我用的时候

$id = PDO :: lastInsertId();

我明白了

Fatal error: Non-static method PDO::lastInsertId() cannot be called statically in C:\xampp\htdocs\createimage.PHP on line 16

这是导致错误的PHP：

<?PHP 

$title = $_POST['title'];
$caption = $_POST['caption'];

$conn = new PDO('MysqL:host=localhost;dbname=imagesite','root','');

$stmt = $conn->prepare('INSERT INTO images (id,link,title,caption) VALUES (NULL,:link,:title,:caption)');

$stmt->execute(array(
    'link' => 'fake','title' => $title,'caption' => $caption
    ));

$id = PDO::lastInsertId();

header("Location: localhost/image?id=$id");

有谁能说出出了什么问题？或者另一种实现我想做的方式？