Question

我编辑代码后转为$_SESSION['just'] = $username;

解决了以下问题

我正在尝试创建一个注册系统，一旦用户注册就会生成OTP，我正在测试localhost上的网站。一旦用户点击注册。我的activateaccount.php无效，它应该将otp和status字段分别设置为NULL和1。即使用户插入OTP不正确，它仍然会将它们路由到login.html

这是我的signup.php

的代码

<?php
session_start();
require 'php includes/db.ini.php';
$fill = "fill all the feilds";

$random = mt_rand(1000,1000000);

if(isset($_POST['first'])){
    $first = trim(mysqli_real_escape_string($connection,$_POST['first']));
    }
    if(isset($_POST['last'])){
        $last = trim(mysqli_real_escape_string($connection,$_POST['last']));
        }
        if(isset($_POST['username'])){
            $username = trim(mysqli_real_escape_string($connection,$_POST['username']));
            }
            //invalid email can be submited, and will be accepted but the account won't be activated if the email dosen't get verified
            if(isset($_POST['email'])){
                $email = trim(mysqli_real_escape_string($connection,$_POST['email']));
                }
                if(isset($_POST['password'])){
                    $password = trim(mysqli_real_escape_string($connection,$_POST['password']));
                }

$required = array('first', 'last', 'username', 'email', 'password');

$error = false;

foreach($required as $field) {
    if (empty($_POST[$field])) {
        $error = true;
    }
    }

    if (!$error) {
        //checking if any of the required feilds are empty 

            $sql = "SELECT Email FROM registeredusers WHERE Email='$email'";
            $result = mysqli_query($connection,$sql);
            $emailCheck = mysqli_num_rows($result);
            $structure = "users/$username/profile photos/project photos";

                if ($emailCheck>0){
                    echo "email exists";
                }else{
                    if (file_exists($structure)){
                    echo "username already exists";
                    }
                        else{
                            $hpwd = password_hash($password,PASSWORD_DEFAULT);
                            $username_lower = strtolower($username);
                            $sql = "INSERT INTO registeredusers (FirstName,LastName,UserName,Email,Password,otp,status)
                            VALUES ('$first','$last','$username','$email','$hpwd','$random','0')";
                            mkdir($structure,0777,true);
                            $result = mysqli_query($connection,$sql);
                            $_SESSION['just'] = $username;
                            header("Location:activateaccount.html");

                        }
                    }}else{
                        echo "fill out all the feilds";
                    }

?>

我认为我创建的会话无法正常运行，或者我可能没有正确使用它，这是我的activateaccount.php代码

<?php
session_start();
include 'php includes/db.ini.php';

$username = $_SESSION['just'];
echo $username;

if(isset($_POST['submit'])){

    $submitted_otp = trim(mysqli_real_escape_string($connection,$_POST['otp']));

}

$required = array('otp');

$error = false;

foreach($required as $field) {
    if (empty($_POST[$field])) {
        $error = true;
    }
    }

    if (!$error) {
        $sql = "SELECT UserName FROM registeredusers WHERE UserName='$username'";
        $result = mysqli_query($connection,$sql);
        $row = mysqli_fetch_assoc($result);
        $otp = $row['otp'];
        echo $otp;

        if($otp !== $submitted_otp){

            echo "the otp isn't correct";

        }else{
            $update = "UPDATE registeredusers SET status='1', otp=NULL";
            session_destroy();
            header("Location: login.html");

        }}


?>

我会在一段时间后在activateaccount.php页面上销毁会话。有谁可以帮助我吗。我甚至无法使用其他文件回显会话。

Answer 1

以下条件永远不会传递if条件，它总是执行其他条件。请检查以下更新条件

if($otp != $submitted_otp){

请改变这种情况。会话将永远不会在标题位置之后设置。请将其更改为

$_SESSION['just'] = $row['UserName'];
header("Location:activateaccount.html");

添加mysql_fetch_assoc

$sql = "INSERT INTO registeredusers (FirstName,LastName,UserName,Email,Password,otp,status)
        VALUES ('$first','$last','$username','$email','$hpwd','$random','0')";
        mkdir($structure,0777,true);
        $result = mysqli_query($connection,$sql);
        $row = mysqli_fetch_assoc($result);
        $_SESSION['just'] = $row['UserName'];

无法在mysql列中插入数据

1 个答案: