如何在所有php文件中使用用于登录的用户名

我有一个登录页面，它的数据以表名“users”存储在数据库中。我想使用我存储的“users”中的inputuser来显示我创建并连接到同一数据库的所有php文件

我只想使用我用来登录数据库的“users”值，这样我就可以知道谁登录了。例如，我正在粘贴登录页面和另一个php页面

login.php

   <?php

$inputuser = $_POST["user"];
$inputpass = $_POST["pass"];

$user = "root";
$password = "";
$database = "ABCD";

$connect = mysql_connect("localhost", $user, $password);   // Variable that Initializes connection to database
@mysql_select_db($database) or die("Database not found");

$query = "SELECT * FROM `users` WHERE `user` = '$inputuser'";
$querypass = "SELECT * FROM `users` WHERE `pass` = '$inputpass'";

echo $password;

$result = mysql_query($query) or die(mysql_error());
$resultpass = mysql_query($querypass) or die( mysql_error());

$row = mysql_fetch_array($result);
$rowpass = mysql_fetch_array($resultpass);

$serveruser = $row["user"];
$serverpass = $rowpass["pass"];

 if ($serveruser && $serverpass){
 if(!$result){
    //header('Location: fail.html');
    die("Username Name or Password is invalid");
}

echo "<br><center>Database Output</b> </center><br><br> ";
mysql_close();

echo $inputpass;
echo $serverpass;

if($inputpass == $serverpass){
    header('Location: home.php');
} 
}else {
        header('Location: fail.html');
        echo "Sorry, bad Login";
}


?>

<?php

$taken = "false";
$database = "ABCD";
$password = "";
$username = "root";

// Connect to database
$con = mysql_connect('localhost', $username, $password) or die("Unable to connect database");
@mysql_select_db($database, $con) or die("Unable to connect");

 $paper_title =  mysql_real_escape_string($_GET['id']); 
$query = "SELECT * FROM paper WHERE ptitle = '$paper_title'"; 
$result = mysql_query($query);
$row = mysql_fetch_array($result);
mysql_close();
?>
<div align = "center">
<form method="post" action="updatedata.php" />

 <table>

 <tr>
<td>Title: </td> 
<td><input type="text" name="ptitle" value="<?php echo "$row[ptitle]" ?>">          </td>
</tr>






 </table>
 <input type ="submit" value ="Update this Information">
 </form>
</div>

---

警告：编写您自己的访问控制层并不容易，而且有很多机会使它严重出错。在这个简短的例子中，你有许多危险的原因，都是因为你鲁莽地缺乏自信。请不要编写自己的身份验证系统，当任何现代的同类产品都带有强大的内置功能时。如果可以，您应该这样做。它们不再得到维护，而是在使用。学习一下，然后考虑使用PDO，很抱歉，但是我没有从上面的评论中得到任何东西。请简单具体。我在这个程序中已经2天了。为了存储用户信息，你可以找到一些关于会话的信息。你可以找到关于@tadman警告的进一步解释
    <?php

    $inputuser = $_POST["user"];
    $inputpass = $_POST["pass"];

    $user = "root";
    $password = "";
    $database = "ABCD";

    $connect = mysql_connect("localhost", $user, $password);   // Variable that Initializes connection to database
    @mysql_select_db($database) or die("Database not found");

    $query = "SELECT * FROM `users` WHERE `user` = '$inputuser'";
    $querypass = "SELECT * FROM `users` WHERE `pass` = '$inputpass'";

    echo $password;

    $result = mysql_query($query) or die(mysql_error());
    $resultpass = mysql_query($querypass) or die( mysql_error());

    $row = mysql_fetch_array($result);
    $rowpass = mysql_fetch_array($resultpass);

    $serveruser = $row["user"];
    $serverpass = $rowpass["pass"];

     if ($serveruser && $serverpass){
     if(!$result){
        //header('Location: fail.html');
        die("Username Name or Password is invalid");
    }

    echo "<br><center>Database Output</b> </center><br><br> ";
    mysql_close();

    echo $inputpass;
    echo $serverpass;

    if($inputpass == $serverpass){
          session_start();
          $_SESSION["user"] = $row["user"];
          header('Location: home.php');
    } 
    }else {
            header('Location: fail.html');
            echo "Sorry, bad Login";
    }
?>