信息未插入数据库（我的代码有什么问题？）

Question

我正在尝试制作一个带有验证的注册表单。我在 html 和 css 代码中添加了一些 php 代码。一切都是好的。它验证用户或电子邮件是否已经存在。如果验证通过，则数据将保存在数据库中，但它不起作用。不知道是什么问题

这是我的代码：

<?php

  if (isset($_POST['submit'])) {
    require 'connect.php';

    $username       = ($_POST['username']);
    $email          = ($_POST['email']);
    $password       = ($_POST['password']);
    $passwordconf   = ($_POST['passwordconf']);

    $errorfields    = "<p class='errormsg'>Please fill out all the fields!</p>";

    if (empty($username) || empty($email) || empty($password) || 
    empty($passwordconf))
    {
      echo "$errorfields";
    }

    $check = mysqli_query($con, "SELECT username FROM users WHERE 
    username='$username' ");

    if (mysqli_num_rows($check) >= 1) {
        echo "username already exists"."</br>". "</br>";
    }

    $erroremail     = "<p class='errormsg'>Email is not in name@domain format! </p>";
    $regex          = "/^[a-z0-9]+([_.-][a-z0-9]+)*@([a-z0-9]+([.-][a-z0-9]+)*)+.[a-z]{2,}$/i";

    if(!preg_match($regex, $email))
    {
      echo "$erroremail";
    }

    $errorpassword  = "<p class='errormsg'>You passwords do not match!</p>";

    if ($password != $passwordconf)
    {
      echo "$errorpassword";
    }

    $check = mysqli_query($con, "SELECT email FROM users WHERE email='$email' ");

    if (mysqli_num_rows($check) >= 1) {
      echo "email already exists";
    }

   } else {

    $con = mysqli_connect('localhost' , 'root', '');

    if(!$con) {
      echo "not connected";
    }

    if (!mysqli_select_db($con, "new accounts")) {
      echo "database not selected";
    }  

    $username= (isset($_POST['username']));
    $email= (isset($_POST['email']));
    $password= (isset($_POST['password']));

    mysqli_query($con, "INSERT INTO users (username, email, password) VALUE ('$username', '$email', '$password')") or die ( "cannot insert in databse");
  }

?>

首先，我写了$username = ($_POST['username']) 然后它显示错误，所以我将其更改为此。它没有向我显示错误，但它没有在数据库中插入任何内容。有人可以告诉我如何解决这个问题吗？谢谢

Answer 1

简短的回答是：

您的代码正在插入来自 isset() 的布尔结果值，这当然不是您的意图。删除 isset() 调用并声明提交的值。

但是，您的代码还有很多问题需要解决。

对于有这么多问题的代码块，这是我愿意慷慨的慷慨：

<?php
if (isset($_POST['submit']) {
    if (empty($_POST['username']) || empty($_POST['email']) || empty($_POST['password']) || empty($_POST['passwordconf'])) {
        $error = "<p class='errormsg'>Please fill out all the fields!</p>";
    } elseif ($_POST['password'] !== $_POST['passwordconf']) {
        $error  = "<p class='errormsg'>You passwords do not match!</p>";

    } elseif (!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL))
        $error = "<p class='errormsg'>Email is not in name@domain format!</p>";
    } else {
        require 'connect.php'; // implement object-oriented $con variable
        if ($stmt=$con->prepare("SELECT * FROM users WHERE username=?")) {
            if ($stmt->bind_param('s', $_POST['username']) && $stmt->execute() && $stmt->store_result()) {
                if ($stmt->num_rows) {
                    $error = "<p class='errormsg'>Username already exists</p>";
                }
            } else {
                $error = "<p class='errormsg'>Username Check Statement Error</p>"; // $stmt->error
            }
            $stmt->close();
        } else {
            $error = "<p class='errormsg'>Username Check Prepare Error</p>"; // $con->error;
        }
        if ($stmt=$con->prepare("SELECT * FROM users WHERE email=?")) {
            if ($stmt->bind_param('s', $_POST['email']) && $stmt->execute() && $stmt->store_result()) {
                if ($stmt->num_rows) {
                    $error = "<p class='errormsg'>Email already exists</p>";
                }
            } else {
                $error = "<p class='errormsg'>Email Check Statement Error</p>"; // $stmt->error
            }
            $stmt->close();
        } else {
            $error = "<p class='errormsg'>Email Check Prepare Error</p>"; // $con->error;
        }
    }
    if ($error) {
        echo $error;
    } else {
        // Perform your insert with $_POST['username'], $_POST['email'], $_POST['password'] , but DO NOT EVER, EVER, EVER store raw passwords...
        // This subject is too extensive and gathers too much scrutiny for me to dare to post any hard-fast lines of code on Whosebug
        // Every minute that you spend researching this topic is time well spent.
        // Not learning about cryptography and password security will lead to many, many unfortunate events for you and your users.
    }
}
?>

所以，一般来说：

为了安全起见，使用带占位符的准备好的语句。
实施错误检查以便您可以快速调试问题。
正确地标记您的代码，以便遵循代码逻辑。
为了稳定性，使用电子邮件验证器而不是正则表达式。
在更昂贵的验证检查（如查询调用）之前执行更简单的验证检查
仅在必要时连接到您的数据库，并且re-use所有查询都使用相同的连接。
一个login/registration系统，如果操作得当，绝非易事；研究许多资源并确定适合您的 php 版本的最佳方法，并实施现有的最佳加密和存储技术。

信息未插入数据库（我的代码有什么问题？）

Information is not inserting in Database (What's wrong with my code?)

php

validation

post

mysqli

submission