matt.sisto Posted May 2, 2009 Share Posted May 2, 2009 Hi all, I am stuggling to define the value of a variable by using a form input to select a value form a table, to then INSERT into another table. Any help appreciated <?php session_start(); if (!isset($_SESSION['username'])) { header("Location: login.php"); exit(); } require "dbconn2.php"; $email_address = mysql_real_escape_string($_POST['email_address']); $unit = $_POST['unit']; $service = $_POST['service']; $quantity = $_POST['quantity']; $con_id1 = $_POST['con_id1']; $con_id2 = $_POST['con_id2']; $con_id3 = $_POST['con_id3']; $extra_info = $_POST['extra_info']; $start_time_hr = $_POST['start_time_hr']; $start_time_min = $_POST['start_time_min']; $time_sec = '00'; $address_first_line = mysql_real_escape_string($_POST['address_first_line']); $post_code = $_POST['post_code']; $country = mysql_real_escape_string($_POST['country']); $month = mysql_real_escape_string($_POST["month"]); $day = $_POST["day"]; $year = $_POST["year"]; $event_start = $year."-".$month."-".$day." ".$_POST["event_start"]; $start_time = $start_time_hr.":".$start_time_min.":".$time_sec." ".$_POST["start_time"]; if ($unit == 'Day' && $quantity >= 1 && $quantity <= 7) { $d = (int) $quantity; $event_end = strtotime("+$d Days", strtotime($event_start)); $end_date = date('D j M Y', $event_end); $end_Time =$start_time; } else if ($unit == 'Hour' && $quantity >= 1 && $quantity <= 7) { $h = (int) $quantity; $end_time = strtotime("+$h Hours", strtotime($start_time)); $end_Time = date('G:i:s', $end_time); $end_date = $event_start; } $sql = "SELECT client_id FROM client WHERE email_address= '$email_address'"; $result1= mysql_query($sql, $connection) or die ("Couldn't perform query $sql <br />".mysql_error()); while ($row = mysql_fetch_assoc($result1)) { $client_id = $row; } $sql = "INSERT INTO calendar_events VALUES (0,'".$event_start."','".$end_date."','".$client_id."','".$service."','".$unit."','".$quantity."','".$start_time."','".$end_Time."','".$con_id1."','".$con_id2."','".$con_id3."','".$address_first_line."','".$post_code."','".$country."','".$extra_info."')"; echo $sql; $result = mysql_query ($sql, $connection) or die ("Couldn't perform query $sql <br />".mysql_error()); header("Location: bookingform.php"); exit(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Book</title> </head> <body> </body> </html> Thanks and regs. Quote Link to comment Share on other sites More sharing options...
Ken2k7 Posted May 2, 2009 Share Posted May 2, 2009 You have NO form fields. Quote Link to comment Share on other sites More sharing options...
matt.sisto Posted May 3, 2009 Author Share Posted May 3, 2009 My form is on a previous page which posts the values to the script. Quote Link to comment Share on other sites More sharing options...
fenway Posted May 4, 2009 Share Posted May 4, 2009 Show us the value of $sql. Quote Link to comment Share on other sites More sharing options...
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