retrieving images

[bonito1]

i have this script called view.php

<?php
  include 'db.inc';

  $file = clean($file, 4);

  if (empty($file))
     exit;

  if (!($connection = @ mysql_pconnect($hostName,
                                       $username,
                                       $password)))
     showerror();

  if (!mysql_select_db("Php_test", $connection))
     showerror();

  $query = "SELECT mime, data FROM file 
            WHERE id = $file";

  if (!($result = @ mysql_query ($query,$connection)))
     showerror();  

  $data = @ mysql_fetch_array($result);

  if (!empty($data["data"]))
  {
    // Output the MIME header
     header("Content-Type: {$data["mime"]}");
    // Output the image
     echo $data["data"];
   }
?>

after i have db.inc

<?php

// These are the DBMS credentials
$hostName = "localhost";
$username = "root";
$password = "oxioxi";

// Show an error and stop the script
function showerror()
{
   if (mysql_error())
      die("Error " . mysql_errno() . " : " . mysql_error());
   else
      die("Could not connect to the DBMS");
}

// Secure the user data by escaping characters 
// and shortening the input string
function clean($input, $maxlength)
{
  $input = substr($input, 0, $maxlength);
  $input = EscapeShellCmd($input);
  return ($input);
}

?>

and i have my page witch part of it is  this

   <?php
  include 'db.inc';
  
  $query = "SELECT id, name, mime FROM file";

  if (!($connection = @ mysql_pconnect($hostName, 
                                    $username,
                                    $password)))
     showerror();

  if (!mysql_select_db("php_test", $connection))
     showerror();
        
  if (!($result = @ mysql_query ($query, $connection)))
     showerror();
?>
    <h1>Image database</h1> 


<?php 


  if ($row = @ mysql_fetch_array($result))
  {
?>

    <table>
    <col span="1" align="right">
    <tr>
       <th>Short description</th>
       <th>File type</th>
       <th>Image</th>

    </tr>
<?php
   do 
   {
?>
    <tr>
       <td><?php echo "{$row["name"]}";?></td>         
       <td><?php echo "{$row["mime"]}";?></td>
       <td><?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td>
    </tr>
<?php
   } while ($row = @ mysql_fetch_array($result));
?>
    </table>
<?php
  } // if mysql_fetch_array()
  else
     echo "<h3>There are no images to display</h3>\n";
?>

i have do it right but for some reason it doesnt displays the pictures the type and the name are displayed normally but not the picture!! instead of them there are small icons of not existing pictures.. plz i want help i really need to do it

[spiderwell]

is view.php missing $file = $_GET['file'] ?

[requinix]

So is the database called "php_test" or "Php_test"?

[bonito1]

it is php_test but it isnt that it makes the difference ..any other ideas?

[bonito1]

is view.php missing $file = $_GET['file'] ?

cant understand what r you saying can you explain?

[spiderwell]

the file view.php, at no point in that file do you attempt to put a value into the variable $file, so it is empty

 

a value for this is being passed via a query string in the other script here:

<?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td>

 

so in view.php you need to grab that value being passed to it

so put this line at the top of view.php

<?php  include 'db.inc';  
$file = $_GET['file'];
$file = clean($file, 4);

[bonito1]

thanx very useful:)