Question

    <?php
    session_start();
    if(isset($_POST['submit'])){
    //
    $user = $_POST['username'];
    $pwrd = $_POST['pwrd'];
    //include database connection
    include('../includes/db_connect.php');
    if(empty($user) || empty($pwrd)){
    echo 'Missing Information';
    }else{
    $user = strip_tags($user);
    $user = $db->real_escape_string($user);
    $pwrd = strip_tags($pwrd);
    $pwrd = $db->real_escape_string($pwrd);
    $pwrd = md5($pwrd);
    $query = $db->query("SELECT user_id, username FROM user WHERE username='$user' AND password='$pwrd'");
    if ($query->num_rows === 1){ // line 18
    while($row = $query->fetch_object()){
    $_SESSION['user_id'] = $row->user_id;
    }
    header('Location: index.php');
    exit();
    }else{
    echo 'Missing Information';
    }
    }
    }
    ?>

我在＆＃34中遇到错误; if（$ query-＆gt; num_rows === 1）＆＃34;当正确输入用户名和密码时，无法设置登录页面以将我引导至索引页面。但是如果它被错误地输入，那么else语句就可以了。

Answer 1

如果您正在尝试检查Statement是否等于1，那么最好只使用'=='而不是'==='，因为它也会检查数据类型...

最好的问候。

Answer 2

更新了......完整

<?php
session_start();
if(isset($_POST['submit'])){
  //
  $user = $_POST['username'];
  $pwrd = $_POST['pwrd'];

  if(empty($user) || empty($pwrd)){
    echo 'Missing Information';
  }else{

    //include database connection
    include('../includes/db_connect.php');

    $user = strip_tags($user);
    $user = $db->real_escape_string($user);
    $pwrd = strip_tags($pwrd);
    $pwrd = $db->real_escape_string($pwrd);
    $pwrd = md5($pwrd);
    $sql = "SELECT user_id, username FROM user WHERE username='$user' AND password='$pwrd'";
    if ($db->query($sql)) {
      while($row = $sql->fetch_object()){
        $_SESSION['user_id'] = $row->user_id;
      }
      mysqli_free_result($sql);

      session_write_close();
      header('Location: index.php');
      exit();

    }else{
      echo 'Username or password is incorrect';
    }

  }

}
?>

Answer 3

根据documentation $db->query失败时返回false。在这种情况下，$query将为false，而不是对象，因此$query->num_rows正在尝试将num_rows属性设置为false，从而导致错误。

处理这种情况的一种方法是将调用包装在if：

中

if ($query = $db->query("SELECT user_id, username FROM user WHERE username='$user' AND password='$pwrd'")) {
    if ($query->num_rows === 1){
        while($row = $query->fetch_object()){
        $_SESSION['user_id'] = $row->user_id;
    }
} else {
    // handle the error
}

注意：尝试在第18行的/Applications/XAMPP/xamppfiles/htdocs/Project/admin/login.php中获取非对象的属性

3 个答案: