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Php oop无法从db获取数据

php

Domica · 技术社区 · 6 年前

所以我有一段代码,我只是按照一些指南来创建,

<?php

session_start();

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

include 'db.conf.php';

$uid = mysqli_real_escape_string($conn, $_POST['uid']);
$pwd = mysqli_real_escape_string($conn, $_POST['pwd']);
$_SESSION['uid'] = $uid;

//Error handleri
//Check jesu inputi empty

if (empty($uid) || empty($pwd))
{
    header("Location: ../index.php?login=empty");
    exit();
}
else
{
    $sql = "SELECT * FROM users WHERE user_uid = '$uid' OR user_email = '$uid'";
    $result = mysqli_query($conn, $sql);
    $resultCheck = mysqli_num_rows($result);

    if ($resultCheck < 1)
    {
        header("Location: ../index.php?login=usernamenepostoji");
        exit();
    }
    else
    {
        if ($row = mysqli_fetch_assoc($result)) {
            //Dehashiranje
            $hashedPwdCheck = password_verify($pwd, $row['user_pwd']);
            if ($hashedPwdCheck == false) {
                header("Location: ../index.php?login=invalidpass");
                exit();
            }
            elseif ($hashedPwdCheck == true)
            {
                //Logiranje
                $_SESSION['u_id'] = $row['user_id'];
                $_SESSION['u_first'] = $row['user_first'];
                $_SESSION['u_last'] = $row['user_last'];
                $_SESSION['u_email'] = $row['user_email'];
                $_SESSION['u_uid'] = $row['user_uid'];
                header("Location: ../index.php?login=success");
                exit();
            }

        }
    }
}

}

else
{
    header("Location: ../index.php?login=error");
    exit();
}

?>

它只是简单的错误处理和登录工作。我理解它,想用更多的oop来重新创建它。

<?php 
session_start();

include 'db.conf.php';

class Login
{

public $username;
public $password;


function __construct()
{
    $this->username = $_POST['uid'];
    $this->password = $_POST['pwd'];
    $this->checkinputs();
}


function checkinputs()
{       
        if (empty($this->username) || empty($this->password)) 
        {
            header("Location: ../index.php?login=empty");
            exit();
        }
        else
        {   
            $username = $this->username;
            $sql = "SELECT * FROM users WHERE user_uid =".$username;
            $result = mysqli_query($conn, $sql);
            $resultCheck = mysqli_num_rows($result);

            if ($resultCheck < 1) 
            {
                header("Location: ../index.php?login=usernamenepostoji");
                exit();
            }
            else 
            {
            if ($row = mysqli_fetch_assoc($result)) {
                //Dehashiranje
                $hashedPwdCheck = password_verify($pwd, $row['user_pwd']);
                if ($hashedPwdCheck == false) {
                    header("Location: ../index.php?login=invalidpass");
                    exit();
                }
                elseif ($hashedPwdCheck == true) 
                {
                    //Logiranje
                    $_SESSION['u_id'] = $row['user_id'];
                    $_SESSION['u_first'] = $row['user_first'];
                    $_SESSION['u_last'] = $row['user_last'];
                    $_SESSION['u_email'] = $row['user_email'];
                    $_SESSION['u_uid'] = $row['user_uid'];
                    header("Location: ../index.php?login=success");
                    exit();
                }
            }
        }
    }
}
}


?>

这就是我得到的,它是字面上相同的代码只是使用函数和其他东西来'分离'成块。它不起作用。我一直在if$resultCheck<1头上卡住,这意味着用户名不存在。不过我肯定是的,因为数据库里什么都没变。所以它让我想到是$conn,它只是没有连接到数据库。我已经将$username变量转储到一个文件中,以检查它是否有效。我只是不知道该怎么办。

1 回复 | 直到 6 年前

Jeff 6 年前

$conn 方法中不存在 checkinputs()

要么全球化:

function checkinputs()
{ 
    global $conn;
    ...
}

我不推荐(因为 using globals is disencouraged

或者把它传给 Login::_construct() $this->conn (然后用作 $this->连接 $result = mysqli_query($this->conn, $sql); ):

function __construct($conn)
{
    $this->conn = $conn; // maybe also check if you have a valid connection!
    $this->username = $_POST['uid'];
    $this->password = $_POST['pwd'];
    $this->checkinputs();
}

function checkinputs()
{       
// no global now!
        ....    
        $result = mysqli_query($this->conn, $sql);
        ....
}

但是 :请切换到 prepared stements . 此代码易受sql注入攻击!

Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?