Php Mysqli导致登录表单根本无法工作_Php_Mysqli

Php Mysqli导致登录表单根本无法工作

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Php Mysqli导致登录表单根本无法工作,php,mysqli,Php,Mysqli,我的登录表单有点问题。我认为问题在于mysqli代码，因为当代码是旧的mysql代码时，登录工作正常，但自从我将其更改为mysqli后，它就不能正常工作假设用户将在登录表单中输入用户名和密码，当用户单击“登录”按钮时，它将在数据库中检查用户名和密码是否正确。如果正确，则导航到menu.php页面；如果登录不正确，则显示一条消息，说明登录不正确，然后重试相反，下面的代码所做的是，当用户输入用户名和密码并单击“登录”按钮时，无论用户名和密码是否正确，它只是刷新from，它不会导航到menu.ph

我的登录表单有点问题。我认为问题在于mysqli代码，因为当代码是旧的mysql代码时，登录工作正常，但自从我将其更改为mysqli后，它就不能正常工作

假设用户将在登录表单中输入用户名和密码，当用户单击“登录”按钮时，它将在数据库中检查用户名和密码是否正确。如果正确，则导航到menu.php页面；如果登录不正确，则显示一条消息，说明登录不正确，然后重试

相反，下面的代码所做的是，当用户输入用户名和密码并单击“登录”按钮时，无论用户名和密码是否正确，它只是刷新from，它不会导航到menu.php页面或显示登录错误的消息

所以我的问题是，为什么会发生这种情况，为什么登录后它不会导航用户或显示错误的登录消息

更新代码以显示当前代码：

<?php

ini_set('display_errors',1); 
 error_reporting(E_ALL);

    // PHP code
    session_start(); 

    $username="xxx";
    $password="xxx";
    $database="mobile_app";

    $mysqli = new mysqli("localhost", $username, $password, $database)or die( "Unable to select database");

    foreach (array('teacherusername','teacherpassword') as $varname) {
            $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : '';
          }

    // move this outside the condiitonal
    $loged = false;

    if (isset($_POST['submit'])) {

    $query = $mysqli->prepare("
    SELECT * FROM Teacher t  
    WHERE 
    (t.TeacherUsername=?)
    AND
    (t.TeacherPassword=?)
    ");

   $stmt=$mysqli->prepare($query);
   $stmt->bind_param("s",$teacherusername);
   $stmt->bind_param("s",$teacherpassword);

   $stmt->execute(); 

   $stmt->bind_result($TeacherId,$TeacherForename,$TeacherSurname,$TeacherUsername,$TeacherPassword);

    while($row=$stmt->fetch())
      {

          if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword']))
          {
              $loged = true;
          }

$_SESSION['teacherforename'] = $row['TeacherForename'];
$_SESSION['teachersurname'] = $row['TeacherSurname'];
$_SESSION['teacherusername'] = $row['TeacherUsername'];

      }

      if ($loged == true){
      header( 'Location: menu.php' ) ;
    }
    }

     ?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Teacher Login </title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="LoginStyle.css">
</head>
<body>
<?php if ($loged == false && $_POST) {
  echo "The Username or Password that you Entered is not Valid. Try Entering it Again.";
  }
?>
<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm">        
    <p>Username</p><p><input type="text" name="teacherusername" /></p>      <!-- Enter Teacher Username-->
    <p>Password</p><p><input type="password" name="teacherpassword" /></p>  <!-- Enter Teacher Password--> 
    <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p>
    </form>

</body>
</html>

这些意味着什么？有人知道如何修复它们吗？我更新了上面的代码以显示当前代码。

您需要移动HTML。在html之后，输入表单有一个IF。如果如下所示，请将html添加到else：

session_start(); 

$username="xxx";
$password="xxx";
$database="mobile_app";

$mysqli = new mysqli("localhost", $username, $password, $database)or die( "Unable to select database");

foreach (array('teacherusername','teacherpassword') as $varname) {
        $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : '';
      }

if (isset($_POST['submit'])) {

$query = $mysqli->prepare("
SELECT * FROM Teacher t  
WHERE 
(t.TeacherUsername=?)
AND
(t.TeacherPassword=?)
");

$loged = false;

$stmt=$mysqli->prepare($query);
$stmt->bind_param("s",$teacherusername);
$stmt->bind_param("s",$teacherpassword);

$stmt->execute(); 



$stmt->bind_result($TeacherId,$TeacherForename,
                    $TeacherSurname,$TeacherUsername,$TeacherPassword);

while($row=$stmt->fetch())
  {

      if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword']))
      {
          $loged = true;
      }

   $_SESSION['teacherforename'] = $row['TeacherForename'];
   $_SESSION['teachersurname'] = $row['TeacherSurname'];
   $_SESSION['teacherusername'] = $row['TeacherUsername'];

  }

  if ($loged == true){
  header( 'Location: menu.php' ) ;
}else{
  echo "The Username or Password that you Entered is not Valid. Try Entering it Again.";
}


}else{
?>
<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm">        
<p>Username</p><p><input type="text" name="teacherusername" /></p>      <!-- Enter Teacher Username-->
<p>Password</p><p><input type="password" name="teacherpassword" /></p>  <!-- Enter Teacher Password--> 
<p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p>
</form>
<?php
}
 ?>

session_start（）；
$username=“xxx”；
$password=“xxx”；
$database=“移动应用程序”；
$mysqli=newmysqli（“localhost”、$username、$password、$database）或die（“无法选择数据库”）；
foreach（数组（'teacherusername'，'teacherpassword'）为$varname）{
$$varname=（isset（$\u POST[$varname]）？$\u POST[$varname]：“”；
}
如果（isset（$_POST['submit']））{
$query=$mysqli->prepare（“
从教师t中选择*
哪里
（t.TeacherUsername=？）
及
（t.TeacherPassword=？）
");
$loged=false；
$stmt=$mysqli->prepare（$query）；
$stmt->bind_参数（“s”，$teacherusername）；
$stmt->bind_参数（“s”，$teacherpassword）；
$stmt->execute（）；
$stmt->bind_result（$TeacherId、$TeacherForename、，
$teacherName、$TeacherUsername、$TeacherPassword）；
而（$row=$stmt->fetch（））
{
如果（$\u POST['teacherusername']==（$row['teacherusername']）和&$\u POST['teacherpassword']==（$row['teacherpassword']））
{
$loged=true；
}
$\会话['teacherforename']=$row['teacherforename']；
$\会话['teacher姓氏]=$row['teacher姓氏']；
$\会话['teacherusername']=$row['teacherusername']；
}
如果（$loged==true）{
标题（'Location:menu.php'）；
}否则{
echo“您输入的用户名或密码无效。请重试输入。”；
}
}否则{
?>
更新/编辑：
根据您的评论，我再次查看了您的mysqli代码，并注意到您犯了一些错误（多个prepare和bind_result语句以及其他一些事情）。我已经更新了下面的代码，这些代码做了一些改进/应该可以解决您的问题。我不太确定您的db架构，但最好实际说明您想要的列，而不是使用（*）如果您将来添加另一列，并且不更新代码，它将中断mysqli语句，因为bind_结果与您的查询不匹配
<?php
  // PHP code
  session_start(); 

  $username="xxx";
  $password="xxx";
  $database="mobile_app";

  $mysqli = new mysqli("localhost", $username, $password, $database);

  /* check connection */
  if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    die();
  }

  // required variables (make them explciit no need for foreach loop)
  $teacherusername = (isset($_POST['teacherusername'])) ? $_POST['teacherusername'] : '';
  $teacherpassword = (isset($_POST['teacherpassword'])) ? $_POST['teacherpassword'] : '';
  $loggedIn = false;

  if (isset($_POST['submit'])) {

    // don't use $mysqli->prepare here
    $query = "SELECT * FROM Teacher WHERE TeacherUsername = ? AND TeacherPassword = ? LIMIT 1";
    // prepare query
    $stmt=$mysqli->prepare($query);
    // You only need to call bind_param once
    $stmt->bind_param("ss",$teacherusername,$teacherpassword);
    // execute query
    $stmt->execute(); 
    // get result and assign variables (prefix with db)
    $stmt->bind_result($dbTeacherId,$dbTeacherForename,$dbTeacherSurname,$dbTeacherUsername,$dbTeacherPassword);

    while($stmt->fetch()) {
      if ($teacherusername == $dbTeacherUsername && $teacherpassword == $dbTeacherPassword) {
        $loggedIn = true;
      }
    }

