PHP - Imagepng() -> Mysql Blob

Hi All,

This is my first post on here. I'd really appreciate any help with this, it's driving me mad.

I'm trying to create a form to save a image into a mySQL database.

I have a website hosted by UK2.net which has a mysql db with a table called gallery(name (varchar 30), size (int), type varchar(30), thePic(mediumBlob)).

I have a form with this:

Code: [Select]
<td><p><label>Pic: </label></td><td><input name="userfile" type="file" id="userfile" /></td></p>
Which when submitted actions addImage.php. The code for this looks like this:

Code: [Select]
<?php


$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));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}

$con = mysql_connect("localhost", "userName", "password") or die(mysql_error()); 

mysql_select_db("dbName", $con) or die(mysql_error()); 

$query = "INSERT INTO gallery (name, size, type, thePic) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

if (!mysql_query($query,$con))
  {
  die('Error: ' . mysql_error());
  }

echo "<br>File $fileName uploaded<br>";


?>
I get the following error message:

Quote

Warning: fopen() [function.fopen]: Filename cannot be empty in /home/hiddenje/public_html/addImage.php on line 13


Does anyone know what this is about? I've been on out friend google and a lot of people seem to be pointing to permissions but I can't seem to apply it to my scenario and just can't get it to work. Im a developer by trade, but this is my first step into the...interesting world of PHP and mySQL.

Again, I'd appreciate any help with this. I'd love someone to talk me through exactly what I'm missing or doing wrong.

Thanks in advance.

this is my 3 files which is i am using to show data of customer
but i am not able to see them their logo image please see them & help me guys

i dont know whats wrong in this.

these files i uploaded please find attachment to see them.

thnx in advanced

[attachment deleted by admin]

Okay, so here is the deal.

Have a table which stores image as blob files. Now i want to read the image width and height directly from the blob field.
Is this possible and if yes, how?

Things i tried so far;
list($size[0],$size[1],$type, $attr) = getimagesize('image.php?i=26ddd45b02859e836d13d4b9fde34281');      
print_r($size);

$img = 'image.php?i=26ddd45b02859e836d13d4b9fde34281';
echo imagesy($img);

image.php grabs the image from DB and show's it with header("Content-type: image/jpg");   
It works for just showing the images with the <img> tag.

Any ideas of help would be great!

Hi,
I'm trying to convert a charting app to  OOP.
So here I've simplified it down to just the bare bones issue of drawing a line to a png image and displaying it.

Code: [Select]

<?php
 class chart {
  var $im;
  var $backgroundColor;
  var $textColor;
  function chart() {
    $this->im= @ImageCreate (100, 200) or die ("Cannot Initialize new GD image stream");
    $this->backgroundColor = ImageColorAllocate ($this->im, 224, 234, 234);
    $this->textColor = ImageColorAllocate($this->im, 233, 14, 91);
  }
 
  function display() {
    imageline ($this->im,0,0,100,200,$this->backgroundColor);
    imageline ($this->im,100,0,0,200,$this->textColor);

    ImagePng ($this->im);
  }
}
?>


Here is the code which instantiates the class.

Code: [Select]
<?php
  include "simple_draw.php";

$mychart = new chart();
$mychart->chart();
$mychart->display();

?>



Basically depending on how I rework it, one of two things happens: a)it generates a bunch of little boxes which say fffd, or b) I get a message saying the resource is invalid or nonexistent (this error isn't coming up right now so I can't be precise.)

If this is run outside the OOP environment it works fine. No sure. I'm really trying to work with classes but this kind of problem is frustrating and admittedly makes it tempting to just go back to regular functions.

Any ideas??

Thanks,
Jeff

<?php

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

include 'helpers.php';

$foldername = uniqid();
 
$destination = "vierkant/$foldername";
$locatie = "circles/$foldername";


createDirectory($locatie);
createDirectory($destination);






    $upload_files = [];

    for ($i = 0; $i < count($_FILES['file']['name']); $i++) {
        move_uploaded_file(
            $_FILES['file']['tmp_name'][$i],
            $destination . '/' . $_FILES['file']['name'][$i]
        );

        $target_name = $_FILES['file']['name'][$i];

        // Remove the extension and replace it with .png.

         imagepng(imagecreatefromstring(file_get_contents($_FILES["file"]["tmp_name"] )), "output.png");




       // One method for this is to use the "explode" function with a dot as a delimiter.


// variable we are going to split


  print_r (explode(".",$target_name));



        // Then replace the last part with png and then stick the string back together with "implode"

        $upload_files[] = [
            'square' => $destination . '/' . $_FILES['file']['name'][$i],
            'circle' => $locatie . '/' . $_FILES['file']['name'][$i],
        ];
    }







foreach ($upload_files as $loopdata) {
         print_r($loopdata['circle']);

        toCircle($loopdata['square'],$loopdata['circle'],'450');


}

 

I have to do this but i get this error: https://ibb.co/mFzxqXL

can anyone help me?

// Remove the extension and replace it with .png.

// One method for this is to use the "explode" function with a dot as a delimiter.

