PHP - Simple Remote Image Display Script Help

Well the subject line is pretty explicit.  I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. 
Code: [Select]




<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 







// Check to see if the type of file uploaded is a valid image type

function is_valid_type($file)

{

     // This is an array that holds all the valid image MIME types

     $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif");

 

     if (in_array($file['type'], $valid_types))

         return 1;

     return 0;

}

 

// Just a short function that prints out the contents of an array in a manner that's easy to read

// I used this function during debugging but it serves no purpose at run time for this example

function showContents($array)

{

     echo "<pre>";

     print_r($array);

     echo "</pre>";

}

 

// Set some constants

 

// This variable is the path to the image folder where all the images are going to be stored

// Note that there is a trailing forward slash

$TARGET_PATH = "images/";

 

// Get our POSTed variable

$image = $_FILES['image'];

 

// Sanitize our input

$image['name'] = mysql_real_escape_string($image['name']);

 

// Build our target path full string.  This is where the file will be moved to

// i.e.  images/picture.jpg

$TARGET_PATH .= $image['name'];

 

// Make sure all the fields from the form have inputs

if ( $image['name'] == "" )

{

     $_SESSION['error'] = "All fields are required";

     header("Location: member.php");

     exit;

}

 

// Check to make sure that our file is actually an image

// You check the file type instead of the extension because the extension can easily be faked

if (!is_valid_type($image))

{

     $_SESSION['error'] = "You must upload a jpeg, gif, or bmp";

     header("Location: member.php");

     exit;

}

 

// Here we check to see if a file with that name already exists

// You could get past filename problems by appending a timestamp to the filename and then continuing

if (file_exists($TARGET_PATH))

{

     $_SESSION['error'] = "A file with that name already exists";

     header("Location: member.php");

     exit;

}

 

// Lets attempt to move the file from its temporary directory to its new home

if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))

{

     // NOTE: This is where a lot of people make mistakes.

     // We are *not* putting the image into the database; we are putting a reference to the file's location on the server

     $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')";

     $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());

     header("Location: images.php");

echo "File uploaded";

     exit;

}

else

{

     // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to

     // Make sure you chmod the directory to be writeable

     $_SESSION['error'] = "Could not upload file.  Check read/write persmissions on the directory";

     header("Location: member.php");

     exit;

}







        } //End of if session variable id is not present.
 


?> 


The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database.
Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work.


Code: [Select]

//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

$sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' ");
$row = mysql_fetch_assoc($sql);
$imagebytes = $row['image'];
header("Content-type: image/jpeg");
print $imagebytes;







Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which.  Can anyone help??

I came up with the following script for displaying an uploaded picture on a webpage but for some reason I can't pinpoint, the picture won't be displayed.  I checked my database from php myadmin and sure enough, the picture was successfully uploaded but somehow, the picture won't be displayed.  So here is the display script.  I named it display_pic.php

Code: [Select]

<?php


//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

                             //address error handling

                             ini_set ('display_errors', 1);
                             error_reporting (E_ALL & ~E_NOTICE);

                             //include the config file
                              require('config.php');

                             $image = stripslashes($_REQUEST[imname]);
                             $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' ");
                             $row = mysql_fetch_assoc($rs);
                             $imagebytes = $row[image];
                             header("Content-type: image/jpeg");
                             print $imagebytes;

                            
 ?>


The tag on the html page that's supposed to display the picture reads something like this
<img src="display_pic.php"   width="140" height="140">



And just in case this might help, I will include the image upload script below, which I think worked just fine because my values were successfully inserted into the database.
Code: [Select]
<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 
// Make sure the user actually 
// selected and uploaded a file
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

      // Temporary file name stored on the server
      $tmpName  = $_FILES['image']['tmp_name'];  
       
      // Read the file 
      $fp      = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);
      

      // Create the query and insert
      // into our database.
      $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')";
      $results = mysql_query($query);
      
      // Print results
      print "Thank you, your file has been uploaded.";
      
}
else {
   print "No image selected/uploaded";
}

// Close our MySQL Link
mysql_close();


} //End of if statmemnt.
 


