Re: sending a form feed to a printer

form feeds don't work in html. It is interpreted as a space. Arthur Chevalier wrote:

Hello, I am having problems getting the \f (form feed) expression to work. After a certain amount of lines print I want to start printing the next line on a new sheet of paper. Any ideas of what I could be doing wrong. Below is a sample piece of my code. The bold code is where I am trying to issue the form feed. while($row = mysql_fetch_array($dbResult)) { print("<TR>\n"); print("<TD class=\"tdborder\"><A HREF=\"show_single_server.php?server_name=$row[0]\" TARGET=\"text_float\">" . $row[0] . "</TD>\n"); print("<TD class=\"tdborder\">". $row[1] . "</TD>\n");

print("<TD class=\"tdborder\">". $row[2] . "</TD>\n");    print("<TD class=\"tdborder\"><A

HREF=\"$PHP_SELF?mid_tier=$row[3]&db_rel=0\" TARGET=\"text_float\">". $row[3] . "</TD>\n"); print("<TD class=\"tdborder\"><A HREF=\"$PHP_SELF?db_rel=$row[4]&mid_tier=0\" TARGET=\"text_float\">". $row[4] . "</TD>\n");

          $rowdata = $row[5];
          if($rowdata == "")
          {
          print("<TD class=\"tdborder\">&nbsp;</TD>\n");
          }
          else
          {                  print("<TD class=\"tdborder\">". $row[5] .  "</TD>\n");
          }
          print("<TD class=\"tdborder\">". $row[6] .  "</TD>\n");
          //print("</TR>\n");
                    if($counter == "20")
            {
            $counter = "0";
            print("</TR>\f");
            continue;
            }
            else
            {
            $counter ++;
            print("</TR>");
            continue;
            }
          }