Hello, I have a very simple routine for inserting blob data (a jpg) into a MySQL database. I have this page deployed on both my local - WINDOWS server, and my production FreeBSD UNIX server. Both of these implementations point at the same production MySQL server. Running the routine from my local server inserts the BLOB with no problems into the DB, however runing the same exact script from the production server produces the following results:

Warning: fopen(C:\test\test.jpg): failed to open stream: No such file or directory in /urs/www/users/millersh/admin/Edit.php on line 25

The strange this is how this works great on my local windows server. Obviously there is a php value or something that needs to be configured on the production (UNIX) server - I have already tried the following:

php_value magic_quotes_gpc "1" and also "On"
php_value upload_max_filesize "10M"
php_value file_uploads "1" and also "On"
php_value upload_tmp_dir "/usr/www/users/millersh/temp/" (this directory has write access on the server)

The windows server PHP info is as follows:
http://sheahome1.no-ip.com/PHPInfo

The UNIX php info is as follows:
http://www.millershea.com/php_info

PLEASE HELP!!!! I thank you kindly in advance.

Not trying to state the obvious, but your message states that it is trying to open c:\test\test.jpg where the Unix environment doesnt use drive letters for partitions. Without seeing the section of code from your Edit.php file, thats what it looks like to me. Is this file being submitted from a form, or called directly? This doesnt sound like a server config problem to me.

Thanks for the reply - I'm somewhat of a PHP newbie - so you may not be stating the obvious. Do I need to first upload the file to a temp directory on the unix server, and THEN pipe the file into the MySQL DB? The file is being submitted from a file browser control on the page and the a postback - the file comes in as one of the $HTTP_POST_VARS[] Is this not what I want to be doing here? Here is the relavant code from Edit.php:

if (IsSet($HTTP_POST_VARS["ImageUpload"])) {
        $EmployeeID=$HTTP_POST_VARS["EmployeeID"];
                //following is the var from the <input type="file">
        $file=$HTTP_POST_VARS["file"];
        $filetyp = substr($file, -3);
        if ($filetyp == 'gif' OR $filetyp == 'jpg') {
            $handle = fopen($file,'r');
            $file_content = fread($handle,filesize
                            ($file));
            fclose($handle);
            $encoded = chunk_split(base64_encode
                            ($file_content));
            $sql = "Update Employees SET Pic = '$encoded'
                            WHERE EmployeeID = '$EmployeeID";
            mysql_query($sql);
        }
        mysql_close($connection);
        exit;

The section of Edit.php that contains the file upload control is basically as follows:

<input type="file" name="file">&nbsp;&nbsp;<input type="submit" name="ImageUpload" value="Upload">

Thanks again for the help - I think you could be right here, maybe you can be certain with this additional info. Thanks!