[phalbachjr]

Ok, I added the query to my php page. Below is a copy of the what it looks like. When I run it the database creates a new record but does not enter any information into it. Every suggestion has moved us closer to the solution but I'm still a few steps away.

<?php
$username = "a5134213_pth";
$password = "****";
$hostname = "mysql13.000webhost.com";

//connection to the database
$conn=mysql_connect ("mysql13.000webhost.com", "a5134213_pth", "****") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("a5134213_emad");

//select a database to work with
$selected = mysql_select_db("a5134213_emad")
or die("Could not select that database");


$data = "INSERT INTO Email (Name,Address) VALUES ('$Name', '$Address')";
$result = mysql_query($data) or die(mysql_error());


echo "Your Information Has Been Added";

$insert = "INSERT INTO Email (Name, Address) VALUES ('".$Name."','".$Address."')";
$result = mysql_query($insert) or die(mysql_error());


?>

[Bad Karma[CORE]]

Here´s an example that is working for me. Very basic, but does the job. You just need to replace the values for your fields/variables/mysql to match yours

PHP Code:

<?php
$name = $_POST['name']; //pass the name from your HTML form
$email = $_POST['email']; //pass the email from your HTML form

//connect to the database
$conn=mysql_connect ("server", "user", "password") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db("name_of_your_database");

//prepare and execute the query
$data = "INSERT INTO details (name,email) VALUES ('$name', '$email')";
$result = mysql_query($data) or die(mysql_error());

//ouput information after adding
echo "Your Information Has Been Added";
?>

[phalbachjr]

THANK YOU SO VERY MUCH!!! It worked with modifying your code to match my info!!

THANK YOU ALL FOR TAKING TIME OUT TO HELP ME!!!! YOU ARE ALL AWESOME!!!

[Bad Karma[CORE]]

anytime