PHP+Apache: Problem trying to display BLOB field pictures

matiasar · 05-09-2011, 10:17 AM

I've been stuck for someday trying to solve this problem. Still I doesn't find the clue.

I'm writing a simple blog with PHP and mysql. I'm uploading images files through form and storing them in a BLOB field:

This is the main fragment, I'm using PDO extension.

Code:

 if ( $_FILES["archivo"]["error"] === 0 && $_FILES["archivo"]["size"] > 10 && $_FILES["archivo"]["size"] < 900000 && strpos($_FILES["archivo"]["type"], "image") !== false  )
                                             {
                                                // opening temp file
                                                $fh = fopen($_FILES["archivo"]["tmp_name"],"rb") or die ("Error no pudo abrirse el archivo temporal.<br />");
                                                $imagen = addslashes(fread($fh,filesize($_FILES["archivo"]["tmp_name"])));
                                                //$imagen = fread($fh,filesize($_FILES["archivo"]["tmp_name"]));
                                                fclose($fh);

                                                try
                                                {
                                                  $stmt = $dbh->prepare('INSERT INTO sblog_post_pics SET blog_post_id=?, filename=?, filesize=?, imgdata=?, type=?');
                                                  $stmt->bindParam(1, $lastInId, PDO::PARAM_INT);
                                                  $stmt->bindParam(2, $_FILES["archivo"]["name"], PDO::PARAM_STR, 255);
                                                  $stmt->bindParam(3, $_FILES["archivo"]["size"], PDO::PARAM_INT);
                                                  $stmt->bindParam(4, $imagen, PDO::PARAM_LOB);
                                                  $stmt->bindParam(5, $_FILES["archivo"]["type"], PDO::PARAM_STR, 50);
                                                  $stmt->execute();
                                                }
                                                catch (Exception $e)
                                                {
                                                        echo "Se ha producido un error al insertar la imagen <br />";
                                                        // echo "verbose: " .$e->getmessage() ."<br/>";
                                                }
                                                unset($imagen);
                                            }
                                           else
                                            {
                                                echo "Error: hubo algun problema con la imagen suministrada.<br />Recuerde que el archivo debe ser menor a 900 kb.<br />Los tipos de archivos permitidos son gif, jpeg y png.";
                                            }

This is the table structure where image data is stored:

Code:

mysql> describe sblog_post_pics;
+--------------+--------------+------+-----+-------------------+-------+
| Field        | Type         | Null | Key | Default           | Extra |
+--------------+--------------+------+-----+-------------------+-------+
| blog_post_id | int(11)      | NO   |     | 0                 |       | 
| filename     | varchar(255) | NO   |     |                   |       | 
| filesize     | int(14)      | NO   |     | 0                 |       | 
| imgdata      | mediumblob   | NO   |     | NULL              |       | 
| type         | varchar(50)  | NO   |     |                   |       | 
| timestmp     | timestamp    | NO   |     | CURRENT_TIMESTAMP |       | 
+--------------+--------------+------+-----+-------------------+-------+

imgdata is the blob field, and type is the mime-type, which I use later in header.

Code for storing the image seems to work ok. Put I can't get image shown. The code I used for this is the following:

displaypic.php

Code:

<?php
     require '../../include/blg_conn_params.php';
     $inId = $_GET["imgid"];
     
        $dbh = new PDO('mysql:host=' .$blg_host .';dbname=' .$blg_bd, $blg_usuario, $blg_sena) or die ("No se pudo conectar a BD! de pics");

               $stmt = $dbh->prepare('SELECT type, imgdata FROM sblog_post_pics WHERE blog_post_id=? ORDER BY blog_post_id DESC');
               $stmt->bindParam(1,$inId, PDO::PARAM_INT);
               $stmt->execute();

               $stmt->bindColumn(1, $header_type, PDO::PARAM_STR);
               $stmt->bindColumn(2, $image, PDO::PARAM_LOB);
               $stmt->fetch(PDO::FETCH_BOUND);
                header("Content-Type: ".$header_type);
                echo $image;

                // Debug
                $fh = fopen("/tmp/testimg.png", "wb");
                fwrite( $fh, stripslashes($image) );
                fclose($fh);
                $fh = fopen("/tmp/displaypic.log", "w");
                fwrite($fh, $header_type."\n");
                fwrite($fh, "image data size strlen=" .strlen($image) ."\n");
                fclose($fh);

              $stmt = null;
              $dbh = null;
?>

Then to try to show the pic e call displaypic.php passing through GET the image's id:

Code:

<?php
$id="35";
?>
<html>
<head>
<title>Testing show pic</title>
</head>
<body>
<p>Just a test:</p>
<p><img src="displaypic.php?imgid=<?php print $id;?>" /></p>
</body>
</html>

I found within a lot of blogs, and code is very similar... But I can't get picture showed.

I was wondering if could be an specific apache issue... Something extra needed to be sent within http header? I don't know... Any suggestion will be very appreciated.

I'm using Debian (Lenny), PHP 5.3 and Apache 2 (apache2 - 2.2.9-10+lenny8 ).

Regards,
Matías

Guttorm · 05-09-2011, 01:05 PM

Hi

What is "/tmp/testimg.png" and "/tmp/displaypic.log"? It should be a PNG file and the MIME type.

Edit, also:

PHP Code:



$imagen = addslashes(fread($fh,filesize($_FILES["archivo"]["tmp_name"])));

Drop the addslashes:

PHP Code:



$imagen = file_get_contents($_FILES["archivo"]["tmp_name"]);

matiasar · 05-09-2011, 02:08 PM

Guttorm,

Thanks for your reply.
Really, I tried with and without addslashes function with the same result...
Both files testimg.png and displaypic.log are just for testing porpouses, testimg.png is just a file written with the image data from the blob field, just to check if the data stored is ok. And that file seems to be ok, image is good.

Guttorm · 05-09-2011, 03:01 PM

Hmm. Did you empty the browser cache? What is the output of this command:

wget --server-response http/localhost/displaypic.php?imgid=35

matiasar · 05-10-2011, 08:48 AM

Guttorm,

Thanks!
At job, I swithed my application to work storing pics in filesystem just for not to delay development. But at home I left my old version, tonight I'll try the wget test you passed me, and I'll let you know.

I had made some tests with wget, but I didn't include --server-response that's interesting.

matiasar · 05-10-2011, 08:33 PM

Guttorm,

Well, this is the result of:

wget --server-response http://localhost/.../displaypic.php?imgid=5

Code:

Petición HTTP enviada, esperando respuesta... 
  HTTP/1.1 200 OK
  Date: Wed, 11 May 2011 01:12:33 GMT
  Server: Apache/2.2.16 (Debian)
  X-Powered-By: PHP/5.3.3-7
  Connection: close
  Content-Type: image/jpeg
Longitud: no especificado [image/jpeg]
Saving to: `displaypic.php?imgid=5'

Content-Type seems to be ok, right?

If I do file command to the data saved by wget I get:

Code:

matias@retux:~$ file displaypic.php\?imgid\=5 
displaypic.php?imgid=5: data

And if I try to open that file with firefox it says that image contains error. The same occurs if I do: http://localhost/.../displaypic=imgid=5 from Firefox.

I've just found that downloaded file through wget is 1 byte bigger than the original image. I compared both files with: od -t x1 and in the beggining of downloaded file (the one using Blob) appears 0x0a as first character. Is it a LF character?
Really, that could be the reason, but still I don't find from where that character could come.