如何从数据库中获取php会话的用户类型

我正在构建一个网站，我想检索数据库中定义的用户类型，并将登录的用户重定向到特定页面。例如，如果用户是admin，它会将该用户重定向到administration页面，如果用户类型是X，它会将该用户重定向到用户类型X的特定页面

现在，我正在尝试一个临时解决方案，因为我在截止日期前（明天）没有足够的时间。我试图通过运行数据库和php查询从数据库中获取用户类型，如下所示：

             $_SESSION['login'] = true;
             $_SESSION['username']=$username;
             $query="select user_type from ".$table_name." where username='$username'";
             $result=mysqli_query($con,$query) or die('error');
             if(mysqli_num_rows($result))
             {
                $_SESSION['user_type']=$user_type;
             }

             echo json_encode( array('result'=>1));

<?php
session_start();
    include("db.php");
    $con=mysqli_connect($server, $db_user, $db_pwd,$db_name) //connect to the database server
    or die ("Could not connect to mysql because ".mysqli_error());

    mysqli_select_db($con,$db_name)  //select the database
    or die ("Could not select to mysql because ".mysqli_error());

    //prevent sql injection
    $username=mysqli_real_escape_string($con,$_POST["username"]);
    $password=mysqli_real_escape_string($con,$_POST["password"]);
    //$user_type=mysqli_real_escape_string($con,$_POST["user_type"]);
        //decrypt password


    //check if user exist already
    $query="select * from ".$table_name." where username='$username'";
    $result=mysqli_query($con,$query) or die('error');
    if (mysqli_num_rows($result)) //if exist then check for password
        {

        //Pickup password to compare with encrypted password
        $query="select password from ".$table_name." where username='$username'";
        $result=mysqli_query($con,$query) or die('error');
        $db_field = mysqli_fetch_assoc($result);
        $hashed_password=crypt($password,$db_field['password']);

         $query="select * from ".$table_name." where username='$username' and password='$hashed_password'";
         $result=mysqli_query($con,$query) or die('error');
         if (mysqli_num_rows($result))  //if passwords match then check actvation status
         {
             $query="select * from ".$table_name." where username='$username' and password='$hashed_password' and activ_status in(1)";
             $result=mysqli_query($con,$query) or die('error');
             if(mysqli_num_rows($result))
             {  


                 $_SESSION['login'] = true;
                 $_SESSION['username']=$username;
                 $query="select user_type from ".$table_name." where username='$username'";
                 $result=mysqli_query($con,$query) or die('error');
                 if(mysqli_num_rows($result))
                 {
                    $_SESSION['user_type']=$user_type;
                 }

                 echo json_encode( array('result'=>1));
             }
             else
             {
             echo json_encode( array('result'=>"$msg_email_1 <br /><a href=\"".$url."\\resend_key.php?user=".$username."\">$msg_email_2</a>."));
                // echo "User Account not yet activated.Check your mail for activation details.";
             }

         }
         else
         {
         echo json_encode( array('result'=>$msg_pwd_error));
        //   echo trim("password incorrect");
         }
        }   
    else
    {
    echo json_encode( array('result'=>$msg_un_error));
    //  die("Username Doesn't exist");
    die();
    }

?>

我可以获取用户名和登录状态，但当我试图获取用户类型时，我在_SESSIONS数组中得到一个空结果

整个php代码如下所示：

             $_SESSION['login'] = true;
             $_SESSION['username']=$username;
             $query="select user_type from ".$table_name." where username='$username'";
             $result=mysqli_query($con,$query) or die('error');
             if(mysqli_num_rows($result))
             {
                $_SESSION['user_type']=$user_type;
             }

             echo json_encode( array('result'=>1));

<?php
session_start();
    include("db.php");
    $con=mysqli_connect($server, $db_user, $db_pwd,$db_name) //connect to the database server
    or die ("Could not connect to mysql because ".mysqli_error());

    mysqli_select_db($con,$db_name)  //select the database
    or die ("Could not select to mysql because ".mysqli_error());

    //prevent sql injection
    $username=mysqli_real_escape_string($con,$_POST["username"]);
    $password=mysqli_real_escape_string($con,$_POST["password"]);
    //$user_type=mysqli_real_escape_string($con,$_POST["user_type"]);
        //decrypt password


    //check if user exist already
    $query="select * from ".$table_name." where username='$username'";
    $result=mysqli_query($con,$query) or die('error');
    if (mysqli_num_rows($result)) //if exist then check for password
        {

        //Pickup password to compare with encrypted password
        $query="select password from ".$table_name." where username='$username'";
        $result=mysqli_query($con,$query) or die('error');
        $db_field = mysqli_fetch_assoc($result);
        $hashed_password=crypt($password,$db_field['password']);

         $query="select * from ".$table_name." where username='$username' and password='$hashed_password'";
         $result=mysqli_query($con,$query) or die('error');
         if (mysqli_num_rows($result))  //if passwords match then check actvation status
         {
             $query="select * from ".$table_name." where username='$username' and password='$hashed_password' and activ_status in(1)";
             $result=mysqli_query($con,$query) or die('error');
             if(mysqli_num_rows($result))
             {  


                 $_SESSION['login'] = true;
                 $_SESSION['username']=$username;
                 $query="select user_type from ".$table_name." where username='$username'";
                 $result=mysqli_query($con,$query) or die('error');
                 if(mysqli_num_rows($result))
                 {
                    $_SESSION['user_type']=$user_type;
                 }

                 echo json_encode( array('result'=>1));
             }
             else
             {
             echo json_encode( array('result'=>"$msg_email_1 <br /><a href=\"".$url."\\resend_key.php?user=".$username."\">$msg_email_2</a>."));
                // echo "User Account not yet activated.Check your mail for activation details.";
             }

         }
         else
         {
         echo json_encode( array('result'=>$msg_pwd_error));
        //   echo trim("password incorrect");
         }
        }   
    else
    {
    echo json_encode( array('result'=>$msg_un_error));
    //  die("Username Doesn't exist");
    die();
    }

?>

---

您有
$\u会话['user\u type']=$user\u type
，但什么是
$user\u type
？您没有在任何地方初始化它，因此在
$\u会话['user\u type']
中不会存储任何内容

您必须首先从数据库获取结果，并将结果的值分配给
$user\u type
变量，然后才能存储它

$row = mysqli_fetch_assoc($result);
$user_type = $row['user_type'];

哦，是的！！这是一个很大的错误！我的大脑在这么晚的时间里几乎不工作：）然而，我仍然得到一个空的结果。我已将代码更改为以下类型：1）$query=“从“$table\u name.”中选择用户类型，其中username='$username'；$result=mysqli_query（$con，$query）或die（'error'）；虽然（$row=mysqli\u fetch\u assoc（result））{$user\u type=$row['user\u type']；$\u SESSION['user\u type']=$user\u type；}2）我将$\u SESSION部分从while条件中移除。3） 我只在while条件中放置了$\u会话部分。它们都没有工作：/Is
$user\u type
也是空的，或者只是
$\u会话['user\u type']
是空的？user\u type是空的。我找到了一个变通方法。我添加了一个虚拟页面，它将从$\u会话中检索用户名，并使用mysql命令请求用户类型，并根据用户类型立即重定向页面。我不知道这东西怎么会在同一页上不起作用，而是在虚拟页上起作用。但是，只要它起作用，而且我离最后期限非常近，我现在就使用这种方法，以后再处理。然而，你的评论使我意识到我的主要错误。非常感谢你的帮助，我真的很感激！：）