这个PHP代码有什么问题，它不会插入数据？

Question

I ran this on Xamp 5.538. The name of the file is application.php. The form appears normally and I get no error messages. However, it does not insert data into the MYSQL database. So far I have only been trying to get the top portion of the form to work; once it does I fill in the rest in the code. My database is named employees; the table’s name is table1. It contains the following fields so far: First, last, mi, address, city, state, zip, phone, email, ss, dob, security I tryed to post the entire form but wasn't smart enough to figure out how to do it or maybe it's just that I hate puzzles. Please figure this out for me. I’ve come along way, but still have further to go. Thank You; Eric Here’s the code:

<?php
// Your database info
    $db_host = 'localhost';
    $db_user = 'root';
    $db_pass = '""';
    $db_name = 'employees';
if (!empty($_POST))
{
    $con = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
    if ($con‐>connect_error)
die('Connect Error (' . mysqli_connect_errno() . ') '.
mysqli_connect_error());
    $sql = "INSERT INTO table1 (first, last, mi, address, city, state, zip, phone, email, ss, dob, security) VALUES (?,?,?,?,?,?,?,?,?,?,?,?)";
    if (!$stmt = $con‐>prepare($sql))
die('Query failed: (' . $con‐errno . ') ' . $con‐error);
    if (!$stmt‐    >bind_param('ssi',$_POST['fist'],$_POST['last'],$_POST['mi'],$_POST['address'],$ _POST['city'],$_POST['state'],$_POST['zip'],$_POST['phone'],$_POST['email'],$_PO    ST['ss'],$_POST['dob'],$_POST['security']))
die('Bind Param failed: (' . $con‐errno . ') ' . $con‐error);
    if (!$stmt‐>execute())
die('Insert Error ' . $con‐>error);
echo "Record added";
$stmt‐>close();
$con‐>close();

}
?>

Answer 1

if (!empty($_POST['sub'])) {
$father = $_POST['father'];
$con = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if (!$con) {
    die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
}
echo $sql = "INSERT INTO `father`( `FirstName`) VALUES ('" . $father . "')";
if (!$stmt = mysqli_prepare($con, $sql)) {
    echo "done";
}
if ($stmt->execute()) {
    // it worked
} else {
    // it didn't
}}

use this code it is working