HI,我是一个PHP新手，今天一个问题折腾了一上午，请大家帮我看一下，万分感谢
- PHP code
Book-O-Rama Book Entry Results Book-O-Rama Book Entry Result
<?php //create short variable names
$isbn = $_POST['isbn'];
$author = $_POST['author'];
$title = $_POST['title'];
$price = $_POST['price'];
if ( !$isbn || !$author || !$title || !$price ) {
echo "You have not entered all the required details.
"."Please go back and try again.";
exit;
}
if ( !get_magic_quotes_gpc() ) {
$isbn = addslashes($isbn);
$author = addslashes($author);
$title = addslashes($title);
$price = doubleval($price);
}
@ $db = new mysqli('localhost', 'bookorama', 'bookorama123', 'books');
if ( mysqli_connect_errno() ) {
echo "Error: Could not connect to database. Please try again later.";
exit;
}
$query = "insert into books values ('$isbn', '$author', '$title', '$price')or die(mysql_error())";
;
echo $query;
$result = $db->query($query);
if ( $result ) {
echo mysqli_affected_rows." book inserted into database.";
} else {
echo mysql_errno()."".mysql_error();
echo "An error has occurred.The item was not added.";
}
$db->close();
?>
这是来自PHP和 MySQL web开发（第四版）中的一段代码。
我想问的是为什么最后总是输出An error has occurred.The item was not added
而且插入结果$result = -1
请大家指点一下
------解决方案--------------------
$query = "insert into books values ('$isbn', '$author', '$title', '$price')";
echo $query;
$result = $db->query($query) or die($db->error);