Re: Frustrated, puzzled and confused: outputting images from a database

Here's what I did. for PHP, I have only Magic-Quotes-gpc ON, the other two are off. The following takes image from form and uploads to teh DB: $temp_pic = fopen($form_data, 'r'); $temp_pic2 = fread($temp_pic, filesize($form_data)); $bin_pic1 = addslashes($temp_pic2); $bin_pic = addslashes($bin_pic1); $add_to_db = add_to_db( $_fname, $_lname, $_email, $_uname, $_pword, $_sex, $_age, $_city, $_state, $_orientation, $_status, $_race, $_religion, $_height, $_weight, $_hair, $_eyes, $_education, $_occupation, $_interests, $_match, $_other, $_date, $bin_pic, $form_data_name, $form_data_type, $dberror ); to pull from the database, I do this: if ( ! $pic == "" ) { $bgcolor4 = "gray"; //table background color print "<td><table border=\"0\" cellpadding=\"5\" bgcolor=\"$bgcolor4\">\n\n<tr>\n<td bgcolor=\"$bgcolor2\"><center>Picture</center></td>\n</tr>\n<tr>\n<td bgcolor=\"$bgcolor\">"; print "<a href=\"" . $home_url . "pullimg.php?uname=$_uname\" target=\"_blank\"><img src=\"pullimg.php?uname=$_uname\" width=\"150\"><br />Click to see full size</a></td>\n\n"; print "</tr>\n</table></td>\n"; } And finally, here is the source of pullimg.php if ( ! mysql_select_db( $db, $db_connect) ) { $dberror = mysql_error(); return false; } $search_result = mysql_query( "SELECT bin_pic, filename, filetype FROM $db_table WHERE _uname = '".$uname."'", $db_connect ); $num_rows_returned = mysql_num_rows( $search_result ); if ( $num_rows_returned == 1 ) { $results_returned = mysql_fetch_array( $search_result ); $pic_data = $results_returned['bin_pic']; $pic_name = $results_returned['filename']; $pic_type = $results_returned['filetype']; } else { print "There has been an error!<br />"; exit; } Header("Content/type: $pic_type"); $pic_content = stripslashes($pic_data); echo($pic_content); mysql_close( $db_connect ); hope that helps! Jon ----- Original Message ----- From: "Jason Murray" <Jason.Murray@melbourneit.com.au> To: <php-general@lists.php.net> Sent: Tuesday, December 12, 2000 6:20 PM Subject: [PHP] Frustrated, puzzled and confused: outputting images from a database > Hi folks, > > I'm storing image data (either image/gif or image/[p]jpeg) (and its > mimetype) in a BLOB column of a MySQL database. > > I have written a PHP script to pull the image data and mime data for > a particular ID in the database, and output. This does seem to be > working (touch wood). > > Before someone suggests writing the file out and storing the filename, > this won't work. We're on load-balanced servers without any file sharing > means between them. > > My problem is that when I pull the image data from the database, I'm > not sure what to do in order to output it. I've tried: > > <? > Header("Content-Type: image/gif"); > echo $imagedata; > Exit(); > ?> > > But this only seems to produce a broken image in a browser, and when > I look at it on a command line it *seems* to work. Please excuse crap > below: > > [jason@phpdev htdocs]# curl > > "http://<<hostname>>/dbgfx.php3?affcode=21&logotype=new" > ...$OEr°IE(4'BY$ÿ¼"Hç > É > ')¤oere/L(_-T)³WK 7o?q'Ë Á^­HEUR0°L îà > %vñ¢hh"f-°÷E...f~!?r¸!...¢cÚ:~Xâ?¨¢?,v(¡¾ý"Ø{4 > ō8 > 8ÔB|]ô-8ü~ÙyEUR5
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Ùy÷]\Õ11 > > oef`Hbh>Ó&oeOEÙ5-'Y¦oe^ > > ¨ádã8R1ã > BV#]ÆÒ"1uÈQ0,3¥S~6*ÁÊX > SB > LHc'GåËh½ > ¤»Fé~^10 fTh´²×0e > 76"} > pGPÍÿz > Rúò-¨À-±s8?áZ7...ÑÄ{vÉË1[jason@phpdev htdocs]# > > > Do I need to use image functions? I thought I would be able to get by > without them. Can I just echo the binary data to the browser? > > I'm completely puzzled. This *should* be working. Please help... > > Jason > > -- > Jason Murray > jasonm@melbourneit.com.au > Web Design Team, Melbourne IT > Fetch the comfy chair! > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >