Question

这看起来应该很简单，但我已经尝试过各种方式，而且似乎无法让它发挥作用。

我有这个登录脚本，我从在线教程改编。我想要做的是让用户使用用户名和密码登录，如果这些是正确的，请将他们的实验室结果发送到另一个表（相同的数据库）并显示它们。我可以让它登录，但就是这样。这是登录代码：

<?php  //Start the Session
session_start();
 require('connect.php');
//3. If the form is submitted or not.
//3.1 If the form is submitted
if (isset($_POST['username']) and isset($_POST['password'])){
//3.1.1 Assigning posted values to variables.
$username = $_POST['username'];
$password = $_POST['password'];
//3.1.2 Checking the values are existing in the database or not
$query = "SELECT * FROM `user` WHERE username='$username' and password='$password'";

$result = mysql_query($query) or die(mysql_error());
$count = mysql_num_rows($result);
//3.1.2 If the posted values are equal to the database values, then session will be created for the user.
if ($count == 1){
$_SESSION['username'] = $username;
}else{
//3.1.3 If the login credentials doesn't match, he will be shown with an error message.
echo "Invalid Login Credentials.";
}
}
//3.1.4 if the user is logged in Greets the user with message
if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "Hi " . $username . "! ";
echo "This is the results of your inquiry.<br><br>";




/*This is where I'm assuming the new query needs to go.
Query a different table named "data"  and pick out information according to 
$username which was put in earlier */




echo "<br><a href='logout.php'>Logout</a>";

}else{
//3.2 When the user visits the page first time, simple login form will be displayed.
?>
<!DOCTYPE html>

<html>
<head>
  <title>Lab Sign In Page</title>
  <link rel="stylesheet" type="text/css" href="style.css">
</head>

<body>
  <!-- Form for logging in the users -->

  <div class="register-form">
    <?php
        if(isset($msg) & !empty($msg)){
            echo $msg;
        }
     ?>

    <h1>Login</h1>

    <form action="" method="post">
      <p><label>User Name :</label> <input id="username" type="text" name="username" placeholder=
      "username"></p>

      <p><label>Password   :</label> <input id="password" type="password" name="password"
      placeholder="password"></p><a class="btn" href="register.php">Signup</a> <input class=
      "btn register" type="submit" name="submit" value="Login">
    </form>
  </div><?php }


   ?>
</body>
</html>

A＆＃34;加入＆＃34;当我谷歌它时我得到的但是这看起来并不正确。有人可以帮忙吗？

Answer 1

您可以使用其他查询来简化：

$sql = mysql_query("SELECT * FROM data WHERE user = '" . mysql_real_escape_string($username) . "'");

然后你可以用这个来处理。

请注意，您不应该使用MySQL驱动程序，因为它已被弃用，而是使用MySQLi（mproved）。你应该逃避POSTed值，这非常重要！

Answer 2

您的上一个else语句正在尝试回显HTML。

您应该使用mysqli或PDO，因为不推荐使用mysql。

<?php  //Start the Session
session_start();
// require('connect.php');
// Establish connection with database
$con=mysqli_connect("localhost","root","","test");
// Check connection
if (mysqli_connect_errno()){ echo "Failed to connect to MySQL: " . mysqli_connect_error(); }
//3. If the form is submitted or not.
//3.1 If the form is submitted
if (isset($_POST['username']) and isset($_POST['password'])){
  //3.1.1 Assigning posted values to variables.
  $username = $_POST['username'];
  $password = $_POST['password'];
  //3.1.2 Checking the values are existing in the database or not
  $query = "SELECT * FROM `people` WHERE username='$username' and password='$password'";

  $result = mysqli_query($con,$query) or die(mysqli_error());
  $count = mysqli_num_rows($result);
  //3.1.2 If the posted values are equal to the database values, then session will be created for the user.
  if ($count == 1){
    $_SESSION['username'] = $username;
      echo "Valid";
  }else{
    //3.1.3 If the login credentials doesn't match, he will be shown with an error message.
    echo "Invalid Login Credentials.";
  }
}

//3.1.4 if the user is logged in Greets the user with message
if (isset($_SESSION['username'])){
  $username = $_SESSION['username'];
  echo "Hi " . $username . "! ";
  echo "This is the results of your inquiry.<br><br>";

echo＆＃34;用户名：$ username
＆＃34 ;; echo＆＃34;会话用户名：＆＃34;。$ _ SESSION [＆＃39;用户名＆＃39;];

//这是我假设新查询需要去的地方。 //查询名为＆＃34; data＆＃34;的另一个表并根据之前放入的$ username选择信息

回声＆＃34;

注销＆＃34 ;;

}else{
  //3.2 When the user visits the page first time, simple login form will be displayed.
}
?>
<!DOCTYPE html>

<html>
<head>
  <title>Lab Sign In Page</title>
  <link rel="stylesheet" type="text/css" href="style.css">
</head>

<body>
  <!-- Form for logging in the users -->

  <div class="register-form">
    <?php
        if(isset($msg) & !empty($msg)){
            echo $msg;
        }
     ?>

    <h1>Login</h1>

    <form action="test.php" method="post">
      <p><label>User Name :</label> <input id="username" type="text" name="username" placeholder=
      "username"></p>

      <p><label>Password   :</label> <input id="password" type="password" name="password"
      placeholder="password"></p><a class="btn" href="register.php">Signup</a> <input class=
      "btn register" type="submit" name="submit" value="Login">
    </form>
  </div>
</body>
</html>

登录后查询另一个表

2 个答案: