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How to Connect html inquiry form That Stores Data in a MySQL Database Using PHP

i am using AppServ v2.5.9 i already done making form the probleme is i dont know how to connect ...i tired this code


define('DB_NAME', 'inquiry_box');
define('DB_USER', 'root');
define('DB_PASSWORD', '*****');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
    die('Could not connect: ' .mysql_error());

$db_selected = mysql_select_db(DB_NAME, $link);

if  (!$db_selected) {
die('Cant\'t use ' . DB_NAME . ': ' . mysql_error());
 $value = $_POST['name'];
 $value = $_POST['email'];
 $value = $_POST['contact_no'];
 $value = $_POST['comment'];

$sql ="INSERT INTO contact_form (name, email, contact_no, comment) VALUES ('$value','$value2','$value3','$value4')";

if (!mysql_query($sql)) {
die('Error: ' . mysql_error());


then what happen is it connect but when i type in the form data doesn't insert in the mysql but it actually show the id only but then no name email conatct and comment appears please help please

3 Answers

Chris Shaw
Chris Shaw
26,676 Points

Hi Maie,

Currently you're overwriting the variable $value with each value from your $_POST array and then looking for variables such as $value2 which isn't defined, what you want to do is assign each value to their variable instead and also escape the string to prevent XSS injection.


$name       = mysql_real_escape_string($_POST['name']);
$email      = mysql_real_escape_string($_POST['email']);
$contact_no = mysql_real_escape_string($_POST['contact_no']);
$comment    = mysql_real_escape_string($_POST['comment']);

$sql = sprintf(
    "INSERT INTO contact_form (name, email, contact_no, comment) VALUES ('%s', '%s', '%s', '%s')",

Also the MySQL module as of PHP 5.5 no longer exists and has been deprecated since PHP 5.4, from now on it's recommended that you use either MySQLi or PDO.



// Check connection
if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); }

// escape variables for security
$name = mysqli_real_escape_string($con, $_POST['name']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$contact_no = mysqli_real_escape_string($con, $_POST['contact_no']);
$comment = mysqli_real_escape_string($con, $_POST['comment']);

$sql="INSERT INTO form (name, email, contact_no, comment) VALUES ('$name', '$email', '$contact_no', '$comment')";

if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); }
echo "1 record added";



is this right???

but why still data does not enter to mysql ?? but it shows id then blank blank blank for example 64 then no name email contact_no comment

Chris Shaw
Chris Shaw
26,676 Points

Hi Maie,

Everything above looks correct, if you're still receiving empty values then it means your $_POST data either has different key names than what you have set above or nothing is been submitted from the form using a method type of POST.

i want to ask how can we make form validation using Java Script and ajax form content is name email contact_no comment i use this code above but i found out the problem is in the js where the code is making send to mail i want it send to mysql


    // hide messages 

    // on submit...
    $("#contactForm #submit").click(function() {


        var name = $("input#name").val();
        if(name == ""){
            $("#error").fadeIn().text("Name required.");
            return false;

        // email
        var email = $("input#email").val();
        if(email == ""){
            $("#error").fadeIn().text("Email required");
            return false;

        // contact_no
        var contact_no = $("input#contact_no").val();
        if(contact_no == ""){
            $("#error").fadeIn().text("Contact number required");
            return false;

        // comments
        var comments = $("#comments").val();

        // data string
        var dataString = 'name='+ name
                        + '&email=' + email        
                        + '&contact_no=' + contact_no
                        + '&comments=' + comments

        // ajax

    // on success...
     function success(){

    return false;