Re: sending a form feed to a printer

[Jason S Motes <jmotes@luhcares.org>]

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;
>             }
>           
> }
>  


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