output table content in listbox

[Anon]

hi,

i wanna output some table content in a listbox. i tried something..but it doesn't work. some help would be great.

\Date and Text should be side by side in the listbox
$result = mysql_query("SELECT DISTINCT Text, Date FROM table ORDER BY Date");

<form name="form1" method="post" action="">
<?
$arr = mysql_affected_rows

  print "<SELECT name='output' size='10'>\n";
  foreach ($arr as $value) {
    print ". htmlentities(stripslashes(mysql_result($result, $arr, "Text"))) .";       
  }
  print "</SELECT>"\n;

  ?>

  </form>

the selection from the listbox i wanna use with form-action to generate a result page. what is the best way to get the selectet value?

regards,
christian

[Anon]

I'm using Informix, but you should be able to get the idea from this code snippet.

Hank

++++++++++++++++++++++++

print("<tr align=center> <td><select size=1 name=pool_no>\n");
$pool_array = ifx_fetch_row($poolCR, "next");
while (is_array($pool_array))
{
print("<option value=\"");
print($pool_array["pool_no"]);
print("\">");
print($pool_array["pool_no"]);
if ( ifx_num_rows($poolCR) == 6) / Stop after getting 6 pools /
{
break;
}
$pool_array = ifx_fetch_row($poolCR, "next");
} / end while /
print("</select>\n");