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Php 登录突然停止工作_Php_Login_Mysqli - Fatal编程技术网

Php 登录突然停止工作

Php 登录突然停止工作,php,login,mysqli,Php,Login,Mysqli,我正在做我的学校项目,我需要一个简单的登录功能。20分钟前它还在工作,但后来我可能犯了一些错误。它不会显示任何错误消息。数据库似乎没问题 “jmeno”=名称,“heslo”=密码 我想你把帖子的价值观混为一谈了。尝试: $username = $_POST['jmeno']; $password = $_POST['heslo']; 我认为你混合了后价值观。尝试: $username = $_POST['jmeno']; $password = $_POST['heslo']; 我认为你混

我正在做我的学校项目,我需要一个简单的登录功能。20分钟前它还在工作,但后来我可能犯了一些错误。它不会显示任何错误消息。数据库似乎没问题

“jmeno”=名称,“heslo”=密码


我想你把帖子的价值观混为一谈了。尝试:

$username = $_POST['jmeno'];
$password = $_POST['heslo'];

我认为你混合了后价值观。尝试:

$username = $_POST['jmeno'];
$password = $_POST['heslo'];

我认为你混合了后价值观。尝试:

$username = $_POST['jmeno'];
$password = $_POST['heslo'];

我认为你混合了后价值观。尝试:

$username = $_POST['jmeno'];
$password = $_POST['heslo'];

我建议调试如下:

<?php $mysqli = new mysqli("localhost","admin","admin","uzivatele");

    if(isset( $_POST['heslo']) && isset($_POST['jmeno'])){
        $username = $_POST['heslo'];
        $password = $_POST['jmeno'];
        /* defends SQL injection */
       // $username = stripslashes($username);
        //$password = stripslashes($password);
        //$password = mysqli_real_escape_string($mysqli, ($_POST['heslo']));
        //$username = mysqli_real_escape_string($mysqli, $_POST['jmeno']);

         $sqllogin = "SELECT * FROM prihlaseni WHERE jmeno = '".$username."'  AND heslo = '".$password."' LIMIT 1";


        echo $sqllogin; //check the sql query string
        $result = mysqli_query($mysqli, $sqllogin);
        print_r($result);
        if (!$result) {
        die(mysqli_error($mysqli));
        }
        $count = mysqli_num_rows($result);       

        if ($count == 1) {
        session_start();
        $_SESSION['loggedin'] = true;
        header('Location: home.php');

        }else {

        echo "<script language='javascript'>alert('Wrong password!');</script>";
    }
    }
?>

我建议调试如下:

<?php $mysqli = new mysqli("localhost","admin","admin","uzivatele");

    if(isset( $_POST['heslo']) && isset($_POST['jmeno'])){
        $username = $_POST['heslo'];
        $password = $_POST['jmeno'];
        /* defends SQL injection */
       // $username = stripslashes($username);
        //$password = stripslashes($password);
        //$password = mysqli_real_escape_string($mysqli, ($_POST['heslo']));
        //$username = mysqli_real_escape_string($mysqli, $_POST['jmeno']);

         $sqllogin = "SELECT * FROM prihlaseni WHERE jmeno = '".$username."'  AND heslo = '".$password."' LIMIT 1";


        echo $sqllogin; //check the sql query string
        $result = mysqli_query($mysqli, $sqllogin);
        print_r($result);
        if (!$result) {
        die(mysqli_error($mysqli));
        }
        $count = mysqli_num_rows($result);       

        if ($count == 1) {
        session_start();
        $_SESSION['loggedin'] = true;
        header('Location: home.php');

        }else {

        echo "<script language='javascript'>alert('Wrong password!');</script>";
    }
    }
?>

我建议调试如下:

<?php $mysqli = new mysqli("localhost","admin","admin","uzivatele");

    if(isset( $_POST['heslo']) && isset($_POST['jmeno'])){
        $username = $_POST['heslo'];
        $password = $_POST['jmeno'];
        /* defends SQL injection */
       // $username = stripslashes($username);
        //$password = stripslashes($password);
        //$password = mysqli_real_escape_string($mysqli, ($_POST['heslo']));
        //$username = mysqli_real_escape_string($mysqli, $_POST['jmeno']);

         $sqllogin = "SELECT * FROM prihlaseni WHERE jmeno = '".$username."'  AND heslo = '".$password."' LIMIT 1";


        echo $sqllogin; //check the sql query string
        $result = mysqli_query($mysqli, $sqllogin);
        print_r($result);
        if (!$result) {
        die(mysqli_error($mysqli));
        }
        $count = mysqli_num_rows($result);       

        if ($count == 1) {
        session_start();
        $_SESSION['loggedin'] = true;
        header('Location: home.php');

        }else {

        echo "<script language='javascript'>alert('Wrong password!');</script>";
    }
    }
?>

我建议调试如下:

<?php $mysqli = new mysqli("localhost","admin","admin","uzivatele");

    if(isset( $_POST['heslo']) && isset($_POST['jmeno'])){
        $username = $_POST['heslo'];
        $password = $_POST['jmeno'];
        /* defends SQL injection */
       // $username = stripslashes($username);
        //$password = stripslashes($password);
        //$password = mysqli_real_escape_string($mysqli, ($_POST['heslo']));
        //$username = mysqli_real_escape_string($mysqli, $_POST['jmeno']);

         $sqllogin = "SELECT * FROM prihlaseni WHERE jmeno = '".$username."'  AND heslo = '".$password."' LIMIT 1";


        echo $sqllogin; //check the sql query string
        $result = mysqli_query($mysqli, $sqllogin);
        print_r($result);
        if (!$result) {
        die(mysqli_error($mysqli));
        }
        $count = mysqli_num_rows($result);       

        if ($count == 1) {
        session_start();
        $_SESSION['loggedin'] = true;
        header('Location: home.php');

        }else {

        echo "<script language='javascript'>alert('Wrong password!');</script>";
    }
    }
?>


它不显示任何错误消息。
--您有吗?错误日志中有错误吗?如果没有错误,您登录后是否在同一页面上?显示您的
home.php
code@AmalMurali是的,它已启用。
它不显示任何错误消息。
--是否?错误日志中是否有错误?如果没有错误,登录后您是否在同一页面上?显示您的
home.php
code@AmalMurali是的,它已启用。
它不显示任何错误消息。
--您有吗?错误日志中有错误吗?如果没有错误,您登录后是否在同一页面上?显示您的
home.php
code@AmalMurali对它已启用。
它不显示任何错误消息。
--您有吗?错误日志中有错误吗?如果没有错误,登录后您是否在同一页面上?显示您的
home.php
code@AmalMurali是的,它已启用。