Re: image upload/display

<? echo "<form method=post enctype='multipart/form-data' action='$PHP_SELF'>\n"; echo "<input type=file name=file>\n"; echo "<input type=submit>\n"; echo "</form>\n"; if (isset($file) AND $file AND $file != "none") { $file_data = fread( fopen($file, 'r'), filesize($file) ); $result = mysql_db_query("some_db", "INSERT INTO some_table VALUES ( '$file_name', '$file_type', '$file_data' )" ); } ?> I tested this on my box, works fine, its simple. don't worrk about $file_name or $file_type whatever you have for a name in your <input type=file name=chrislee> php will make for you. ie. $chrislee_type $chrislee_name php is great here :) Chris Lee Mediawaveonline.com In article <Pine.LNX.4.21.0012211713030.15077-100000@pshg.edu.ee>, "Siim Einfeldt aka Itpunk" <siim_e@pshg.edu.ee> wrote: > Hey folks, > > I`m having hard time on getting the picture upload/displaying to work. > Actually it seems to write all the needed things to the database, but it > doesn´t show the picture. If you would, check the following(all the > files that work with the upload) and see if you find any problems with > it. > > > <?php > /* This is the most important part of the upload form *//*form.php3*/ > > echo ('<INPUT TYPE="file" NAME="form_data">'); > ?> > > <?php > /* This is the actual uploader *//*form_upload.php3*/ > include("connection.php3"); > $pildityyp="$form_data_type"; > $pildistaatus="fill the fields. "; > > if (eregi("image",$pildityyp)) { > $pildistaatus="OK"; > > } Else { > $pildistaatus="You can upload only pictures, but > you tried to upload '$pildityyp'."; > } > > if($img1=="none") { > $pildistaatus="The file field was left empty."; > > } > > echo "Picture status is $pildistaatus"; > > if ($pildistaatus=="OK") { > > $data = addslashes(fread(fopen($form_data, "r"), > filesize($form_data))); > > mysql_query("INSERT INTO > pildid (tyyp,subid,pic,picname,picnr,content_type) VALUES ('".$tyyp."', > '".$subid."', '".$data."', > '".$picname."', > '".$picnr."','".$form_data_type."')"); > if(mysql_error()==""){ > $mes="The picture has been added!"; > > }else{ > $mes="Couldn`t change the picture"; > } > > echo > "<SCRIPT>location.href='admin.php3?mes=$mes'</SCRIPT>"; > die; > > } > > } > ?> > > <?php > /* THIS GETS THE DATA FOR DISPLAYING THE PICTURE *//*showpic.php */ > > > include ("connection.php3"); > > $query = "select pic,content_type from pildid WHERE id='31'"; > $result = @MYSQL_QUERY($query); > > $data = @MYSQL_RESULT($result,0, "pic"); > $type = @MYSQL_RESULT($result,0, "content_type"); > > Header("Content-type: $type"); echo $data; > > > > ?> > > <!-- This is the file that should display the image//--> > <HTML> > <HEAD><TITLE></TITLE></HEAD> > <BODY> > <?php > echo('<IMG SRC="showpic.php">'); > ?> > </BODY> > </HTML> > > Thanks in advance, Siim Einfeldt > > >