PHP - Using File Put Contents With Blob Image Data

Hi, I wonder whether someone may be able to help me please.

Through articles I've read on the Interent I've put together the code shown below which allows a user to upload, view and delete image files from a mySQL database.

Code: [Select]
if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

// This function makes usage of
// $_GET, $_POST, etc... variables
// completly safe in SQL queries
function sql_safe($s)
{
    if (get_magic_quotes_gpc())
        $s = stripslashes($s);
 
    return mysql_real_escape_string($s);
}
 
// If user pressed submit in one of the forms
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
    if (!isset($_POST["action"]))
    {
        // cleaning title field
        $title = trim(sql_safe($_POST['title']));
 
        if ($title == '') // if title is not set
            $title = '(No title provided';// use (empty title) string
 
        if (isset($_FILES['photo']))
        {
            @list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']);
            // Get image type.
            // We use @ to omit errors

            if ($imtype == 3) // cheking image type
                $ext="png";   // to use it later in HTTP headers
            elseif ($imtype == 2)
                $ext="jpeg";
            elseif ($imtype == 1)
                $ext="gif";
            else
                $msg = 'Error: unknown file format';
 
            if (!isset($msg)) // If there was no error
            {
                $data = file_get_contents($_FILES['photo']['tmp_name']);
                $data = mysql_real_escape_string($data);
                // Preparing data to be used in MySQL query
 
                mysql_query("INSERT INTO {$table}
                                SET ext='$ext', title='$title',
                                    data='$data'");
 
                $msg = 'Success: Image Uploaded';
            }
        }
        elseif (isset($_GET['title']))      // isset(..title) needed
            $msg = 'Error: file not loaded';// to make sure we've using
                                            // upload form, not form
                                            // for deletion

elseif($_FILES["fileupload"]["size"]/1024000 >= 10) // 10mb
{     
$msg = "<br />Your uploaded file size:<strong>[ ". $_FILES["fileupload"]["size"]/1024000  . " MB]</strong> is more than allowed 10MB Size.<br />";       
}

        if (isset($_POST['del'])) // If used selected some photo to delete
        {                         // in 'uploaded images form';
            $imageid = intval($_POST['del']);
            mysql_query("DELETE FROM {$table} WHERE imageid=$imageid");
            $msg = 'Photo deleted';
        }
 
        if (isset($_POST['view'])) // If used selected some photo to delete 
        { // in 'uploaded images form'; 
            $imageid = intval($_POST['view']); 
            mysql_query("SELECT ext, data FROM {$table} WHERE imageid=$imageid"); 
 
            if(mysql_num_rows($result) == 1) 
            { 
                $image = $row['myimage']; 
                header("Content-type: image/gif"); // or whatever 
                print $image; 
                exit; 
            } 
        } 
    }
    else
    {
        $imageid = intval($_POST['del']);
 
        if ($_POST["action"] == "view")
        {
            $result = mysql_query("SELECT ext, UNIX_TIMESTAMP(imagetime), data
                                     FROM {$table}
                                    WHERE imageid=$imageid LIMIT 1");
 
            if (mysql_num_rows($result) == 0)
                die('no image');
 
            list($ext, $imagetime, $data) = mysql_fetch_row($result);
 
            $send_304 = false;
            if (php_sapi_name() == 'apache') {
                // if our web server is apache
                // we get check HTTP
                // If-Modified-Since header
                // and do not send image
                // if there is a cached version
 
                $ar = apache_request_headers();
                if (isset($ar['If-Modified-Since']) && // If-Modified-Since should exists
                    ($ar['If-Modified-Since'] != '') && // not empty
                    (strtotime($ar['If-Modified-Since']) >= $imagetime)) // and grater than
                    $send_304 = true;                                     // imagetime
            }
 
            if ($send_304)
            {
                // Sending 304 response to browser
                // "Browser, your cached version of image is OK
                // we're not sending anything new to you"
                header('Last-Modified: '.gmdate('D, d M Y', $ts).' GMT', true, 304);
 
                exit(); // bye-bye
            }
 
            // outputing HTTP headers
            header('Content-Length: '.strlen($data));
            header("Content-type: image/{$ext}");
 
            // outputing image
            echo $data;
            exit();
        }
        else if ($_POST["action"] == "delete")
        {
            $imageid = intval($_POST['del']);
            mysql_query("DELETE FROM {$table} WHERE imageid=$imageid");
            $msg = 'Photo deleted';
        }
    }
}
?>

The problem I'm having is around the error message shown if the File Size is over the prescribed limit.

The part of the script that deals with this starts with the line: Code: [Select]
elseif($_FILES["fileupload"]["size"]/1024000 >= 10) // 10mb
Even though the file upload may fail because of the size of the file I receive the 'Error: unknown file format' message, and I'm not sure why.

I'm certainly no expert when it comes to PHP, so perhaps my lack of knowledge is letting me down.

But I just wondered if someone could perhaps take a look at this please and let me know where I'm going wrong.

