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;
}
------解决思路----------------------
$mysqli 不是 mysqli 对象
------解决思路----------------------
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 (?, ?, ?, ?, ?, ?, ?))
------解决思路----------------------
mysqli_stmt_bind_param($stmt,'ssssssd', $bookuname, $bookdep, $bookduty, $bookphone, $booktel, $bookemail, $bookgid);
http://php.net/manual/zh/mysqli-stmt.bind-param.php
the number of variables and length of string types must match the parameters in the statement.