Re: image upload/display

Oh and to read the file is even easier. dsp_file.php <? $query = "SELECT * FROM some_table WHERE file_name = '$file_name' "; $result = mysql_db_query("some_db", $query); $r = mysql_fetch_array($result); Header("Content-Type: {$r["file_type"]}"); echo $prod_r["file_data"]; exit(); ?> somefile.html <html> <body> <img src='dsp_file.php?file_name=chrislee.jpg'> </body> </html> Chris Lee Mediawaveonline.com In article <91tfvl$osi$1@toye.p.sourceforge.net>, "none" <root@nowhere.com> wrote: > <? > echo "<form method=post enctype='multipart/form-data' > action='$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 >> >> >> > >