Question

在我的login.php中，我将用户名和用户ID存储在会话中。登录后，用户选择他们的页面，一旦他们通往他们的页面，他们可以选择一个只需要他们的名字的讲师姓名，而不是其他讲师。我知道所选的讲师姓名需要存储在会话中。之后，我必须与用户ID或用户名匹配，以便控制用户可以看到的内容。问题是如何匹配login.php和`lecturer.php中的这些会话。我应该为会话创建单独的文件吗？

的login.php

<?php

 require ('connect.php');

  $username = $_POST['username'];
  $password = $_POST['password'];


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

   if ($username && $password) {
       $check = mysql_query("SELECT * FROM users WHERE username='".$username."' AND password= '".$password."'");
       $rows = mysql_num_rows($check);


    if(mysql_num_rows($check) != 0){

      session_start();
      $run_login =mysql_fetch_array($check);
      $uid = $run_login['id'];
      $_SESSION['uid'] = $_POST['uid'];
      $_SESSION['username']=$_POST['username'];
      header("location:../../statistics/home.php");

    }
    else{
      die("Could not find the Username or password.");
   }
 }
  else {
     echo "Please fill all the fields.";
 }
}
?>

lecturer.php

 <?php 

   include 'connect.php';

   $year = mysql_real_escape_string($_POST['year']);
   $lecturer = mysql_real_escape_string($_POST['lecturer']); // Don't forget to handle the SQL Injections ...
   $years     = array(
        2005,
        2006,
        2007
   );
   $lecturers = array(
        'dimopoulos',
        'lagkas',
        'kehagias',
        'chrysochoou'
   );
  if(isset($_POST['submit'])){

        if (in_array($lecturer, $lecturers) && in_array($year, $years)) {

                 $sql = "SELECT unit_name,a1,a2,a3,l1,l2,l3,l4,l5,l6,l7,lavg,r1,r2,u1,u2,u3 FROM $lecturer WHERE year=$year";

            $result = mysql_query($sql);
        }

        else {
            echo "No data found";
        }

  }
  else{
      echo "Please select";
  }

  ?>
 <html>
 <head>
    <link rel="stylesheet" type="text/css" href="../../statistics/style.css">
 </head>
 <body>
   <div id="container">
   <table id="table" width="900" border="1" cellspacing="1">
   <tr>
    <td>Unit Name</td>
    <td>A1 </td>
    <td>A2 </td>
    <td>A3 </td>
    <td>L1 </td>
    <td>L2 </td>
    <td>L3 </td>
    <td>L4 </td>
    <td>L5 </td>
    <td>L6 </td>
    <td>L7 </td>
    <td>LAVG </td>
    <td>R1 </td>
    <td>R2 </td>
    <td>U1 </td>
    <td>U2 </td>
    <td>U3 </td>


   </tr>

   <?php
       while($unit=mysql_fetch_assoc($result)){
        echo "<tr>";
        echo "<td>".$unit['unit_name']."</td>";
        echo "<td>".$unit['a1']."</td>";
        echo "<td>".$unit['a2']."</td>";
        echo "<td>".$unit['a3']."</td>";
        echo "<td>".$unit['l1']."</td>";
        echo "<td>".$unit['l2']."</td>";
        echo "<td>".$unit['l3']."</td>";
        echo "<td>".$unit['l4']."</td>";
        echo "<td>".$unit['l5']."</td>";
        echo "<td>".$unit['l6']."</td>";
        echo "<td>".$unit['l7']."</td>";
        echo "<td>".$unit['lavg']."</td>";
        echo "<td>".$unit['r1']."</td>";
        echo "<td>".$unit['r2']."</td>";
        echo "<td>".$unit['u1']."</td>";
        echo "<td>".$unit['u2']."</td>";
        echo "<td>".$unit['u3']."</td>";
        echo "</tr>";    
    }
?>
  </table>
  </div>
  </body>
  </html>



    lecturerForm.php


    <form name="myform" action="lecturer.php" method="POST" >
<b>Lecturers:<b/>
<select name="lecturer">  
<option value="Choose">Please select..</option>
<?php
    $sql=mysql_query("SELECT lec_name FROM lecturer");

    while($row=mysql_fetch_array($sql)){

        echo "<option value='".$row['lec_name']."'>".$row['lec_name']."</option>";
    }
    ?> 
</select><br/><br/>

<b>Year:<b/>
<select name="year"> 
<option value="Choose">Please select..</option>
<option value="2005">2005</option> 
<option value="2006">2006</option>
<option value="2007">2007</option></select><br/><br/>


<br/>
<input type="submit" name="submit" value="Submit">
<input type="reset" name="reset" value="Clear">

</form>

Answer 1

将session_start()放在lecturer.php page的开头。

注意：
您在login.php中有一个错误的var名称，其中设置了$_SESSION['uid']：

$uid = $run_login['id']; $_SESSION['uid'] = $uid; // not $_POST['uid'];

对用户的php会话感到困惑？

1 个答案: