PHP代码不会产生错误代码，但它不会嵌入到SQL表中

Question

这是我的代码，我之前使用的几乎完全一样，并且有效。我已经查看了其他已回答的问题而没有让它工作。

＆＃39; template.php＆＃39;保持访问已在所有其他页面上工作的sql数据库。最后是表格的图像

<?php
include_once('template.php');
if (isset($_POST['username']) and isset($_POST['password'])) {
    $name = $mysqli->real_escape_string($_POST['username']);
    $pwd = $mysqli->real_escape_string($_POST['password']);
    $query = <<<END
INSERT INTO outlets(device_name,description,id_r)
VALUES('{$_POST['device_name']}','{$_POST['description']}','{$_POST['id_r']}')
END;
    if ($mysqli->query($query) !== TRUE) {
        die("Could not query database" . $mysqli->errno . " : " . $mysqli->error);
        header('Location:index.php');
    }
}
$content = <<<END
<form method="post" action="add_device.php">
<input type = "text" name="device_name" placeholder="Device Name"><br>
<input type="text" name="description" placeholder="Description"><br>
<select name="id_r">
  <option value="1">Living Room</option>
  <option value="2">Bedroom</option>
  <option value="3">Kitchen</option>
  <option value="4">Bathroom</option>
</select>
<input type="submit" value="Register">
<input type="Reset" value="reset">
</form>
END;
echo $content;
?>