    /* close statement */
    $stmt->close();

    /* close connection */
    $mysqli->close();

    if ($loggedIn == true){
      // left your session code as is - but think wisely about using
      // the Username as a session variable (security risk)
      $_SESSION['teacherforename'] = $dbTeacherForename;
      $_SESSION['teachersurname'] = $dbTeacherSurname;
      $_SESSION['teacherusername'] = $dbTeacherUsername;
      header( 'Location: menu.php' ) ;
      die();
    }
  }
?>
<html>
<head></head>
<body>
  <?php if ($loggedIn == false && $_POST) {
    echo "The Username or Password that you Entered is not Valid. Try Entering it Again.";
    }
  ?>
  <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm">        
  <p>Username</p><p><input type="text" name="teacherusername" /></p>      <!-- Enter Teacher Username-->
  <p>Password</p><p><input type="password" name="teacherpassword" /></p>  <!-- Enter Teacher Password--> 
  <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p>
  </form>
</body>
</html>

首先，您似乎正在尝试混合使用mysql和mysqli函数。对于准备好的语句，您不需要使用_real_escape_字符串，而是使用？'s作为占位符。有关更多信息，请尝试：但正如afuzzyllama所说，您的主要问题是没有将重定向/会话内容置于html之上。@SteveH我更新了代码，因此它使用了prepare语句，因此不会混淆mysql和mysqli功能。它仍然不起作用，但您是否有错误报告？@user1421767-是的，请查看我的更新答案以解决您的问题我已打开错误报告，使用您的代码我收到警告和致命错误。我已更新问题以显示当前代码（哪是您的代码）我得到的两个php错误很抱歉，我没有太注意您的mysqli代码，因为我认为它是正确的。基本上，您的mysqli语法是错误的，您试图将一个prepare语句传入另一个prepare语句。我将在回答中更新代码以更正您的查询。@user1421767请查看我更新的代码-s你应该纠正你的问题。
<?php
  // PHP code
  session_start(); 

  $username="xxx";
  $password="xxx";
  $database="mobile_app";

  $mysqli = new mysqli("localhost", $username, $password, $database);

  /* check connection */
  if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    die();
  }

  // required variables (make them explciit no need for foreach loop)
  $teacherusername = (isset($_POST['teacherusername'])) ? $_POST['teacherusername'] : '';
  $teacherpassword = (isset($_POST['teacherpassword'])) ? $_POST['teacherpassword'] : '';
  $loggedIn = false;

  if (isset($_POST['submit'])) {

    // don't use $mysqli->prepare here
    $query = "SELECT * FROM Teacher WHERE TeacherUsername = ? AND TeacherPassword = ? LIMIT 1";
    // prepare query
    $stmt=$mysqli->prepare($query);
    // You only need to call bind_param once
    $stmt->bind_param("ss",$teacherusername,$teacherpassword);
    // execute query
    $stmt->execute(); 
    // get result and assign variables (prefix with db)
    $stmt->bind_result($dbTeacherId,$dbTeacherForename,$dbTeacherSurname,$dbTeacherUsername,$dbTeacherPassword);

    while($stmt->fetch()) {
      if ($teacherusername == $dbTeacherUsername && $teacherpassword == $dbTeacherPassword) {
        $loggedIn = true;
      }
    }

    /* close statement */
    $stmt->close();

    /* close connection */
    $mysqli->close();

    if ($loggedIn == true){
      // left your session code as is - but think wisely about using
      // the Username as a session variable (security risk)
      $_SESSION['teacherforename'] = $dbTeacherForename;
      $_SESSION['teachersurname'] = $dbTeacherSurname;
      $_SESSION['teacherusername'] = $dbTeacherUsername;
      header( 'Location: menu.php' ) ;
      die();
    }
  }
?>
<html>
<head></head>
<body>
  <?php if ($loggedIn == false && $_POST) {
    echo "The Username or Password that you Entered is not Valid. Try Entering it Again.";
    }
  ?>
  <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm">        
  <p>Username</p><p><input type="text" name="teacherusername" /></p>      <!-- Enter Teacher Username-->
  <p>Password</p><p><input type="password" name="teacherpassword" /></p>  <!-- Enter Teacher Password--> 
  <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p>
  </form>
</body>
</html>