javascript - 当我刷新页面时，数据自动存储在数据库中

Question

我用 php 制作了注册页面。当我在填写所有内容后注册一个用户时，数据已正确存储在数据库中，但之后如果我刷新注册页面而不填写任何内容，那么最近的用户数据将再次存储在数据库中。为什么会这样？最后的脚本也不起作用

下面是我的php代码

<?php 

if (isset($_POST['register'])) {

    $ip = getIp();

    $c_name = $_POST['c_name'];
    $c_email = $_POST['c_email'];
    $c_pass = $_POST['c_pass'];
    $c_image = $_FILES['c_image']['name'];
    $c_image_tmp = $_FILES['c_image']['tmp_name'];
    $c_country = $_POST['c_country'];
    $c_city = $_POST['c_city'];
    $c_contact = $_POST['c_contact'];
    $c_address = $_POST['c_address'];

    move_uploaded_file($c_image_tmp, "customer/customer_images/$c_image");

      $insert_c = "insert into customers (customer_ip,customer_name,customer_email,customer_pass,customer_country,customer_city,customer_contact,customer_address,customer_image) values ('$ip','$c_name','$c_email','$c_pass','$c_country','$c_city','$c_contact','$c_address','$c_image')";
}
 $run_c = mysqli_query($con, $insert_c);

$sel_cart = "select * from cart where ip_add='$ip'";

$run_cart = mysqli_query($con, $sel_cart);
$check_cart = mysqli_num_rows($run_cart);
if ($check_cart==0) {

    $_SESSION['customer_email']=$c_email;
    echo "<script>alert('Accound has been created successfully')</script>";
    echo "<script>window.open('customer/my_accont.php'.'_self')</script>";
}
else {
    $_SESSION['customer_email']=$c_email;
    echo "<script>alert('Accound has been created successfully')</script>";
    echo "<script>window.open('checkout.php'.'_self')</script>";
}   

 ?>

Answer 1

当您刷新已发布表单(method="POST")的页面时，您将重新发送数据。