[SOLVED] Small insert problem

[waynew]

Okay, basically, I'm allowing users to insert orders into a database. Problem is, the price variable doesn't seem to end up in the database, even though it is coming through the POST and making its way into the query. The table column amountPaid is set to double.

 

$session_id = session_id();
$data_ok = true; //assumed true unless proven otherwise

if (isset($_POST['book']) && $data_ok){
$event = mysql_real_escape_string($_POST['event']);
$date = mysql_real_escape_string($_POST['date']);
$time = mysql_real_escape_string($_POST['time']);
$datetime = $date.' '.$time;
$num_children = mysql_real_escape_string($_POST['num_children']);
$num_adults = mysql_real_escape_string($_POST['num_adults']);
$price = mysql_real_escape_string($_POST['price']); 
echo $price;

$insert = mysql_query("
INSERT INTO Orders (sessionId,eventId,noChildren,noAdults,dateTime,amountPaid)
		VALUES ('$session_id','$event','$num_children','$num_adults','$datetime',$price)") or die(mysql_error());


}

[MasterACE14]

	

$insert = mysql_query("
INSERT INTO Orders (sessionId,eventId,noChildren,noAdults,dateTime,amountPaid)
VALUES ('$session_id','$event','$num_children','$num_adults','$datetime','$price')") or die(mysql_error());

 

Should work now. You were missing the single quotes around $price in the query.

 

Regards, ACE

[waynew]

Sorry, I had removed them when trying to debug. No, I'm just after finding out that the problem doesn't rest in the insert; it rests in the output. Whoops. Thanks anyway.

[MasterACE14]

Well your script looks fine.

 

Try adding error_reporting(E_ALL); to the top of the script and see what comes up.

[waynew]

It's ok, I found the problem. Basically, the column is called amountpaid and not amountPaid. The programmer before me had kept with the camelback naming convention throughout all of his columns... expect for that one column. Thanks!

[MasterACE14]

that certainly explains it then lol.

 

Hit topic solved (bottom left)