我正在尝试为php表单编写一个简单的代码。我在与html相同的页面中编写了php代码。 php页面的名称是signup.php 但是我得到这样的错误：
注意：未定义的索引：nm in 第5行的C：\ xampp \ htdocs \ mini_project \ signup2.php
注意：未定义的索引：电子邮件 第6行的C：\ xampp \ htdocs \ mini_project \ signup2.php
如果我将php和html代码分成两个不同的页面，我的代码就没有问题。
为什么我不能在同一页面上编写php和html？我做错了什么？
我正在使用php 7.
<?php
include 'connect.php'
$nm = $_POST['nm'];
$email = $_POST['email'];
$password = $_POST['password'];
$cpassword = $_POST['cpassword'];
$type = $_POST['type'];
$remember = $_POST['remember'];
$signup = $_POST['signup'];
if (isset($signup))
{
$query = "INSERT INTO user_details (nm,email,pass,user_type) VALUES ('$nm','$email','$password','$type')";
$result = mysqli_query($conn,$query);
}
?>
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link href="css/bootstrap.min.css" rel="stylesheet" media="screen">
<script src="js/jquery.min.js"></script>
<script src="js/bootstrap.min.js"></script>
<body>
<div class="container">
<div class="row">
<div class="col-md-4 col-md-offset-4">
<div class="login-panel panel panel-default">
<div class="panel-heading">
<h3 class="panel-title">Please Sign In</h3>
</div>
<div class="panel-body">
<form role="form" method="post" action="signup.php">
<fieldset>
<div class="form-group">
<input class="form-control" placeholder="Please Enter Name" name="nm" id="nm" type="text" autofocus>
</div>
<div class="form-group">
<input class="form-control" placeholder="Please Enter E-mail" name="email" id="email" type="email">
</div >
<div class="form-group">
<input class="form-control" placeholder="Please Enter Password" name="password" id="password" type="password" value="">
</div>
<div class="form-group">
<input class="form-control" placeholder="Please Confirm Password" name="cpassword" id="cpassword" type="password" value="">
</div>
<div class="form-group">
<select class="form-control" align="center" name="type">
<option value="0">Please Select User Type</option>
<option value="1">I want to Hire</option>
<option value="Second Year">I want to Work</option>
</select>
</div>
<div class="checkbox">
<label>
<input name="remember" id="remember" type="checkbox" value="Remember Me">Remember Me
</label>
</div>
<!-- Change this to a button or input when using this as a form -->
<input type="submit" name="signup" id="signup" value="Signup" class="btn btn-lg btn-primary btn-block">
</fieldset>
</form>
</div>
</div>
</div>
</div>
</div>
</body>
</html>
答案 0 :(得分：5)
原因是因为您尚未提交表单，但您已尝试根据$_POST值定义变量。您需要检查是否先设置了值：
if (isset($_POST['nm'])) $nm = $_POST['nm'];
if (isset($_POST['email'])) $email = $_POST['email'];
或使用三元运算符：
$nm = isset($_POST['nm']) ? $_POST['nm'] : "";
$email = isset($_POST['email']) ? $_POST['email'] : "";
或在PHP 7中，简单地说：
$nm = $_POST['nm'] ?? "";
$email = $_POST['email'] ?? "";
答案 1 :(得分：1)
这是因为$ _POST数组中的各个键尚未设置 - 它们是在表单提交时设置的。
在尝试访问变量之前，您应首先测试变量是否已设置（使用isset（）函数）; e.g。
<?php
include 'connect.php'
if (isset($_POST['signup')) {
$nm = $_POST['nm'];
$email = $_POST['email'];
// The rest of your code here...
