feeding variable back into drop down

andynightingale

See: http://www.ipixels.co.uk/swift_pumps/phpbuilder.php
How can I feed the variable created by the drop down back into the <option selected> tag so that after the page refreshes back to itself your choice is still showing in the drop down?

<?php
session_start();
session_register($row["chem_description"]);
$Host = "xxxxx";
$User = "xxxxxx";
$Password = "xxxxx";
$DBName = "xxxxxx";
$TableName = "chemicals";
$Link = mysql_connect($Host, $User, $Password) or die(mysql_error());
?>

 <?php
  $qry1 = "SELECT chem_description FROM $TableName"; 
  $res1 = mysql_db_query($DBName,$qry1,$Link); 
  for ($i=0; $i <mysql_num_rows($res1); $i++) { 
  $row = mysql_fetch_array($res1); 
  printf("<option value=\"%s\">
%s 
</OPTION>
",$row["chem_description"],$row["chem_description"]); } mysql_free_result($res1); 
?>

</SELECT>
</TD>
</TR>
</TABLE>
</FORM>
</BODY>
</HTML>

sarahk

try

<?php 
  $qry1 = "SELECT chem_description FROM {$TableName}";  

  $res1 = mysql_db_query($DBName,$qry1,$Link);  

  while( $row = mysql_fetch_array($res1))
  {
     $selected = ($chemical == $row['chem_description'])? 'selected' : '';
     echo "<option value='{$row['chem_description']}' {$selected}>{$row['chem_description']}</OPTION>"
  } 
  mysql_free_result($res1);  

?>

In this the variable $selected will be an empty string unless the values match. If the tag selected is given then the page will show that value selected.

andynightingale

Thanks SarahK, that works beautifully, see:

http://www.ipixels.co.uk/swift_pumps/phpbuilder2.php