Hi ben,
You are creating the same table each time you run the code which would throw an
error the second time you run the code since the table is already there.
You have two choices here:
1. remove the table creation script and the call to create it since the table
only needs to be created
Ben:
First, using a $_POST value directly into a MySQL query is EXTREMELY
unsafe. Always filter data from any source to make sure it's what you
expect. SQL injection is one of the easiest ways to cause real damage
to a website. http://en.wikipedia.org/wiki/SQL_injection
Check out this fuction
Thanks all for your replies. Much appreciated. I have edited the code and
took points into account:
$con = mysql_connect(localhost,ben_test,removed) or die(con);
$db = mysql_select_db(ben_test) or die(db);
$sql1 = mysql_query(INSERT INTO `comments` (`messages`) VALUES
($comments)) or
Ben Stones wrote:
Thanks all for your replies. Much appreciated. I have edited the code and
took points into account:
$con = mysql_connect(localhost,ben_test,removed) or die(con);
$db = mysql_select_db(ben_test) or die(db);
$sql1 = mysql_query(INSERT INTO `comments` (`messages`) VALUES
