我是编码新手，在这里问我的第一个问题。 我尝试使用 Php 和 MySQL 工作台创建 HTML 注册表单。
根据网上教程， 我提交表单的输出必须是“注册成功。”
但是，它显示：
此页面无法正常工作。
如果问题仍然存在，请联系网站所有者。
HTTP 错误 405
这是我的 HTML 代码：
<!DOCTYPE html>
<html lang="en">
<head>
<title>Document</title>
</head>
<body bgColor="yellow">
<form method="post" action="connect.php">
<fieldset>
<legend>Details</legend>
<p>
<label>User Type: </label>
<select name="user_type">
<option>Choose a user type</option>
<option>Admin</option>
<option>Teacher</option>
<option>Student</option>
</select>
</p>
<p>
<label>First Name: </label>
<input type="text" name="first_name" />
</p>
<p>
<label>Last Name: </label>
<input type="text" name="last_name" />
</p>
<p>
<label>Gender: </label>
<br/>
<input type="radio" name="gender" value="Female">Female<br/>
<input type="radio" name="gender" value="Male">Male<br/>
<input type="radio" name="gender" value="Other">Other
</p>
<p>
<label>Phone number: </label>
<input type="text" name="phone" />
</p>
<p>
<label>Email: </label>
<input type="text" name="email" />
</p>
<p>
<label>Password: </label>
<input type="password" name="pass" />
</p>
<input type="submit">
</fieldset>
</form>
</body>
</html>
这是我的 PHP 代码：
<?php
//Based on the name of input and select
$user_type = $_POST['user_type'];
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$gender = $_POST['gender'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$pass = $_POST['pass'];
// Database connection
$conn = new mysqli('local host','root','college');
if($conn->connect_error){
die('Connection Failed: '.$conn->connect_error);
}
else
{
$stmt = $conn->prepare("insert into users(user_type, first_name, last_name, gender, phone, email, pass)
values(?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param("ssssiss", $user_type, $first_name, $last_name, $gender, $phone, $email, $pass)
$stmt->execute();
echo "Signed up successfully.";
$stmt->close();
$conn->close();
}
?>
我应该使用 Xampp 而不是 MySQL Workbench 还是这段代码有问题？
<?php
$user_type = $_POST['user_type'];
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$gender = $_POST['gender'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$pass = $_POST['pass'];
$conn = new mysqli('0.0.0.0:3306','root','root', 'ast');
if($conn->connect_error){
die('Connection Failed: '.$conn->connect_error);
}
else
{
$stmt = $conn->prepare("insert into users(user_type, first_name, last_name, gender, phone, email, pass) values(?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param("ssssiss", $user_type, $first_name, $last_name, $gender, $phone, $email, $pass);
$stmt->execute();
echo "Signed up successfully.";
$stmt->close();
$conn->close();
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Document</title>
</head>
<body bgColor="yellow">
<form method="post" action=<?=$_SERVER['PHP_SELF']?>>
<fieldset>
<legend>Details</legend>
<p>
<label>User Type: </label>
<select name="user_type">
<option>Choose a user type</option>
<option>Admin</option>
<option>Teacher</option>
<option>Student</option>
</select>
</p>
<p>
<label>First Name: </label>
<input type="text" name="first_name" />
</p>
<p>
<label>Last Name: </label>
<input type="text" name="last_name" />
</p>
<p>
<label>Gender: </label>
<br/>
<input type="radio" name="gender" value="Female">Female<br/>
<input type="radio" name="gender" value="Male">Male<br/>
<input type="radio" name="gender" value="Other">Other
</p>
<p>
<label>Phone number: </label>
<input type="text" name="phone" />
</p>
<p>
<label>Email: </label>
<input type="text" name="email" />
</p>
<p>
<label>Password: </label>
<input type="password" name="pass" />
</p>
<input type="submit">
</fieldset>
</form>
</body>
</html>
试试这个。