Question

我有这个登录代码：

$con = new mysqli($host, $user, $password, $database);
if(isset($_POST['login'])){
 $username = $_POST['username'];
 $password = md5($_POST['password']);
 $stmt = $con->prepare("SELECT username, password FROM user WHERE username=? AND   password=? LIMIT 1");
 $stmt->bind_param('ss', $username, $password);
 $stmt->execute();
 $stmt->bind_result($username, $password);
 $stmt->store_result();
 if($stmt->num_rows == 1)  //To check if the row exists
 {
  $result=$con->query($stmt);
  $row=$result->fetch_array(MYSQLI_ASSOC);

  $status = $row['status'];
  if($status=='d') {
   echo "YOUR account has been DEACTIVATED.";
  } else {
   $_SESSION['Logged'] = 1;
   $_SESSION['user_id'] = $user_id;
   $_SESSION['username'] = $username;
   echo 'Success!';
   exit();
  }
 }
 else {
  echo "INVALID USERNAME/PASSWORD Combination!";
 }
 $stmt->close();
}
else 
{   
}
$con->close();

当按下登录按钮时显示此错误：

Warning: mysqli::query() expects parameter 1 to be string, object given in /nginx/html/user/demo/login.php on line 93    
Fatal error: Call to a member function fetch_array() on a non-object in /nginx/html/user/demo/login.php on line 94

有这些错误的行：

第93行是：$ result = $ con＆gt; query（$ stmt）;
第94行是$ row = $ result-＆gt; fetch_array（MYSQLI_ASSOC）;

Answer 1

当您使用预准备语句时，不要致电query或fetch_array。您调用fetch，并将结果放入与bind_result绑定的变量中。

您还引用了未在SELECT子句中列出的值。

$con = new mysqli($host, $user, $password, $database);
if(isset($_POST['login'])){
    $username = $_POST['username'];
    $password = md5($_POST['password']);
    $stmt = $con->prepare("SELECT username, status, user_id FROM user WHERE username=? AND  password=? LIMIT 1");
    $stmt->bind_param('ss', $username, $password);
    $stmt->execute();
    $stmt->bind_result($username, $status, $user_id);
    $stmt->store_result();
    if($stmt->num_rows == 1)  //To check if the row exists
        {
            $stmt->fetch();

            if($status=='d'){
                echo "YOUR account has been DEACTIVATED.";
            }else{
                $_SESSION['Logged'] = 1;
                $_SESSION['user_id'] = $user_id;
                $_SESSION['username'] = $username;
                echo 'Success!';
                exit();
            }

        }
    else {
        echo "INVALID USERNAME/PASSWORD Combination!";
    }
    $stmt->close();
}
else 
    {   

    }
$con->close();

错误登录脚本mysqli

1 个答案: