请高手指导关于Mysqli的prepare出错问题

很久没写程序了，由于php5.0升级的mysqli类操作数据库有很多优点，本人想尝试用mysqli的方法操作数据库，经过从网上查资料、看手册，浏览、删除都能正常实现，唯独插入功能总是提示错误：Fatal error: Call to a member function prepare() on null in D:\xampps\htdocs\txl\addbook.php on line 20 。
快愁死我了，难道这个问题解决不了我要放弃Mysqli吗？请高手指点我的代码错在哪里？还是服务器的问题？（服务器我用的是Xampps1.9.7集成环境）

if($_POST['contacts'] == "true")
{
$stmt = $mysqli->prepare("INSERT INTO contactsuser(bookUname, bookDep, bookDuty,bookphone,bookTel,bookEmail,bookGid) VALUES (?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param('sssd', $bookUname, $bookDep, $bookDuty, $bookphone, $bookTel, $bookEmail, $bookGid);

$bookUname = $_POST['bookUname'];
$bookDep = $_POST['bookDep'];
$bookDuty = $_POST['bookDuty'];
$bookphone = $_POST['bookphone'];
$bookTel = $_POST['bookTel'];
$bookEmail = $_POST['bookEmail'];
$bookGid = $_POST['bookGid'];

/* execute prepared statement */
$stmt->execute();
printf("%d Row inserted.\n", $stmt->affected_rows);
/* close statement and connection */
$stmt->close();
exit;
}

回复讨论(解决方案)

$mysqli 不是 mysqli 对象

试试面向过程风格的方法：

$stmt = mysqli_prepare($link, "INSERT INTO contactsuser(bookUname, bookDep, bookDuty,bookphone,bookTel,bookEmail,bookGid) VALUES (?, ?, ?, ?, ?, ?, ?)")

我刚才的程序确实疏忽忘了包含数据库链接文件了，但包含之后，包括改成面向过程风格后还是有错误：
Warning: mysqli_stmt_bind_param(): Number of elements in type definition string doesn't match number of bind variables in D:\xampps\htdocs\txl\addbook.php on line 24

程序如下：
$link = mysqli_connect('localhost','root','root','addressbook') or die('Unale to connect');
$stmt = mysqli_prepare($link, "INSERT INTO contactsuser(bookUname, bookDep, bookDuty,bookphone,bookTel,bookEmail,bookGid) VALUES (?, ?, ?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($stmt,'sssd', $bookUname, $bookDep, $bookDuty, $bookphone, $bookTel, $bookEmail, $bookGid);
$bookUname = $_POST['bookUname'];
$bookDep = $_POST['bookDep'];
$bookDuty = $_POST['bookDuty'];
$bookphone = $_POST['bookphone'];
$bookTel = $_POST['bookTel'];
$bookEmail = $_POST['bookEmail'];
$bookGid = 1;
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);

数据库的字段类型我也检查了，除了主键bookid是自动增加的int型，bookGid是int型，其他都是varchar类型

VALUES (?, ?, ?, ?, ?, ?, ?)
$stmt,'sssd', $bookUname, $bookDep, $bookDuty, $bookphone, $bookTel, $bookEmail, $bookGid
七个问号对应着八个参数 。。。。。加一个问号再试试

mysqli_stmt_bind_param($stmt,'ssssssd', $bookUname, $bookDep, $bookDuty, $bookphone, $bookTel, $bookEmail, $bookGid);

http://php.com/manual/zh/mysqli-stmt.bind-param.php

The number of variables and length of string types must match the parameters in the statement.

楼上说的对，应该是ssssssd

太好了！感谢各位高手帮忙！特别是saint_leer和misakaqunianx...说的“ssssssd”的问题说到点子上了，一直没搞懂这传字符串啥意思！