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My code is, <?php // This block allows our program to access the MySQL database. // Stores your login information in PHP variables require_once '223login.php'; // Accesses the login information to connect to the MySQL server using your credentials and database $db_server = mysql_connect($db_hostname, $db_username, $db_password); // This provides the error message that will appear if your credentials or database are invalid if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); // Store the query string from 2.2.3.A Step 17 $query = "SELECT * FROM requests WHERE model_id=1"; // Searches the database returning results that match the query // Results come in a table stored in $requests_for_model $requests_for_model = mysql_query($query); // The mysql_num_rows function returns an integer representation of number of rows for the table passed as an argument $number_of_requests = mysql_num_rows($requests_for_model); for ($current_row = 0; $current_row < $number_of_requests; $current_row++) { // walker variable (through rows in the table) $request = mysql_fetch_row($requests_for_model); // query directed to get a table containing the one row that has a store matching the store ID in the current request $query = "SELECT * FROM store_info WHERE store_id='" . $request[1] . "'"; $store_requesting = mysql_query($query); // A table containing the store $store = mysql_fetch_row($store_requesting); // Get the only row in the table (unique store ID number guarantees this) echo $store[2]; // output the 2nd item in the row for that store, which is the city if ($current_row < $number_of_requests - 1) // Makes sure commas are only printed after results that are not the last { echo ", "; } } ?> I do not know how to fix this. Help much appreciated.