Hello, my name is Ben Stones. I am quite a beginner to PHP, and as a new
years resolution I am going to learn PHP (finally!)
Cut to the chase I have created a basic looping script that would display
anything submitted in a form, on seperate lines; here is the PHP code:
$con =
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
[EMAIL PROTECTED] wrote:
I'm getting a health insurance quote engine going...
http://www.insuranceshoppers.net/getquotes/short-term-health.php
... it sends me an email with the info, and enters the info into a leads
table in mysql.
Then I want to have it show the rates (on a new page
Ron Piggott wrote:
I am trying to query my rate_plan table and then display the results in
a table format. I need help developing then SELECT syntax and how I am
able to echo the results in a table. I want to display the results 7
columns, 4 rows. There are 7 tenures:
I have rate plans set
