我正在尝试用PHP和HTML制作一个示例寄存器表单。
我有HTML页面:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Register</title>
</head>
<body>
</br></br></br></br>
<form method="post" action="register.php">
<label>Username:</label>
<input type="text" id="uname" name="uname" placeholder="insert username" required></input></br>
<label>Password:</label>
<input type="password" id="pwd" name="pwd" placeholder="insert password" required></input></br>
<label>Retype Password:</label>
<input type="password" id="cpwd" name="cpwd" placeholder="confirm password" required></input></br>
<input type="submit" name="submit" value="Registrami"></input>
</form>
</body>
</html>
以及被调用的PHP操作:
<?php
$username="def";
$password="def";
if(isset($_POST['submit'])) {
if (isset($_POST['uname'])) {
echo "yes";
$username = $_POST['uname'];
} else {
echo "no";
}
if (!empty($_POST['pwd'])) {
$password = $_POST['pwd'];
}
if (!empty($_POST['cpwd'])) {
$cpassword = $_POST['cpwd'];
}
}
echo $username;
echo $password;
require('connection.php');
?>
$ _POST中没有任何内容传递。当我尝试分配$username=$_POST['uname']
时,它不会更改值。
我也试过使用$_REQUEST.
有任何想法吗?
感谢您对此事感兴趣!
答案 0 :(得分:0)
$post_data = file_get_contents("php://input"); //getting post like this `