Question

我开发了一个登录表单，用户需要填写用户名＆amp;我存储在（table1）中的密码，还需要输入我存储在（table2）中的密钥。如果是用户名，密码和密码。密钥是正确的，它将显示SUCCESSFULLY LOGIN TO USER PROFILE PAGE ...＆＃34;如果错了，它会显示＆＃34;抱歉......你输错了ID和密码......请重试......＆＃34;。下面是我的登录表单和connectivity.php的代码..现在，如果我输入正确的用户名，密码和正确的密钥，它将始终显示错误的消息。

＆＃13;

<?php 
define('DB_HOST', 'localhost'); 
define('DB_NAME', 'login');
define('DB_USER','root'); 
define('DB_PASSWORD',''); 
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error()); $db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error()); 
$Username = $_POST['user']; $Password = $_POST['pass'];  $Captcha = $_POST['captcha'];
function SignIn() { session_start(); 

//starting the session for user profile page 
if(!empty($_POST['user'])) 
//checking the 'user' name which is from Sign-In.html, is it empty or have some text 
{

$query = mysql_query("SELECT username.userName, username.pass, ,secret.key where userName = '$_POST[user]' AND pass = '$_POST[pass]' AND key = '$_POST[captcha]'") or die(mysql_error());


$row = mysql_fetch_array($query); 


if(!empty($row['userName']) AND !empty($row['pass']) AND !empty($row['captcha'])) 
{ 
$_SESSION['userName'] = $row['pass'] = $row['captcha']; 
 echo "SUCCESSFULLY LOGIN TO USER PROFILE PAGE...";
 } 
 else 
 { 
 echo "SORRY... YOU ENTERED WRONG ID AND PASSWORD... PLEASE RETRY...";
  } 
  } 
  } if(isset($_POST['submit'])) { SignIn(); 
  } 
  ?>

＆＃13;

<form method="POST" action="connectivity.php"> 
User <br><input type="text" name="user" size="40"><br> 
Password <br><input type="password" name="pass" size="40"><br> 
Captcha <br><input type="password" name="captcha" size="40">

<input id="button" type="submit" name="submit" value="Log-In"> 
</form>

＆＃13;

Answer 1

替换此行

$query = mysql_query("SELECT username.userName, username.pass, ,secret.key where userName = '$_POST[user]' AND pass = '$_POST[pass]' AND key = '$_POST[captcha]'") or die(mysql_error());

与

$user = $_POST['user'];
$pass = $_POST['pass'];
$captcha = $_POST['captcha'];

$query = mysql_query("SELECT username.userName, username.pass ,username.key from username where userName = '".$user."' AND pass = '".$pass."' AND key = '".$captcha."'") or die(mysql_error());

你在查询中犯了错误，在用脚本编写查询之前尝试进行虚拟查询并使用sql ui或phpmyadmin在mysql中运行而不是在脚本中实现

Php登录表单，包含2个表的数据

1 个答案: