At 09:10 08/09/01 +0700, Ahmad Gunawan wrote:
>1. Kalau saya punya input type BUTTON, gimana caranya supaya
>kalau di-click bisa jump ke whatever.asp misalnya, atau kalau
>bisa sekalian jump ke whatever.asp?x=v1&y=v2 ???
Input type image kalau di click memang langsung submit formnya kok
Jadi biasa aja <input type="image" border="0" name="Search" 
src="file_buitton.gif" width="25" height="25">

>2. Kalau saya punya sebuah field (kolom) dgn tipe image (blob)
>pada suatu table dalam SQL Server 7.0, bisa enggak ditampilkan
>dalam document.html, kalau bisa gimana caranya ???
>
>3. Input type apa yg digunakan pada form supaya user bisa meng-
>entry kan gambar/foto ???
>
>4. Kalau nomor 3 tidak bisa, mungkin dgn cara input type TEXT,
>dgn yg di-entry-kan berupa nama file gambar (bmp/gif/jpg), lalu
>proses upload file tadi ke database bagaimana ???

Mmmmm..... kalau saya dng MySQL/Perl logikanya begini (pakai CGI.pm):
* User input nama file yg akan diupload (Form content type=multipart)
$CONTENT .=
                 $q->start_multipart_form().
                 "Enter the file to process:<br>"
                 .$q->filefield('image_to_upload','',25)
                 .$q->reset.$q->submit('submit','Process File');

* Script Perl A :
         * upload file itu ke suatu direktori tertentu di server

sub upload {
         my($file,$server_dir, $nama_file_upload) = @_;
         # $file = $q->param('filename') i.e. MUST BE a filehandle
         my($CONTENT);
         $nama_file_upload = $file unless ($nama_file_upload);
         $nama_file_upload =~ s!^.*(\\|\/)!!;
         print $q->h5($nama_file_upload );
         open (FILE,">$server_dir/$nama_file_upload");
         binmode(FILE);
         while (read($file, my $buffer,1024))
         {
                 print FILE $buffer;
         }
         close (FILE);
         #$CONTENT .= $q->p('File uploaded as ',$q->b($nama_file_upload));
         return $nama_file_upload;
}
         $image_file=$q->param('image_to_upload');
         $file_name = &upload($image_file,'/home/your_server/www/images');

         * Menyimpan nama file itu di database
         $dbh->do(qq~insert into $sql_table set file_image='$file_name'~);

* Script Perl B :
         * mengambil nama file dr database & tampilkan sebagai code HTML
         $sth = $dbh->prepare( qq~select file_image from $sql_table ~);
         $sth->execute();
         $sth->bind_columns( undef, \$file_image);
         while($sth->fetch) {
                 print $q->img({-src=>$file_image}),'<br>';
         }

Coba ambil aja logikanya & konversi sintax2nya buat ASP/SqlServer


Best Regards,
Ferry I



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