Many thanks

Chris

Hello,
I am storing files that my users upload to the website as a blob in mysql database. Here is the code that does uploading:
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = mysql_real_escape_string($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
   $fileName = addslashes($fileName);
}
$query = mysql_query("INSERT INTO documents (idapplicant, fileType, fileSize, fileData, fileName)
         VALUES ('$idapplicant','$fileType','$fileSize', '$content','$fileName')",$db);

Here is the code for download:
echo "File found:".@mysql_result($result, 0, "fileID"); //always correct
$data = @mysql_result($result, 0, "fileData");
$name = @mysql_result($result, 0, "fileName");
$size = @mysql_result($result, 0, "fileSize");
$type = @mysql_result($result, 0, "fileType");

header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $data;

I can upload/download files no problem. However, I noticed problem with GIF files. All GIF files are just dark screen after downloading. All other files (PDF, JPEG) looked just fine. I compared files (before upload and after download) in HEX editor and found that the first byte on all files is set to A0 (it is added in front of all the data) and the last byte is deleted!
I have tried:
$content = addslashes($content);

instead of:
$content = mysql_real_escape_string($content);
with the same results
I also tried removing this line of code and in that case no file would be uploaded at all!
Any idea?
Thank you

Hi!

I am building a little PHP/MySQL application where pictures are uploaded and stored in MySQL in a longblob. These are special circumstances and storing images in the database in an absolute must.

Uploading is working fine. The upload script inserts the following into the field `image_data`:
base64_encode(file_get_contents($_FILES['image']['tmp_name']))

The problem is displaying the images. I cannot for the life of me make it work. I've tried the code on multiple systems with various versions of PHP and MySQL.

I have two files: one called view.php and one called show.php.

# view.php
# It gets $info['id'] from a query getting the ID of the recent-most image uploaded. 

echo '<img src="show.php?id='.$info['id'].'" alt="" />';

# show.php
# $id is determined by $_GET['id'] and passes through security checks I've omitted here.
# There is zero (not even a whitespace) output before the header()s are sent.

$query = "SELECT `image_data`,`image_mime`,`image_size` FROM `upload`.`files` WHERE `id` = '$id';";

$sql = mysql_query($query) or die(mysql_error());

$image = mysql_fetch_assoc($sql);

header('Content-Type: ' . $image['image_mime']);
header('Content-Length: ' . $image['image_size']);

echo base64_decode($image['image_data']);

The problem is that no image is displayed either in view.php or when I call show.php directly with a valid ID. I have verified that $image['image_mime'] and $image['image_size'] contain the right data.

However, if I download show.php and change extension to for example .jpg, the image is there. So the image is stored correctly in the database and $image['image_data'] is outputting the right data.

I even compared checksums for the image before and after and they're identical, so I would conclude that the error is in the outputting of the image - but I can't figure out what.

Error_report is set to E_ALL but there's nothing useful coming out.

Any ideas?

Hi everyone, i am just trying to learn php for a bit of fun really and started making a sort of 'facebook' website. I am having trouble however trying to display different users images, for example when trying to find a correct 'friend' only the image of the last result is being shown for all people with the same name... here is my code below, if anyone can help me out that would be great

file 1

$count=1; 

while ($numids>=$count){
  echo "<form method=\"post\" action=\"friendadded.php\">";
  $frienduserid=$_SESSION["passedid[$ii]"];
  $friendfirstname=$_SESSION["passedfirstname[$ff]"];
  $friendlastname=$_SESSION["passedlastname[$ll]"];
  $_SESSION['friendsuserpicid'] = $frienduserid;
  echo "<table width=\"700\" height=\"50\" border=\"1\" align=\"center\">";
  echo "<tr>";
  echo "<th></th>";
  echo "<th>First Name</th>";
  echo "<th>Last Name</th>";
  echo "</tr>";
  echo "<tr>";
  echo "<td><center>";
  echo "<img border=\'0\' src=\"frienduserpic.php\" width=\"80\" height=\"80\" align=\"middle\"/>";
  echo "</center></td>";
  echo "<td><center>";
  echo $friendfirstname;
  echo "</center></td>";
  echo "<td><center>";
  echo $friendlastname;
  echo "</center></td>";
  echo "</tr>";
  echo "</table>";
  echo "<center><input type=\"submit\" value=\"Add this friend\" name=\"Add Friend\"></center><br/>";  

  $ii=$ii+1;
  $ff=$ff+1;
  $ll=$ll+1;
  $count=$count+1; 
  
  echo "</form>";
}




file 2

session_start();
$passeduserid=$_SESSION['friendsuserpicid'];
$timespost=$_SESSION['postednum'];
$host= 
$username=
$password=
$db_name= 
$tbl_name=
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$query = mysql_query("SELECT * FROM $tbl_name WHERE picid='".$passeduserid."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header("Content-type: image/jpeg");
echo $content;


Thanks in advance

it should display the image but nothing appears. what have i done wrong?

Code: [Select]
echo "<img src=\"photo.php?id=$studentid\" alt='photo' />";
photo.php
Code: [Select]
<?php

$id= $_GET['id'];

 $getphoto= mysql_query("SELECT photo FROM Student WHERE SID=$id LIMIT 1");

$row5 = mysql_fetch_assoc($getphoto);

$photogot = $row5['PHOTO'];
header("Content-type: image/gif");

   print $photogot;
 


?>