Jump to content

[SOLVED] Calender script not working with session


matt.sisto

Recommended Posts

Does anybody know why my cal.php does not work with the session, it just redirects to the loginform1.php.

 

cal.php
<?php
session_start();

  if (!isset($_SESSION['username']))
    {
   header("Location: loginform1.php");
   echo ("Welcome $session[username]");
   exit();
    }
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>

<body>

<div align="center">
   <?php
      if ((!isset($_GET["Month"])) && (!isset($_GET["Year"]))) {
       $Month = date("m");
       $Year = date("Y");
      } else {
       $Month = $_GET["Month"];
       $Year = $_GET["Year"];
      } 
      $Timestamp = mktime(0,0,0,$Month,1,$Year); 
      $MonthName = date("F", $Timestamp); 
   ?>

   <?php
      
      echo "<table border=\"0\" cellspacing=\"0\" cellpadding=\"3\">";
      echo "<tr><td colspan=\"7\" align=\"left\">Calendar Table for $MonthName, $Year</td></tr>";
      echo "<tr bgcolor=\"#999999\">"; 
      
      $daysOfWeek = array("Sun","Mon","Tue","Wed","Thu","Fri","Sat");
      
      foreach ($daysOfWeek as $value) {
        echo "<td align=\"center\"><strong><font color=\"#ffffff\">$value</font></strong></td>";
      }
      echo "</tr>"; 
      
      $MonthStart = date("w", $Timestamp);
      if ($MonthStart == 0) { 
        // if the month starts on a Sunday
        $MonthStart = 7;
      } 
      

      $LastDay = date("d", mktime(0,0,0,$Month+1, 0, $Year)); 
      $StartDate = -$MonthStart; 


      for ($k=1;$k<=6;$k++){  //print six rows for six possible weeks
        echo"<tr>"; 
        for ($j=1;$j<=7;$j++){ //seven columns per row 
          $StartDate++;
          if($StartDate < $LastDay) { //blank calendar space
            if($StartDate > 0) {
              echo"<td>$StartDate</td> \n";  
            } else {
            echo"<td bgcolor=\"#eeeeee\"></td> \n";  
           }
          } elseif (($StartDate <=1) && ($StartDate >= $LastDay)) { //date goes here
            if($StartDate >= 0) {
              echo"<td>$StartDate</td> \n";  
            }   
          }
      } 
      echo"</tr>"; 
      } //End Table Row 
      
      echo "</table>";
   ?>
   <hr width="200">
   <form action="cal.php" accept-charset="UNKNOWN" enctype="application/x-www-form-urlencoded" method="GET" >
      <?php
         echo "<select name=\"Month\">";
         for($m=1;$m<=12;$m++){  
           $selected = "";
           $longDate = date("F", mktime(0,0,0,$m,1,$Year));
           if ($Month==$m){ 
            $selected = "selected ";
           }
           echo "<option value=\"$m\" $selected>$longDate</option> \n";
         }
         echo "</select>";
         echo "<select name=\"Year\">";
         for($y=date;$y<=date+1;){  
           $selected = "";
           $longDate = date("Y", mktime(0,0,0,1,1,$y));
           if ($Year==$y){ 
            $selected = "selected \n";
           }
           echo "<option value=\"$y\" $selected>$longDate</option> \n";
         }
         echo "</select>";
      ?>
      <input type="submit" value="go">
   </form>
</div>

</body>
</html>

 

appreciate any help

Obviously, you're not properly logged in.

 

Just for the moment, copy and paste this data at the VERY TOP of your cal.php script.

<?php
session_start();
echo "<PRE>"; var_dump($_SESSION); echo "</PRE>";
exit();
?>

 

What output does it give?

Where are you setting the session?

Also, there's no need to echo stuff to the user, or exit() after you've redirected them to login.php. They won't even see it so there's not much point...

 

Unless of course the re-direct fails..

I'm setting the session in the logincheck1.php script

 

<?php
  
session_start();
require "dbconn2.php";

//Using the function mysql_real_escape_string() AFTER a connection
//has been established will clean incoming variables and prevent
//users from tampering with your SQL by inserting some of their own
$email_address = mysql_real_escape_string($_POST['email_address']);
$passwd = mysql_real_escape_string($_POST['passwd']);
$id = mysql_real_escape_string($_POST['id']);


if($id == 'client') {
   $sql = "SELECT * FROM client WHERE email_address='".$email_address
        ."' AND passwd='".$passwd."'";
   $result = mysql_query ($sql, $connection) or die ("Could not perform query $sql <br />".mysql_error());
   $row = mysql_fetch_row($result);
}
else if($id == 'consultant') {
   $sql = "SELECT * FROM consultant WHERE email_address='".$email_address
        ."' AND passwd='".$passwd."'";
   $result = mysql_query ($sql, $connection) or die ("Could not perform query $sql <br />".mysql_error());
   $row = mysql_fetch_row($result);
}
else if($id == 'organisation') {
    $sql = "SELECT * FROM organisation WHERE email_address='".$email_address
        ."' AND passwd='".$passwd."'";
   $result = mysql_query ($sql, $connection) or die ("Could not perform query $sql <br />".mysql_error());
   $row = mysql_fetch_row($result);
}
else{
   echo 'Incorrect type';
}

if ($row != null)

{
      $_SESSION['username'] = $row['first_name'];
      header("Location: index.php");
      exit();
}

else
{
	header("Location: loginform1.php");
	exit();
}
     
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>logincheck1.php</title>
</head>

<body>
</body>
</html>

 

 

if ($row != null)

{
      $_SESSION['username'] = $row['first_name'];
      header("Location: index.php");
      exit();
}

 

You might change these lines of code to this (just for testing purposes):

if ($row != null)

{
      $_SESSION['username'] = $row['first_name'];
echo "<PRE>" . var_dump($row) . "</PRE>";
/*      header("Location: index.php");
      exit();*/
}

 

Post the output of that.

array(10) { [0]=>  string(1) "7" [1]=>  string(1) "1" [2]=>  string(7) "Malcolm" [3]=>  string(1) "X" [4]=>  string(0) "" [5]=>  string(0) "" [6]=>  string(19) "[email protected]" [7]=>  string(0) "" [8]=>  string(10) "0000-00-00" [9]=>  string( "password" } 

Here's the answer:

$row = mysql_fetch_row($result);

 

If you change that to mysql_fetch_array, you should get your desired results.

 

You can also change this line:

      $_SESSION['username'] = $row['first_name'];

to this:

      $_SESSION['username'] = $row[2]; // assuming Malcolm is in the first_name column per your example

 

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.