我在使用用户名和密码进行简单注册以插入mysql数据库时遇到了简单错误。
注意:未定义的变量:第42行的C:\ xampp \ htdocs \ x \ index.php中的用户名
注意:未定义的变量:第42行的C:\ xampp \ htdocs \ x \ index.php中的密码 注册成功
<!DOCTYPE html>
<html>
<head>
<title>THIS IS MY FIRST LOGIN PAGE</title>
<link rel="stylesheet" type="text/css" href="style.css"></link>
<link rel="stylesheet" type="text/css" href="">
</head>
<body>
<div class="container">
<img src="men.png" alt="missing">
<form method="post" >
<div class="font-input">
<input type="text" name="name" id="name" placeholder="Enter Name">
</div class="font-input">
<div>
<input type="password" name="pass" id="pass" placeholder="Enter Password">
</div>
<div>
<input type="button" name="login" value="Login">
</div>
</form>
</div>
</body>
</html>
<?php
$conn = mysqli_connect('localhost','root','','y');
if(!$conn)
{
die('Connection failed!'.mysqli_error($conn));
}
if (isset($_POST['name'], $_POST['pass']))
{
$username = $_POST['name'];
$password = $_POST['pass'];
}
$sql = "INSERT INTO hack(myusername,mypassword) VALUES('$username', '$password')";
if(mysqli_query($conn,$sql))
{
echo "Registerd Successfully";
}
else
{
echo mysqli_error($conn);
}
?>
答案 0 :(得分:0)
尝试一下:
我们检查用户名和密码是否与表单一起发送,如果是,则在没有错误的情况下将它们插入数据库。
代码:
if((isset($_POST['name'])) && (isset($_POST['pass'])))
{
$username = $_POST['name'];
$password = $_POST['pass'];
$sql = "INSERT INTO hack(myusername, mypassword) VALUES('$username', '$password')";
if(mysqli_query($conn,$sql))
{
echo "Registerd Successfully";
}
else
{
echo mysqli_error($conn);
}
}