登錄後會話輸出用戶名和用戶類型PHP

Question

好吧，我堅持這一點。我嘗試了isset功能，但沒有任何反應...

登錄後，用戶將被重定向到特定頁面。

如果主持人是她USER_TYPE她就會被重定向如果代理的是她，她USER_TYPE就會被重定向到agent.php

到moderator.php 頁面我在這裏的index.php文件，其中登錄表單包括

<form action="index.php" method=get> 
    <?php 
    session_start(); 
    if ($_SESSION["logging"] && $_SESSION["logged"]) { 
     printme(); 
    } 
    else { 
     if (!$_SESSION["logging"]) { 
      $_SESSION["logging"] = true; 
      loginform(); 
     } 
     else if ($_SESSION["logging"]) { 
      $number_of_rows = checkpass(); 
      if ($number_of_rows == 1) { 
       $_SESSION[user] = $_GET[userlogin]; 
       $_SESSION[logged] = true; 
       echo "<h1>You have logged in successfully</h1><br/>"; 
       echo "<a href='logout.php'>Logout</a> | <a href='users.php'>Click to proceed</a>"; 
      } 
      else { 
       loginform(); 
      } 
     } 
    } 

    function loginform() { 
     print ("<center><div id='login_header'><b><font face='Arial Black' color='black' size='4px'>Sign in to Minquep!</font></b></div></cen     ter>"); 
     print("<br/><br/>"); 
     print ("<center><label>Username:</label><input type='text' name='userlogin' size='20'><br/><label>Password:</label><input type='    password' name='password' size='20'></center>"); 
     print "<br/><input type='submit' value='Submit' name='submit' class='submit'>"; 
    } 

    function checkpass() { 
     $dbHost = 'localhost'; 
     $dbUser = 'root'; 
     $dbPass = ''; 
     $dbname = 'minquep_test'; 
     $conn = mysql_connect($dbHost, $dbUser, $dbPass); // Connection Code  mysql_select_db($dbname, $conn); // Connects to database  
     $sql = "select * from users where login='$_GET[userlogin]' and password='$_GET[password]'"; 
     $result = mysql_query($sql, $conn) or die(mysql_error()); 
     $fetched = mysql_fetch_array($result); 
     if ($fetched['user_type'] == "moderator") { 
      echo '<script type="text/javascript">window.alert("You have logged in successfully!\n")</script>'; 
      print("<b><h1>hi mr.$_SESSION[user]</h1>"); 
      echo "<meta http-equiv=\"refresh\" content=\"0;URL=pages/moderator.php\">"; 
     } 
     else if ($fetched['user_type'] == "agent") { 
      echo '<script type="text/javascript">window.alert("You have logged in successfully!\n")</script>'; 
      echo "<meta http-equiv=\"refresh\" content=\"0;URL=pages/agent.php\">"; 
     } 
    } 

    function content() { 
     print("<b><h1>hi mr.$_SESSION[user]</h1>"); 
     print "<br><h2>only a logged in user can see this</h2>"; 
    } 

    function printme() { 
     echo '<script type="text/javascript">window.alert("You have logged in successfully!\n")</script>'; 
    } 

    ?> 

</form> 

從上面的代碼，這是我如何根據用戶的user_type重定向用戶到他們的特定頁面。

if ($fetched['user_type'] == "moderator") { 
    echo '<script type="text/javascript">window.alert("You have logged in successfully!\n")</script>'; 
    print("<b><h1>hi mr.$_SESSION[user]</h1>"); 
    echo "<meta http-equiv=\"refresh\" content=\"0;URL=pages/moderator.php\">"; 
} 
else if ($fetched['user_type'] == "agent") { 
    echo '<script type="text/javascript">window.alert("You have logged in successfully!\n")</script>'; 
    echo "<meta http-equiv=\"refresh\" content=\"0;URL=pages/agent.php\">"; 
} 

現在我moderator.php 裏面我只是打電話，我應該打印登錄用戶的用戶名和USER_TYPE的moderator_include.php。

moderator.php

<div id="wrapper"> 
    <div id="container"> 

     <div id="header"> 

      <?php include "moderator_header.php"; ?> 

     </div> 

它包括moderator_header.php這是

<div class="logo"> 
    <a href="moderator.php"><img class="logo_img" src="../images/minquepLOGO.png"/></a> 
</div> 

<div id="title"> 
    <img src="../images/title.gif"/> 

</div> 
<br/> 

<?php 
    session_start(); 
    if ($_SESSION["logged"] = true) { 


     print("<b><h1>hi mr. $_SESSION[user] . You are logged in as /*THE USER_TYPE GOES HERE */ </h1>"); 
    } 
?> 

我試圖輸出作爲用戶名

if (isset($_SESSION['logged'])){ 
    print("<b><h1>hi mr. $_SESSION[user] . You are logged in as /*THE USER_TYPE GOES HERE */ </h1>"); } 

但什麼也沒有發生......

關於如何輸出的用戶的USER_TYPE ......我沒有任何想法如何這一點，因爲它不是會話的一部分的index.php中發生

順便說一下我的logout.php是這樣

<?php 
    session_start(); 
    if (session_destroy()) { 
     print"<h2><B><blink>you have logged out successfully</B></blink></h2>"; 
     print "<h3><a href='index.php'>back to main page</a></h3>"; 
    } 
?> 

請幫我...謝謝

Answer 1

print("<b><h1>hi mr " . $_SESSION['user'] . "You are logged in as" . $userType . "</h1>"); } 

試試吧:)

編輯

編輯變量$ USERTYPE它應該是什麼？

Answer 2

有時PHP變得有點棘手......幾件事情要記住

1）始終啓動anyoutput會議收到的，這意味着在你的代碼的頂部，在開始會話之前，不應該有一個空格或一個空行。

2）當你有一個啓動會話的文件，並且包含另一個文件時，你不必在包含的文件中再次啓動它。

和序跟蹤會話，在任何頁面，你只想添加以下代碼：

<pre><?php print_r($_SESSION); ?></pre> 

，看看結果是什麼。