急请高手指导关于Mysqli的prepare出错有关问题

程序员文章站 2024-01-08 21:48:16

...

急！！！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;
}
------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.

声明：本文内容由网友自发贡献，版权归原作者所有，本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容，请联系admin@php.cn核实处理。