PHP - 运行1个插入查询同时运行两次记录

addcustomer.php

<?php
include_once('functions.php'); 
?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8s" />
<title>Add Customer</title>
<script type="text/javascript">
function CheckFields()
{
    var cusname = document.getElementById("cusname").value;
    var cusadd = document.getElementById("cusadd").value;

    if(cusname == '')
    {
        alert("Please enter your customer name");
        return false;
    }

    if(cusadd == '')
    {
        alert("Please enter your customer address");
        return false;
    }

    return true;
}
</script>
</head>

<body onLoad="focus();add.name.focus()">
<?php
if(array_key_exists('submit_check', $_POST))
{
    if($_POST['submit'] == 'Add')
    {   
        if(!AddCustomer(trim($_POST['cusname']), trim($_POST['cusgender']), trim($_POST['cusadd'])))
        {
            echo "Add Customer Error, Please try it again later";
        }
        else
        {
            echo "Customer Information had been successfully added into the database";
        }
    }   
}
else
{
    AddCustomerForm();
}
?>
</body>
</html>

dbconnect.php

<?php
$host = "localhost";
$user = "root";
$passwd = "";
$dbname = "customertest";
?>

functions.php

<?php
include('dbconnect.php');

function AddCustomerForm()
{
?>
    <form id="add" name="add" method="post" action="<?php echo     $_SERVER['PHP_SELF']; ?>">
    <b><u>Customer Details</u></b><br />
    <input type="hidden" name="submit_check" value="1" />
    Customer Name : <input type="text" id="cusname" name="cusname" maxlength="50" /><br />
    Gender : <input type="radio" name="cusgender" value="M" checked />Male
             <input type="radio" name="cusgender" value="F" />Female<br /><br />
    <b><u>Customer Address</u></b><br />
    Address : <input type="text" id="cusadd" name="cusadd" maxlength="100" size="50" /><br /><br />
    <input type="submit" name="submit" value="Add" onclick="return CheckFields();" />
    <input type="reset" name="reset" value="Reset" />
    </form>
<?php
}

function ConnectToDb()
{
    global $host, $user, $passwd, $dbname;
    @$db = mysql_pconnect($host, $user, $passwd);
    if(!$db)
    {
        echo "Couldn't connect to the database";
        exit;
    }
    mysql_select_db($dbname); 
}

function AddCustomer($cusname, $cusgender, $cusadd)
{
    ConnectToDb();

    $query = "INSERT INTO customer (cusname, cusgender, cusjoindate) VALUES (\"".$cusname."\", '".$cusgender."', NOW())";
    $result = mysql_query($query);

    if($result)
    {
        $cusID = mysql_insert_id();

        $query = "INSERT INTO customer_address (cusID, cusaddress) VALUES ('".$cusID."', '".$cusadd."')";
        $result = mysql_query($query);

        return 1;
    }
    else
    {
        return 0;
    }
}
?>

From above code, when I fill in the customer form and click Add button, I get the double record in my database as shown as image below:
enter image description here

I really don't understand why the database store 2 records at the same time.
Am I writing wrong code? Can someone help me?

Did u assigned the Primary & FOREIGN KEY Properly. Means the cusID in the customer address table that refers the cusID(Primary key) in customer table.I'm tried the same as bellow it works fine.

ALTER TABLE customer_address ADD FOREIGN KEY (cusID) REFERENCES customertest.customer(cusID) ON DELETE RESTRICT ON UPDATE RESTRICT;