php - PHP での検証後にデータベースにデータが挿入されない

Question

HTML および PHP 検証コード: "postresume.php"

<?php
$firstName = $lastName = $emailId = $phoneNo = $qualification = $dob = $totex = $address="";
$firstNameErr = $lastNameErr = $emailIdErr = $phoneNoErr = $qualificationErr = $dobErr = $totexErr = $addressErr = "";

if ($_SERVER['REQUEST_METHOD']== "POST") {
    $valid = true; 

    if (empty($_POST["firstName"])) {
        $firstNameErr = "*First name is required";
        $valid = false; //false
    }
    else {
        $firstName = test_input($_POST["firstName"]);
    }

    //LastName Error
    if (empty($_POST["lastName"])) {
        $lastNameErr = "*Last name is required";
        $valid = false;
    }
    else {
        $lastName = test_input($_POST["lastName"]);
    }

    // validation for,email,phoneno,qualification,dob,totex,address will be same as the above

    if($valid){
        echo '<META HTTP-EQUIV="Refresh" Content="0; URL=datasubmitted.php">';    
        exit;
    }
}

function test_input($data) {
    $data = trim($data);
    $data = stripslashes($data);
    $data = htmlspecialchars($data);
    return $data;
}
?>

これは「postresume.php」の私のフォームです:

<form action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']);?>" method="post">
    <div class="label">*First Name:
        <div class="txtbox">
            <input name="firstName" type="text" id="txt" placeholder="Enter Your First Name." value="<?php echo $firstName; ?>"/>
            <span class="error"><p></p><?php echo  $firstNameErr; ?></span>
        </div>
    </div>
    <div class="label">Last Name:
        <div class="txtbox">
            <input name="lastName" type="text" id="txt" placeholder="Enter Your Last Name." value="<?php echo $lastName; ?>"/>
            <span class="error"><p></p><?php echo $lastNameErr; ?>
        </div>
    </div>
    // email,phoneno,qualification,dob,totex,address will be same as the above

データベース mysql にデータを挿入するための PHP コード: "datasubmitted.php"

<?php
$con=mysqli_connect("localhost","root","geetha@99","test");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO register (fname, lname, emailid, phoneno, qualification, dob,   totalex, address)
    VALUES   ('$_POST[firstname]','$_POST[lastname]','$_POST[email]','$_POST[phoneno]','$_POST[qualifi]','$_POST[dob]','$_POST[totex]','$_POST[address]')";

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

mysqli_close($con);
?> 

私を助けて、私のコードの何が問題なのか教えてください。