// variable we are going to split:

  // Then replace the last part with png and then stick the string back together with "implode"

Hey all.  I'm getting down the basics of the GD Library's image creation processes.  I'm trying to test out creating png images.  However, where there should be transparency there is a black background.  Here's my code mostly based on the image tutorial found here at phpfreaks.

Code: [Select]
<?php
// font
$font = 'c:\windows\fonts\arial.ttf';
$fontsize = 12;

// array of quotes
$quotes = array(
"I like pie.",
"Surf's Up.",
"Smoke em if you got em.",
"Game on.",
"I need TP for my bunghole.");

// select quote
$pos = rand(0,count($quotes)-1);
$quote = $quotes[$pos];

// image
$image = imagecreatefrompng('quote.png');

$white = imagecolorallocate($image, 255, 255, 255);
$black = imagecolorallocate($image, 0, 0, 0);
imagettftext($image, 11, 0, 107, 32, $black, $font, $quote);
imagettftext($image, 11, 0, 105, 30, $white, $font, $quote);


// tell the browser that the content is an image
header('Content-type: image/png');
// output image to the browser
imagepng($image);

// delete the image resource 
imagedestroy($image);
?>
Cheers all, thanks!   

Hi,

I have a MySQL database with BLOB data (MS Word files, Excel, PowerPoint, PDF etc.). I have a show_file function that assembles the blobs to send the file to the browser. It's been working great for a decade. Now, I am looking to filter the data against XSS vulnerabilities, much like I do with strings using htmlentities(). How do you go about doing that with BLOB data? I'm assuming htmlentities() will strip out characters from the BLOB data that will render the file unusable, correct? Here is my function:

function show_file( $fileID ) {
$nodeList = array();
$fileInfo = get_record( 'FileList', 'fileID', $fileID ) or trigger_error( 'Not a valid file ID: ' . $fileID );
// Pull list of inodes
$nodes    = get_recordset( 'FileData', 'fileID', $fileID, 'blobID' );
if ( !$nodes ) { trigger_error( 'Failure to retrieve file inodes: ' . mysql_error() ); }  
while ( $node = mysql_fetch_array( $nodes ) ) { $nodeList[] = $node['blobID']; }
// Send down the header to the client
if ( strpos( $_SERVER['HTTP_USER_AGENT'], 'MSIE' ) ) { header( 'Cache-Control: public' ); }
header( 'Content-Type: ' . $fileInfo['fileType'] );
header( 'Content-Length: ' . $fileInfo['fileSize'] );
header( 'Content-Disposition: attachment; filename=' . $fileInfo['fileName'] );
// Loop thru and stream the nodes 1 by 1
for ( $z = 0; $z < count( $nodeList ); $z++ ) {
$query = 'SELECT fileData FROM FileData WHERE blobID = ' . $nodeList[$z];
if ( $result = mysql_query( $query ) ) {
echo mysql_result( $result, 0 );
} else {
trigger_error( 'Failure to retrieve file node data: ' . mysql_error() );
}
}
}


So, I am looking to do something like echo mysql_result( htmlentities($result), 0 );

Thanks for any help you may provide,
George.

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

Hi I have create a script that adds a image as blob successfully,
however now I want to create a thumbnail, I have the following code; can someone help...

# open and code into blob
$fp = fopen($safename, 'r');
$content = fread($fp, filesize($safename));
$thumb = $content;
$content = addslashes($content);
fclose($fp);

# resize accordingly...
$thumb = new resize($content, $width, $height, 300);

# the class that does the resizing (WHERE I THINK ITS GONE WRONG)
class resize {
 public $new_image_blob = "";
 function __construct($blob, $width, $height, $amount) {
   
  # the maximum width and height as set by the user
  $thumb_height_max   = $amount;
  $thumb_width_max = $amount;
   
  # maintain aspect ratio landscape or portrait
  if($width < $height) {
   $new_width = ($thumb_height_max / $height) * $width;
   $new_height = $thumb_height_max;
   $needtoresize   = ($height < $thumb_height_max);
  } else {
   $new_width = $thumb_width_max;
   $new_height = ($thumb_width_max / $width) * $height;
   $needtoresize = ($width < $thumb_width_max);
  }   
   
  # now that we have the new width and heightwe need to resize the blob
  $im = imagecreatefromstring($blob);
  $thumb = imagecreatetruecolor($new_width, $new_height);
  imagecopyresampled($thumb, $im, 0, 0, 0, 0, $new_width, $new_height, ImageSX($im), ImageSY($im));
  $this->new_image_blob = addslashes($thumb);
 }
}

# then the query below adds the code (the original blob goes in correctly (ablob) but bblob (the resized blob doesn't))
$iS = "INSERT INTO $tableb (pal, afield, bfield, cfield, dfield, efield, ffield, gfield, ablob, bblob, cblob, dblob) VALUES
('6', '$fk', '$filename', '$size', '$fileExtension', '$width', '$height', '$orientation', '$content', '$thumb->new_image_blob', '$four', '$two')";

Hi there,

How could I get the value of every 4 bytes of a BLOB? Additionally, how can I add an item to the end of a BLOB and store it back in the database?

