Question

有人可以向我推进正确的方向并告诉我这个代码有什么问题。我没有收到任何错误，但它没有登录我。我的想法是在sql数据库中有一行名为memb的行，并检查状态是1还是0.如果是1，它将带你到安全页面，如果是0，它会给你子网页

<?php


require_once('config.php');  


if(isset($_POST['Login']) && isset($_POST['uname']) && isset($_POST['pass1'])) {
// Here check if all input are sent 

$uname = $sql-> real_escape_string($_POST['uname']);
$pass1 = $sql-> real_escape_string($_POST['pass1']);


 if ( !empty($uname) && !empty($pass1)) {
  // Check if you input are not empty

  $query= mysqli_query($sql,"SELECT * FROM login WHERE uname='".$uname."' AND pass='".$pass1."' && memb='?'");
  if(!$query){ die(mysqli_error($sql)); }
  $checkuser= mysqli_num_rows($query);


  if($checkuser != 1) {


   $error = "Username doesn't exist in our database!";
  }

  // Change $login with $query
  while ($row = mysqli_fetch_array($query)) {
   $checkpass= $row['pass'];

 while ($row = mysqli_fetch_array($query)) { 
   $checkmemb= $row['memb'];   

   // here i changed $pass1 to $checkpass
   if ($pass1 == $checkpass && $memb == 0){

    setcookie("user", $uname, time()+7200);
    $_SESSION['user'] = $uname;
    $_SESSION['start'] = time();
    $_SESSION['expire'] = $_SESSION['start'] + (60 * 60 * 60);
    header("Location: pay.php");
    exit();
   } elseif($pass1 == $checkpass && $memb == 1) {
     setcookie("user", $uname, time()+7200);
    $_SESSION['user'] = $uname;
    $_SESSION['start'] = time();
    $_SESSION['expire'] = $_SESSION['start'] + (60 * 60 * 60);
    header("Location: main.html");
    exit();
   }
  }
 }
 }
 }

Answer 1

尝试更改此内容：

if(isset($_POST['Login']) && isset($_POST['uname']) && isset($_POST['pass1'])) {
// Here check if all input are sent

为：

if(isset($_POST['uname']) && isset($_POST['pass1'])) {
// Here check if all input are sent

没有测试，这只是一个快速查看的猜测。但它似乎没有做任何事情，因为你的输入检查失败了。

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1 个答案: