Question

因此，我在PHP中使用scracth构建了一个网站，并使用WAMP服务器进行测试。

所有内容都在localhost上运行，但在将代码上传到服务器后，使用数据库连接的所有内容都不起作用。

我已经检查过php中的POST是否在服务器上运行，并且它正常工作，我在谷歌搜索后改变了一些php.ini配置并且我已经完成了这个：

register_globals = off; (was default)

allow_url_fopen = on; (was default)

magic_quotes_gpc = off; (was default)

表单代码

http://i.imgur.com/k5kMx9N.png

类中的登录功能

function LoginUser($UserName, $Password)
{
    //Protect from SQL Injection
    $clientusername = stripslashes(mysql_real_escape_string($UserName));
    $clientpassword = stripslashes(mysql_real_escape_string($Password));

    $clientusername = trim($clientusername);
    $clientpassword = trim($clientpassword);

    if($clientusername != "" && $clientusername != "Username" && $clientpassword != "" && $clientpassword != "Password")
    {           
        $this->ConnectToMySQL();
        $sql = "SELECT * FROM Users WHERE UserName = '" . $clientusername . "' AND Password = '" . $clientpassword . "'";
        $result = mysqli_query($this->DBConnection, $sql);
        mysqli_close($this->DBConnection);      
        // Mysql_num_row is counting table row      
        $count = mysqli_num_rows($result);
        // If result matched $myusername and $mypassword, table row must be 1 row
        if($count == 1){
            if(!isset($_SESSION))
                session_start();
            $row = mysqli_fetch_array($result);
            $_SESSION['ID'] = session_id();
            $_SESSION['timeout'] = time();
            session_write_close();
            return("O seu login foi feito com sucesso.</br></br><a href='privatemenu.php'>Ir para o Menu</a>");
        }
        else return("Não foram encontrados utilizadores com os dados fornecidos.</br></br><a href='login.php'>Voltar á pagina de Login</a>");

    }
    else return("Ocorreu um erro com a entrada dos seus dados.</br></br><a href='login.php'>Voltar á pagina de Login</a>");
}

所以，问题是这个，函数总是在登录功能中返回文本“Ocorreu um erro com a entrada dos seus dados”。

我写了登录数据，按下按钮登录，帖子正在工作我用相同的表格进行了测试，并在另一个页面上回显了表格中的帖子，但是函数永远不会低于第一个if。

任何帮助都将非常感激，因为我认为这不是来自函数，我有另一个只在数据库上插入并完成相同的事情。最奇怪的是数据正确地通过了这些页面。

如果对此有任何疑问，请问我，以便我能更好地解释。

所以感谢Paddyd这里正确的代码完全正常工作：

function LoginUser($UserName, $Password)
{
    //Protect from SQL Injection
            $this->ConnectToMySQL();
    $clientusername = stripslashes(mysqli_real_escape_string($this->DBConnection, $UserName));
    $clientpassword = stripslashes(mysqli_real_escape_string($this->DBConnection, $Password));

    $clientusername = trim($clientusername);
    $clientpassword = trim($clientpassword);

    if($clientusername != "" && $clientusername != "Username" && $clientpassword != "" && $clientpassword != "Password")
    {           
        $sql = "SELECT * FROM Users WHERE UserName = '" . $clientusername . "' AND Password = '" . $clientpassword . "'";
        $result = mysqli_query($this->DBConnection, $sql);
        mysqli_close($this->DBConnection);      
        // Mysql_num_row is counting table row      
        $count = mysqli_num_rows($result);
        // If result matched $myusername and $mypassword, table row must be 1 row
        if($count == 1){
            if(!isset($_SESSION))
                session_start();
            $row = mysqli_fetch_array($result);
            $_SESSION['ID'] = session_id();
            $_SESSION['timeout'] = time();
            session_write_close();
            return("O seu login foi feito com sucesso.</br></br><a href='privatemenu.php'>Ir para o Menu</a>");
        }
        else return("Não foram encontrados utilizadores com os dados fornecidos.</br></br><a href='login.php'>Voltar á pagina de Login</a>");

    }
    else return("Ocorreu um erro com a entrada dos seus dados.</br></br><a href='login.php'>Voltar á pagina de Login</a>");
}

Answer 1

Warning: mysql_real_escape_string(): Access denied for user 'apache'@'localhost' (using password: NO) in /home/httpd/vhosts/jaimevale.com/httpdocs/config/configurations.php on line 461

这些是造成问题的原因：

$clientusername = stripslashes(mysqli_real_escape_string($UserName));
$clientpassword = stripslashes(mysqli_real_escape_string($Password));

您尝试在未建立SQL连接的情况下使用mysql_real_escape_string。在进行这些调用之前建立连接或使用mysql_real_escape_string的替代方法。

尝试在函数开头调用$this->ConnectToMySQL();。

修改：将mysql_real_escape_string更改为mysqli_real_escape_string

在WAMP localhost上运行但在WebServer上不起作用的PHP代码

1 个答案: