anthony-needs-you Posted December 19, 2008 Share Posted December 19, 2008 Can somebody help me to know where to put this mysql_query("SELECT * FROM test WHERE expireDate > CURDATE()"); into this $query = "SELECT id, destination, airport, resort, hotel, board, duration, departureDate, expireDate, price, description, stars, customerRef FROM test ORDER BY price"; $result = mysql_query($query) or die('Error : ' . mysql_error()); while(list($id, $destination, $airport, $resort, $hotel, $board, $duration, $departureDate, $expireDate, $price, $description, $stars, $customerRef) = mysql_fetch_array($result, MYSQL_NUM)) { Thanks everyone Quote Link to comment https://forums.phpfreaks.com/topic/137679-solved-php-mysql_query/ Share on other sites More sharing options...
JonnoTheDev Posted December 19, 2008 Share Posted December 19, 2008 Take the query string out of the mysql_query() function and put it into the $query variable replacing the old query: $query = "SELECT * FROM test WHERE expireDate > CURDATE()"; Quote Link to comment https://forums.phpfreaks.com/topic/137679-solved-php-mysql_query/#findComment-719551 Share on other sites More sharing options...
anthony-needs-you Posted December 19, 2008 Author Share Posted December 19, 2008 Cheers have followed your instruction and it works! Thanks a lot! $query = "SELECT id, destination, airport, resort, hotel, board, duration, departureDate, expireDate, price, description, stars, customerRef FROM test WHERE expireDate > CURDATE()";[code] Quote Link to comment https://forums.phpfreaks.com/topic/137679-solved-php-mysql_query/#findComment-719554 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.