这个问题在这里已有答案：
我已经尝试过这个PHP代码并且出现了很多错误，我无法识别我是初学者的错误在哪里我真的希望你尽快得到你的帮助
错误：
连接成功
注意：未定义的索引：第23行的C：\ xampp \ htdocs \ phpfile.php中的名称
注意：未定义的索引：第24行的C：\ xampp \ htdocs \ phpfile.php中的电子邮件
注意：未定义的索引：第25行的C：\ xampp \ htdocs \ phpfile.php中的电话
注意：未定义的索引：第26行的C：\ xampp \ htdocs \ phpfile.php中的密码错误：INSERT INTO用户（名称，电子邮件，电话，密码）VALUES（''，''，''，''）没有数据库选
HTML表格：
<html>
<head>
<title> event reg </title>
<link rel="stylesheet" href="style.css">
</head>
<body>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<form enctype="from-data" action="phpfile.php" method="post">
<label for ="name"> User Name :</label>
<input type="text" name="name" id="name" maxlength="30" Size="30"/>
<br />
<br />
<label for ="email"> E-mail : </label>
<input type="email" name="email" id="email" maxlength="255" Size="30"/>
<br />
<br />
<label for ="phone"> Phone :</label>
<input type="text" name="phone" id="phone" maxlength="10" Size="30"/>
<br />
<br />
<label for ="password"> password : </label>
<input type="password" name="password" id="password" maxlength="12" Size="30"/>
<br />
<br />
<label for ="event"> Event name: </label>
<select name="event" id="event">
<option value="Inceptionopening"> Inception </option>
</select>
<br />
<br />
<br />
<br />
<input type="submit" name="submit" id="submit" value="submit" />
<input type="reset" name="reset" id="reset" value="Cnacel" />
</form>
</body>
</html>
//连接到它的PHP代码：
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "userdb";
// Create connection
$conn = mysqli_connect($servername, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected Successfully <br>";
$name=$_POST['name'];
$email=$_POST['email'];
$phone=$_POST['phone'];
$password=$_POST['password'];
$sql="INSERT INTO user(name,email,phone,password)VALUES('$name','$email','$phone','$password')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
试试这个
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "userdb";
$con = mysqli_connect($servername,$username,$password,$dbname);