PHP 代码说输入为空但它不是？

Question

我的 php 代码有问题，似乎 returns "Password Should Not Be empty" 当我输入密码时。我认为它可能来自我的 html 代码，因为我让他们输入了两次。我是否需要在我的数据库中创建另一个变量来确认密码？

<?php
$firstname = filter_input(INPUT_POST, 'firstname');
$lastname = filter_input(INPUT_POST, 'lastname');
$email = filter_input(INPUT_POST, 'email');
$password = filter_input(INPUT_POST, 'password');
if (!empty($firstname)) {
    if (!empty($lastname)) {
        if (!empty($email)) {
            if (!empty($password)) {
                $host = "127.0.0.1:3307";
                $dbusername = "root";
                $dbpassword = "";
                $dbname = "register";
                // Create connection
                $conn = new mysqli($host, $dbfirstname, $dblastname, $dbemail,
                    $dbpassword);
                if (mysqli_connect_error()) {
                    die('Connect Error (' . mysqli_connect_errno() . ') '
                        . mysqli_connect_error());
                } else {
                    $sql = "INSERT INTO Signup (firstname, lastname, email, password) values ('$firstname','$lastname','email','password')";
                    if ($conn->query($sql)) {
                        echo "New record is inserted sucessfully";
                    } else {
                        echo "Error: " . $sql . "" . $conn->error;
                    }
                    $conn->close();
                }
            } else {
                echo "Password should not be empty";
                die();
            }
        }
    }
} else {
    echo "Username should not be empty";
    die();
}

Answer 1

您的错误是因为您的 post 值为 $_POST['psw']，而您的代码期望 $_POST['password']。要么更改您的 HTML 表单，要么更改您的 PHP 代码，使值相同。

您的代码非常令人困惑。我把它简化了一点，试试这个，看看你是否仍然得到同样的错误：

# Check to see if you're getting the right variables in the first place:
var_dump($_POST); //Remove once you're sure you're getting the right stuff

if(!$firstname = filter_input(INPUT_POST, 'firstname')){
    die("First name should not be empty");
}
if(!$lastname = filter_input(INPUT_POST, 'lastname')){
    die("Last name should not be empty");
}
if(!$email = filter_input(INPUT_POST, 'email')){
    die("Email should not be empty");
}
if(!$password = filter_input(INPUT_POST, 'password')){
    die("Password should not be empty");
}

$host = "127.0.0.1:3307";
$dbusername = "root";
$dbpassword = "";
$dbname = "register";

$conn = new mysqli($host, $dbfirstname, $dblastname, $dbemail, $dbpassword);
if (mysqli_connect_error()) {
    die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
}

$sql = "INSERT INTO Signup (firstname, lastname, email, password) values ('$firstname','$lastname','email','password')";
if (!$conn->query($sql)) {
    echo "Error: " . $sql . "\r\n" . $conn->error;
    $conn->close();
    exit;
}

echo "New record is inserted successfully";

一般准则

不要嵌套 if() 语句。它使下面的代码非常混乱。
如果布尔结果是交易破坏者，请在 if/else 语句中首先显示该结果。例如，如果 SQL 查询失败，则首先在 if() 语句中获得否定结果（这将停止您的代码），然后一起跳过 else 语句。
评论你的代码。

Answer 2

<form method="post" action="overstack.php"> 
  <div class="container">
    <br> 
    <h1>Sign Up</h1>
    <p>Please fill in this form to create an account.</p>
    <hr>

    <label for="firstname"><b>First Name</b></label>
    <input type="text" placeholder="Enter First Name" name="firstname" required>

    <label for="lastname"><b>Last Name</b></label>
    <input type="text" placeholder="Enter Last Name" name="lastname" required>

    <label for="email"><b>Email</b></label>
    <input type="text" placeholder="Enter Email" name="email" required>

    <label for="psw"><b>Password</b></label>
    <input type="password" placeholder="Enter Password" name="psw" required>

    <label for="psw-repeat"><b>Repeat Password</b></label>
    <input type="password" placeholder="Repeat Password" name="psw-repeat" required>

    <label>
      <input type="checkbox" checked="checked" name="remember" style="margin-bottom:15px"> Remember me
    </label>

    <p>By creating an account you agree to our <a href="#" style="color:dodgerblue">Terms & Privacy</a>.</p>

    <div class="clearfix">
      <button type="button" class="cancelbtn">Cancel</button>
      <button type="submit" class="signupbtn">Sign Up</button>
    </div>
  </div>
</form>

这是我的 html 表单，用于上述 php 代码

PHP 代码说输入为空但它不是？

PHP code saying input is empty but it's not?

html

php

forms

login

rows