help with code for drop down menu?

Question

can anyone please help I'm going crazy I need 2 seperate drop down menus one for quantity upto 5 items and one for size from 16 to 30 can anybody help me with the html code, I really would be eternally gratefull

Answer 1

If your options are "static" (they never change), the code given above is about right, although incomplete (missing a "SELECTED" for one line).
If your options are "dynamic" (changing accoring to a source list), you can't do it with HTML: you need to build your list on the server BEFORE sending it to HTML, ie. using PhP.
Sample of a fixed in PhP
$txt .= "";
$txt .= " Title";
$txt .= " \r\n";
$txt .= " if ($lst[5] == "Mr.")
$txt .= " selected='SELECTED'";
$txt .= ">Mr.\r\n";
$txt .= " if ($lst[5] == "Miss")
$txt .= " selected='SELECTED'";
$txt .= ">Miss\r\n";
$txt .= " if ($lst[5] == "Mrs.")
$txt .= " selected='SELECTED'";
$txt .= ">Mrs.\r\n";
$txt .= " if ($lst[5] == "Dr.")
$txt .= " selected='SELECTED'";
$txt .= ">Dr.\r\n";
$txt .= " if ($lst[5] == "Rev.")
$txt .= " selected='SELECTED'";
$txt .= ">Rev.\r\n";
$txt .= " \r\n";
$txt .= " ";
$txt .= "";

Sample of a Database built (PhP + MySQL)
//=== DROP DOWN MATERIALS ===========
$txt .= " ".htmlentities('Matériel')."
";
$txt .= " ";
$link3 = dbconnect();
$query3 = "SELECT * FROM `frmmat` WHERE 1";
$list3 = mysql_query($query3) or die("Query failed : " . mysql_error());
while ($lst3 = mysql_fetch_array($list3))
{
$txt .= "" . $lst3[0] .
"";
}
mysql_free_result($list3);
mysql_close($link3);
$txt .= " ";
$txt .= " ";

Bit complex, I agree...
Need help? mail me.

Answer 2

1
2


Add as many options as you want.