}
?>
答案 2 :(得分：0)
无论页面是如何请求的，您都在运行代码，但只有在您向页面发布内容时才定义变量。您应该检查它是该页面的POST而不仅仅是常规的GET请求。如果您只是将所有代码移动到检查中以定义注册，那么可能正常工作。
答案 3 :(得分：0)
虽然您的页面在未设置$_POST数据时加载，但它不会呈现您提到的密钥。使用以下任何解决方案：
1）检查设置键：
<?php
include 'connect.php'
$nm = isset($_POST['nm']) ? $_POST['nm'] : "";
$email = isset($_POST['email']) ? $_POST['email'] : "";
$password = isset($_POST['password']) ? $_POST['password'] : "";
$cpassword = isset($_POST['cpassword']) ? $_POST['cpassword'] : "";
$type = isset($_POST['type']) ? $_POST['type'] : "";
$remember = isset($_POST['remember']) ? $_POST['remember'] : "";
$signup = isset($_POST['signup']) ? $_POST['signup'] : "";
if (isset($signup))
{
$query = "INSERT INTO user_details (nm,email,pass,user_type) VALUES ('$nm','$email','$password','$type')";
$result = mysqli_query($conn,$query);
}
?>
2）或者你可以写下如下：
<?php
include 'connect.php'
if (isset($_POST['signup']))
{
$nm = $_POST['nm'];
$email = $_POST['email'];
$password = $_POST['password'];
$cpassword = $_POST['cpassword'];
$type = $_POST['type'];
$remember = $_POST['remember'];
$signup = $_POST['signup'];
$query = "INSERT INTO user_details (nm,email,pass,user_type) VALUES ('$nm','$email','$password','$type')";
$result = mysqli_query($conn,$query);
}
?>
答案 4 :(得分：0)
你的php代码应该这样写：
<?php
if($_REQUEST['method'] == "POST") {
include 'connect.php'
$nm = $_POST['nm'] ?? "";
$email = $_POST['email'] ?? "";
$password = $_POST['password'] ?? "";
$cpassword = $_POST['cpassword'] ?? "";
$type = $_POST['type'] ?? "";
$remember = $_POST['remember'] ?? "";
$signup = $_POST['signup'] ?? "";
if (isset($signup))
{
$query = "INSERT INTO user_details (nm,email,pass,user_type) VALUES ('$nm','$email','$password','$type')";
$result = mysqli_query($conn,$query);
}
}
?>
答案 5 :(得分：0)
正如许多答案中已经说过的那样，您收到通知的原因是因为您正在尝试使用尚未发送到服务器的数据。为了避免这些，你首先要测试你是否确实已经发送了数据，然后才对它采取行动 这就是为什么它将它放入一个单独的文件中的原因，但如果你直接打开表单的目标URL（绕过表单），它仍会给你相同的通知。
请注意，即使已将内容发布到服务器，也并不意味着您期望的所有数据都会存在。有时浏览器的正常功能意味着他们不会发送一些字段（未经检查的复选框等），有时候用户尝试做一些你不希望他们做的事情。
这就是为什么检查每个价值也很重要的原因。不仅如果它存在，而且它是否在可接受的输入值/模式范围内。最后一位称为输入验证，通常使用filter_var()函数和its bretheren完成。
所以，将所有这些组合成一个脚本示例：
include 'connect.php';
if (isset ($_POST['signup'])) {
$error = process_signup ($conn);
}
/**
* Processes the signup form, and saves the registration to the DB if successful.
* Returns an error string detailing what went wrong, in case of errors.
*
* @param PDO $db
* @return void|string
*/
function process_signup (PDO $db) {
// Testing for early exit conditions first, so that we don't do unnecessary work.
$signup = filter_input (INPUT_POST, 'signup', FILTER_VALIDATE_BOOLEAN);
if (!isset ($signup)) {
// Since we're not signing up, just exit early.
return;
}
// Using a variable to keep track of the validation status.
$errors = array ();
// Validate that the name only contains "word characters" as defined by regex.
