I know I have coded this incorrectly, but cannot seem to find a way to correct it. The aim, is to have a dropdown displayed populated with results from mysql database. It is currently displaying a dropdown for every address. I know why this is happening but cannot seem to correct it. Should the echo be out of the while loop? or is the placement correct but it is wrong somewhere else? would be gratful if someone could check it and inform me where i AM GOING WRONG WITH IT. Many thanks.
<?php
$customer = mysql_real_escape_string( $_GET["customer"] );
$con = mysql_connect("localhost","root","");
$db = "sample";
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db, $con);
$query_rs_select_address2 = sprintf("SELECT * FROM company_com where idcode_com = '$customer'");
$rs_select_address2 = mysql_query($query_rs_select_address2, $con) or die(mysql_error());
$row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2);
$totalRows_rs_select_address2 = mysql_num_rows($rs_select_address2);
while ($row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2))
{
$address=$row_rs_select_address2['address1_com']. " ". $row_rs_select_address2['address2_com']. " ". $row_rs_select_address2['address3_com']. " ". $row_rs_select_address2['town_com']. " ". $row_rs_select_address2['postcode_com'];
echo '<select name="customer">'.'<option value="">Select delivery address</option>'.'<option value="address">'.$address.'</option>'.'</select>';
}
?>
Try this: