我在填充数据库表的页面上有一个表单：

<table border="1">
  <tr>
    <td align="center">Form Input Employees Data</td>
  </tr>
  <tr>
    <td>
      <table>
        <form method="post" action="input.php">

        <tr>
          <td>Product Name</td>
          <td><input type="text" name="name" size="20">
          </td>
        </tr>
        <tr>
          <td>Brand</td>
          <td><input type="text" name="brand" size="40">
          </td>
        </tr>
        <tr>
          <td></td>
          <td align="right"><input type="submit" name="submit" value="Sent"></td>
        </tr>
        </table>
      </td>
    </tr>
</table>

这里是输入。php：

<?
//the example of inserting data with variable from HTML form
//input.php
mysql_connect("localhost","xxx","xxx");//database connection
mysql_select_db("xxxxxxx");




$current_user= wp_get_current_user();
$id= $current_user->user_login; 


//inserting data order
$order = "INSERT INTO wp_userdata
            (id, product_name, product_brand)
            VALUES
            (\'$id\',
                        \'$_POST[name]\',
            \'$_POST[brand]\')";

//declare in the order variable
$result = mysql_query($order);  //order executes
if($result){
    echo("<br>Input data is succeed");
} else{
    echo("<br>Input data is fail");
}
echo $id;
?>

我正在尝试将用户名作为“id”传递给数据库。我收到下一个错误：

Fatal error: Call to undefined function wp_get_current_user() in /home/xxx/input.php on line 10

我试图改变现状，但仍然没有奏效。。