可能的重复：
如何使用已弃用的 mysql_* 函数成功重写旧的 mysql-php 代码？
我无法将这些值插入数据库。但是我也没有收到任何错误消息。
<html>
<body>
<form action="database.php" method="post">
Name : <input type ="text" name = "name"/>
Number :<input type ="text" name = "number"/>
<input type ="submit" value = "submit"/>
</form>
</body>
</html>
数据库.php
<?php
class Database
{
var $host;
var $user;
var $pass;
var $data;
var $con;
var $table;
var $db;
public function controls()
{
$this->host="localhost";
$this->user="cgiadmin";
$this->pass="cgi";
$this->data="j2";
}
public function connection()
{
$this->con="mysql_connect($this->host,$this->user,$this->pass)";
}
public function tablename()
{
$this->table="Insert into employee(name,number) values('$_POST[name]','$_POST[number]')";
}
public function databaseconnection()
{
$this->db="mysql_select_db($this->data,$this->con)";
}
}
$name=new Database;
$name->connection();
if(!($name->con))
{
echo "'Error: ' . mysql_error()";
}
$name->databaseconnection();
$name->tablename();
echo "thanks for taking the survey";
?>
尝试这个...
修改表名()函数
public function tablename($nam,$num)
{
$this->table=mysql_query("INSERT INTO employee(name,number) VALUES ('$nam','$num')");
}
获取值并调用 tablename() 函数
$name=new Database;
$name->connection();
if(!($name->con))
{
echo "'Error: ' . mysql_error()";
}
$name->databaseconnection();
$nam=$_POST[name];
$num=$_POST[number];
$name->tablename($nam,$num);
echo "thanks for taking the survey";
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:hwhale#tublm.com(使用前将#替换为@)
