为什么我的$_GET表单不将数据(使用MySql )插入到MySql数据库或表中？

Question

以下是html表单：

<form action="register.php" METHOD=GET align="center">
<table border="2" align="center">
<tr>
<td>Desired Username:</td><td><input name="username" type="text" size"20"></input></td>
</tr>
<tr>
<td>Your Email:</td><td><input name="email" type="text" size"30"></input></td>
</tr>
<tr>
<td>Desired Password:</td><td><input name="password" type="password" size"20"></input>                                       </td>
</tr>
<tr>
<td>Confirm Password:</td><td><input name="password2" type="password" size"20"></input>          </td>
</tr>
</table>
<input type="submit" value="Register!" align="center"></input>
</FORM>

下面是register.php上的PHP代码：

<?php
$myServer = "localhost";
$myUser = "Censored";
$myPass = "Censored";
$myDB = "Censored";

//connection to the database
$conn = "mssql_connect($myServer, $myUser, $myPass)
or die('Couldn't connect to SQL Server on $myServer')"; 

//select a database to work with
$selected = "mssql_select_db($myDB, $conn)
or die('Couldn't open database $myDB')"; 

//add new user to DB
if($_GET["username"] && $_GET["email"] && $_GET["password"] && $_GET["password2"] )
{
if($_GET["password"]==$_GET["password2"])
{
$sql="INSERT INTO Users(Username, Password, Email)              VALUES('$_GET[username]','$_GET[password]','$_GET[email]')";
$result="mysql_query($sql,$conn) or die (mysql_error())";
}

if ($result)
{
echo "<h1 align='center'>Registration Successful!</h1>";

echo "<p align='center'>Click<a href='index.php'> here</a> to log in.</p>";

}
else echo "Oops! Something went wrong. Please <a       href='http://www.canilosa.zxq.net/index.php'>go back</a> and fix it.";
}
?>

我是PHP和MySql的初学者。我检查并重新检查了所有的用户名，密码，数据库等。回显工作，它说注册是成功的，但当我检查我的数据库没有数据。

感谢所有回答的人！它现在运转良好:)。就像我说的，我对所有这些都是陌生的，所以大部分代码都是从教程片段中提取出来的，我只是略读一下，看看它是否有意义。再次感谢！(我正在创建一个包含大量PHP的游戏网站.所以我可能很快就会回来了，xD)

Answer 1

正在使用的Mysql函数不正确.

变化

mssql_connect($myServer, $myUser, $myPass)

至

mysql_connect($myServer, $myUser, $myPass)

和

mssql_select_db($myDB, $conn)

至

mysql_select_db($myDB, $conn)

文档链接mysql.asp

Answer 2

改变这个

<form action="register.php" METHOD=GET align="center">

至

<form action="register.php" METHOD="GET" align="center">
                                   ^   ^

也应该是

$conn = mysql_connect($myServer, $myUser, $myPass) or die("Couldn't connect to SQL Server on $myServer"); 

//select a database to work with
$selected = mysql_select_db($myDB, $conn) or die("Couldn't open database $myDB"); 

Answer 3

您已经使用了mssql_connect和mssql_select_db，但是实际的命令是mysql_connect和mysql_select_db

只需用以下代码替换代码

//connection to the database
$conn = mysql_connect($myServer, $myUser, $myPass)
or die('Couldn't connect to SQL Server on $myServer'); 

//select a database to work with
$selected = mysql_select_db($myDB, $conn)
or die('Couldn't open database $myDB');