Use of undefined constant name - assumed 'name' in …的異常，代碼如下

Use of undefined constant name - assumed 'name' in ……的錯誤，代碼如下：
function add_books(){
include_once("class_books.php");//包含圖書類
if($_POST["submit"]=="添加"){
if($_POST["add_book_name"]=="" || $_POST["add_book_price"]=="" || $_POST["add_book_author"]==""){
echo "添加失敗，請把資訊填寫完整
";
echo "重試";
}
else{
$b=new books;
$b->__set(name,$_POST["add_book_name"]);
$b->__set(price,$_POST["add_book_price"]);
$b->__set(author,$_POST["add_book_author"]);
$b->add();
echo "圖書$_POST[book_name]添加成功！
";
}
}
}
紅字表示的是資料庫中表的欄位名

------解決方案--------------------

PHP code

$b->__set(name,$_POST["add_book_name"]); $b->__set(price,$_POST["add_book_price"]); $b->__set(author,$_POST["add_book_author"]); 