Cheers,
George

Hi All,

Having issues uploading files larger than 1mb. This is what I have currently as default when I ran phpinfo() (working locally on my machine)...

upload_max_filesize: 432M
post_max_size: 432M
memory_limit: 8M
max_input_time: 60
max_execution_time: 30

I'm looking for the file to be converted into a blob, it works perfectly fine for files less than 1mb, but doesn't even run the mysql query above that.

Any Ideas anyone?

include("../../connect.php");
   
# these settings should help
set_time_limit(0);
   
# going in as a blob from now on
$stamp = mktime();
$safename = $_FILES['Filedata']['tmp_name'];
$filename = $_FILES['Filedata']['name'];
$size = $_FILES['Filedata']['size'];
$type = $_FILES['Filedata']['type'];
$fk = $_REQUEST['fk'];      
$sqlname = $stamp . "-" . $_FILES['Filedata']['name'];
   
# open and code in
$fp = fopen($safename, 'r');
$content = fread($fp, filesize($safename));
$content = addslashes($content);
fclose($fp);
   
$insertS = "INSERT INTO $tableb (pal, afield, bfield, cfield, dfield, efield, ffield, ablob) VALUES
('6', '$fk', '$filename', '$size', '$type', '$width', '$height', '$content')";
   
$insertQ = mysql_query($insertS);
   
print "1";   

Hi,

I am having a wierd problem and I dont know what is causing it - PHP/MySql.?

Server  - Apache on Fedora
Php - 5.3.2
MySql - 5.1.47

In the same server config I have implemented storing files into blob fields any type of file by using mysql as well as mysqli.
The one implemented by mysql is working just fine.

But the one using mysqli is not working for anything other than text files.
For image files the following error is coming -
Error interpreting JPEG image file (Not a JPEG file: starts with 0x0a 0xff)
For word/pdf etc the files are getting corrupted in such a way that the concerned application is not recognising it as a valid document.




/*MySQL
using mysql_connect
Storing*/
$name=trim($_POST['fname']);
$desc=trim($_POST['description']);
$file=$_FILES['formfile']['name'];
$filetype=$_FILES['formfile']['type'];
$filename=$_FILES['formfile']['tmp_name'];
$fileerror=$_FILES['formfile']['error'];
$filesize=$_FILES['formfile']['size'];

$fp = fopen($filename, 'r');
$content = fread($fp, filesize($filename));
fclose($fp);
if (!get_magic_quotes_gpc())
{
$content = addslashes($content);
$name=addslashes($name);
}

$query="insert into forms_master (name, description, data, filename, filesize, filetype, entrydate) values (";
$query.="'$name', '$desc', '$content', '$file', '$filesize', '$filetype', now())";

/*Retreiving*/
$id=$_GET[sha1('id')];
$form=mysql_fetch_array(mysql_query("select filename, filesize, filetype, data from forms_master where sha1(id)='$id'", $link));
header("Content-length: ".$form['filesize']);
header("Content-type: ".$form['filetype']);
header("Content-Disposition: attachment; filename=".$form['filename']);
print $form['data'];


Works absolutely fine


/*MySQL using mysqli_connect
Storing
*/
$file=$_FILES['fname'];
$name=$file['name'];
$tmp_name=$file['tmp_name'];
$size=$file['size'];
$type=$file['type'];
$fp = fopen($tmp_name, 'r');
$content = fread($fp, filesize($tmp_name));
fclose($fp);
$content = mysqli_real_escape_string($link, $content);
$name=mysqli_real_escape_string($link, $name);
 
$query="insert into document_master (name, size, filetype, data) values ('$name', '$size', '$type', '$content')";

//Retreiving
$id=$_GET[sha1('id')];
$file=mysqli_fetch_array(mysqli_query($link, "select name, size, filetype, data from document_master where sha1(id)='$id'"));
header("Content-type: ".$file['filetype']);
header("Content-length: ".strlen($file['size']));
header("Content-Disposition: attachment; filename=".$file['name']);
print $file['data'];



Note.
I uploaded an image file through phpMyAdmin.
Then tried to retreive it via the mysqli code above. The same error came.

Any help will be greatly appreciated.

Hey there, need some help badly.

I'm trying to get the size of the BLOB data in mysql. Did a search and there are like no tutorial on this. Only relevant info was from this page, at the far bottom, it just wrote this (below code). But the writer did not explain how to use it.

SELECT OCTET_LENGTH(content) FROM BloBTest WHERE filename='myimage.png'

I just give it a trial an error, wrote this, but did not work.

$LENGTH = mysql_query("SELECT OCTET_LENGTH(blobdata) FROM thedatabase WHERE id = 'selected_id'");
echo $LENGTH;

Please help....

So I have a database with BLOBs (jpeg images). I also have a handy PHP page to display the image given an image ID. This page works fine, by getting the data, setting the content-type header to "image/jpeg", and echoing the data.

Now here's the rub: I need the page to also display other information, in text / html format (i.e. show image, and below it show time the image was taken and other information also stored in the database. Because the content-type header has been set as "image/jpeg", I can't display this additional content. How can I make this work?

Thanks!

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