?> 



So any insights as to why the script fails to display the image? Any help is appreciated.

I had posted a similar post  a few days back, about an display script which is supposed to retrieve a stored image from my database and displays it on an html page, but fails to do so.  Someone suggested that I  upload my image onto a folder on the server and then save the image path name in my database.  I found this script(below)  which does just that.  It uploads the image into a folder on the server called images and stores the image path into a table in my database called images and I know this script works because i saw the file saved in the images folder and the path name inserted into the images table.
Code: [Select]


<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 







// Check to see if the type of file uploaded is a valid image type

function is_valid_type($file)

{

     // This is an array that holds all the valid image MIME types

     $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif");

 

     if (in_array($file['type'], $valid_types))

         return 1;

     return 0;

}

 

// Just a short function that prints out the contents of an array in a manner that's easy to read

// I used this function during debugging but it serves no purpose at run time for this example

function showContents($array)

{

     echo "<pre>";

     print_r($array);

     echo "</pre>";

}

 

// Set some constants

 

// This variable is the path to the image folder where all the images are going to be stored

// Note that there is a trailing forward slash

$TARGET_PATH = "images/";

 

// Get our POSTed variable

$image = $_FILES['image'];

 

// Sanitize our input

$image['name'] = mysql_real_escape_string($image['name']);

 

// Build our target path full string.  This is where the file will be moved to

// i.e.  images/picture.jpg

$TARGET_PATH .= $image['name'];

 

// Make sure all the fields from the form have inputs

if ( $image['name'] == "" )

{

     $_SESSION['error'] = "All fields are required";

     header("Location: member.php");

     exit;

}

 

// Check to make sure that our file is actually an image

// You check the file type instead of the extension because the extension can easily be faked

if (!is_valid_type($image))

{

     $_SESSION['error'] = "You must upload a jpeg, gif, or bmp";

     header("Location: member.php");

     exit;

}

 

// Here we check to see if a file with that name already exists

// You could get past filename problems by appending a timestamp to the filename and then continuing

if (file_exists($TARGET_PATH))

{

     $_SESSION['error'] = "A file with that name already exists";

     header("Location: member.php");

     exit;

}

 

// Lets attempt to move the file from its temporary directory to its new home

if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))

{

     // NOTE: This is where a lot of people make mistakes.

     // We are *not* putting the image into the database; we are putting a reference to the file's location on the server

     $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')";

     $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());

     header("Location: images.php");

echo "File uploaded";

     exit;

}

else

{

     // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to

     // Make sure you chmod the directory to be writeable

     $_SESSION['error'] = "Could not upload file.  Check read/write persmissions on the directory";

     header("Location: member.php");

     exit;

}







        } //End of if session variable id is not present.
 


?> 











Now the display image script accompanying this insert image script is shown below. Code: [Select]
<?php

// Get our database connector

require("config.php");

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"

      "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

 

<html xmlns="http://www.w3.org/1999/xhtml">

    <head>

        <title>Dream in code tutorial - List of Images</title>

    </head>

 

    <body>

     

        <div>

 

            <?php   

                // Grab the data from our people table

                $sql = "SELECT * from images  WHERE  image_cartegory = 'main'";

                $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error());

 

                while ($row = mysql_fetch_assoc($result))

                {

                    echo "<div class=\"picture\">";

                    echo "<p>";

 

                    // Note that we are building our src string using the filename from the database

                     echo "<img src=\"images/ " . $row['filename'] . " \" alt=\"\" />";

                   

                    echo "</p>";

                    echo "</div>";

                }



            ?>

         

        </div>

    </body>

</html>



Well, needless mentioning, the picture isn't displayed.  Just a tiny jpeg icon is displayed at the top left hand corner of the page.  Figuring out that the problem might be the lack of a header that declares the image type, I add this line
header("Content-type: image/jpeg");
but all it does is display a blank white page, without the tiny icon this time.  What am I missing? Any clues??

