declaring variables

I used PHP a long time ago. Now I have to create a new script an have a
problem with the declaration of variables.

Problem:
The purpose of the page is to request datasheets. Which datasheets are
available is stored in a file. After the reader has selected the
datasheets these should be listed for confirmation or correction. if the
selection is confirmed an e-mail should be sent. The basic script is
running, however:

To identify the datasheets each record in the input file has an unique
ID string which I want to use as a variable. Until now this is the code:

// output for one datasheet
function bestellline ($fd, $fn)
{
echo "<tr>\n";
echo " <td>$fd</td>\n";
echo " <td align=center><input type=checkbox name=\"$fn\"
value=\"$fn\"";
if ($$fn!="") { echo "CHECKED"; };
echo ">&nbsp;</td>\n";
echo "</tr>\n";
}

and the calling sequence for this funtion:

$tmparr = array();
$fb = fopen("../daten/sicherh.dat", "r");
while (!feof($fb)) :
$tmp = fgets($fb,4096);
if (empty($tmp)) break;
$tmparr = explode("|", $tmp);
bestellline ($tmparr[1], $tmparr[0]);
endwhile;
fclose ($fb);

Something is wrong with the $$fn variable. What's my error? Or should I
search for another way to solve this?

Lothar Geyer

"Lothar Geyer" <Lothar.Geyer@EDV-Berater-Online.de> wrote in
news:3F09182A.DD2A784C@EDV-Berater-Online.de...
> function bestellline ($fd, $fn)
> {
> echo "<tr>\n";
> echo " <td>$fd</td>\n";
> echo " <td align=center><input type=checkbox name=\"$fn\"
> value=\"$fn\"";
> if ($$fn!="") { echo "CHECKED"; };
Is this a typo? ($$fn means the variable who's name is $fn, and unless
$fn=="fd" there is no $$fn in the function's scope)

Otherwise you will need to provide some more detailed information about your
problem. What is the format of the input-file? What is the unexpected output
(or lack of the same)?


Martin