$options = array ('options' => array ('regexp' => '/^\w+\\z/u'));
$nm = filter_input (INPUT_POST, 'nm', FILTER_VALIDATE_REGEXP, $options);
if (!$nm) {
// Add more descriptive error messages here.
$errors[] = 'Name';
}
$email = filter_input (INPUT_POST, 'email', FILTER_VALIDATE_EMAIL);
if (!$email) {
$errors[] = 'Email';
}
// Create a regex that defines a _minimum_ level of password complexity.
$password = filter_input (INPUT_POST, 'password', FILTER_VALIDATE_REGEXP, $options);
if (!$password) {
$errors[] = 'Password';
}
// If the password confirmation is set, and identical to the validated password, we're good.
$cpassword = isset ($_POST['cpassword']) && $_POST['cpassword'] === $password ? true : false;
if (!$cpassword) {
$errors[] = 'Password confirmation';
}
// These are optional, so no need to add error messages if they're missing/wrong.
$type = filter_input (INPUT_POST, 'type', FILTER_VALIDATE_INT);
$remember = filter_input (INPUT_POST, 'remember', FILTER_VALIDATE_BOOLEAN);
// If some of the validation failed, we need to stop here and tell the user about it.
if (!empty ($errors)) {
// Note that this is VERY rudimentary, you should do something more informative.
$errors = "Following fields failed validation: ".implode (', ', $errors);
return $errors;
}
// ALWAYS remember to hash the password!!!
$password = password_hash ($password, PASSWORD_DEFAULT);
// Using PDO as it's a bit easier than MySQLi.
$query = "INSERT INTO user_details (nm,email,pass,user_type) VALUES (:name, :email, :password, :type)";
$statement = $db->prepare ($query);
$data = array (
':name' => $nm,
':email' => $email,
':password' => $password,
':type' => $type
);
if (!$statement->execute ($data)) {
// Something went wrong with saving the data to the database. Handle it!
return "Someting went wrong!";
}
// Everything is good, so redirect the user to a confirmation page.
// Redirect prevents reloading the page from re-submitting the data.
header ("Location: success.php");
die ();
}
?>
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link href="css/bootstrap.min.css" rel="stylesheet" media="screen">
<script src="js/jquery.min.js"></script>
<script src="js/bootstrap.min.js"></script>
<body>
<h3 class="panel-title">Please Sign In</h3>
<form role="form" method="post" action="signup.php">
<fieldset>
<input class="form-control" placeholder="Please Enter Name" name="nm" id="nm" type="text" autofocus>
<input class="form-control" placeholder="Please Enter E-mail" name="email" id="email" type="email">
<input class="form-control" placeholder="Please Enter Password" name="password" id="password" type="password" value="">
<input class="form-control" placeholder="Please Confirm Password" name="cpassword" id="cpassword" type="password" value="">
<select class="form-control" align="center" name="type">
<option value="0">Please Select User Type</option>
<option value="1">I want to Hire</option>
<option value="2">I want to Work</option>
</select>
<input name="remember" id="remember" type="checkbox" value="Remember Me">
<label for="remember">Remember Me</label>
<input type="submit" name="signup" id="signup" value="Signup" class="btn btn-lg btn-primary btn-block">
</fieldset>
</form>
</body>
</html>
请注意，我已经从中删除了很多（不必要的，恕我直言）HTML，并注意到这是一个相当粗略的例子。足以向您展示您需要采取的步骤，但尚未准备好在生产中使用。错误处理应该更加优雅，并向用户详细解释失败的内容以及这些字段中所需/允许的内容。
我还想指出在这里使用预备陈述，因为它们是正确防御SQL injections的唯一方法。
答案 6 :(得分：-1)
没人知道设置php.ini，.htaccess，connect.php是什么... 她的错误在什么阶段显示规则
为什么不在html文件中使用该设计：
if ($ _ POST) {
$ Nm = isset ($ _ POST [ 'nm']) $ _ POST [ 'nm']: null;?
...
}
或者使用display_errors ...