I have created 5 websites under 5 different domains. Contents of this websites' are similar and using a same template for each one. Now I need to create an admin panel to control these websites. PHP and MySql I will use for this backend.   My problem is how can I manage these five website with one backend? Reason is I will use different domain for my backend. My all 5 client will use this same backend system to manage their own website.   So can I know from the professionals here, is it possible to display mysql data on these 5 website. If it is possible then how? Any links to article or tutorials would be welcome and appreciated.   NOTE: I checked

mysql federated storage engine

but no idea? Is it the way where do I need to go?   Thank you.

Hi all,
I have two issue with script.
1. It works in PhpEd and with apache but doesn't work at remote server with apache.
    Error is well known -  "Warning: Cannot modify header information - headers already sent "

2. when I added more than 20 records like $var = $_POST['var']; it stops work local. Error is same.


adminka_wrapp.php
Code: [Select]
<?php
$addFormatName = $_POST['addFormatName'];
$addFormatDes  = $_POST['addFormatDes'];
$page               = $_GET['page'];


if (isset($addFormatName) && isset($addFormatDes))
  {
      $page = 'add_formats';
  } else if ($page == NULL)
    {
        $page =  "user_access_log";
    }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <link rel="stylesheet" type="text/css" href="../style/adminkaview.css">       
    </head>
    <body>
        <div class="col-wrap1">
            <div class="col-wrap2">
                <div class="col1">
                    <div class="content" id="c1">
                        <p><a href="?page=view_add_format">Format and extensions</a></p>
                    </div>
                </div>
                <div class="col2">
                    <div class="content" id="c2">
<?php require("$page.php");?>
                    </div>
                </div>
                <div class="clear"></div>
            </div>
        </div>
    </body>
</html>

view_add_format.php
Code: [Select]
<?php
require_once "../function.php"; 

$q = connect("SELECT
                      `formats`.*
              FROM
                       `xxx`.`formats` 
            ");
echo "<form method='post' action='adminka_wrapp.php'>
     <table>
            <tr>
                <td>id</td>
                <td>Formats Name</td>
                <td>Description</td>
            </tr>";             
  while($rowResult = $q->fetch_assoc())
  {
    echo "<tr>";  
    $id            = $rowResult["id"];  
    $formatName    = $rowResult["f_name"];  
    $description   = $rowResult["description"];
    echo    "<td>".$id."</td>"
           ."<td>".$formatName."</td>"
           ."<td>".$description."</td>";
    echo "</tr>";
  }
  
echo " <tr>
           <td>
               <input type='submit' value='AddFormat' name='submitAddFormat'>
           </td>
           <td>
                <input type='text' name='addFormatName' maxlength='20' size='5'>
            </td>
            <td>
                <input type='text' name='addFormatDes' maxlength='20' size='5'>
            </td>    
                ";                
echo "</table>
      </form>";          
?>

add_formats.php
Code: [Select]
<?php
require "../function.php"; 
 
  if (isset($addFormatName) && isset($addFormatDes) )
 {
   
        $q  = connect("INSERT INTO  
                                   `xxx`.`formats`     
                                   (`id`
                                   ,`f_name`
                                   ,`description`
                                   )
                       VALUE
                                   (NULL
                                   ,'$addFormatName'
                                   ,'$addFormatDes'
                                   )               
                   ");   
        header("Location: adminka_wrapp.php?page=view_add_format");
        exit();     
 }
?>

function.php
Code: [Select]
<?php
function connect($query)
{
  $db = new mysqli('127.0.0.1', 'xxx', 'xxx', 'xxx');
  if (mysqli_connect_errno()) 
     {
         printf("Connect failed: %s\n", mysqli_connect_error());
         exit();
     }
$result = $db->query($query);
$db->close();
return $result;   
}
?>