On Monday 04 November 2002 16:42, Steve Jackson wrote: > I want to display an address from my database based on an orderid. Where > am I going wrong? I am still unsure how Php interacts with mySQL but am > I right in assuming it should be an array? > > My code... > > function get_shipping_address($address_array) > { > $conn = db_connect(); > $orderid = get_order_id($orderid); > $query = "SELECT * FROM orders WHERE orderid = '$orderid'"; > $row = mysql_fetch_array($query);
You don't need the foreach loop (and in fact you've mixed up its parameters so it wouldn't work anyway). > foreach ($address_array as $row) > { > $ship_name = $row["ship_name"]; > $ship_address = $row["ship_address"]; > $ship_city = $row["ship_city"]; > $ship_zip = $row["ship_zip"]; > $ship_country = $row["ship_country"]; > } Just simply: $ship_name = $row["ship_name"]; $ship_address = $row["ship_address"]; $ship_city = $row["ship_city"]; $ship_zip = $row["ship_zip"]; $ship_country = $row["ship_country"]; Here's how foreach can be used: foreach ($row as $field_name => $field_value) { echo "$field_name contains $field_value<br>"; } -- Jason Wong -> Gremlins Associates -> www.gremlins.com.hk Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * /* QOTD: "It's been Monday all week today." */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php