
    echo "Not found your email in our database";
    }


}

Warning: mysql_affected_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\a\l\forget.php on line 20

Answer 1

$result是 false因为您的查询无效(有语法错误)。采用:

$sql = "UPDATE members SET conf=2 WHERE email = '$email_to';"

(注意 $email_to 周围的引号)

还有 mysql_num_rows()应该用于 SELECT仅限查询。对于 UPDATE , INSERT和 DELETE , 使用 mysql_affected_rows() 反而。

最后，为了将来引用，如果您的查询不起作用，请打印错误和使用的 SQL 查询(类似于 Col Shrapnel 的答案)。它将帮助您了解问题所在。