-1 
 <?php
  session_start();
  ?>
   <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
   "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
   <html xmlns="http://www.w3.org/1999/xhtml">
  <head>
  <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  <title>Member System - Login</title>
   </head>
   <body>
   <?php

 $form = "<form action='login.php' method='post'>
 <table>
 <tr>
     <td>Username:</td>
     <td><input type='text' name='user'  /</td>
 </tr>
 <tr>
     <td>Password:</td>
     <td><input type='password' name='password'  /</td>
 </tr>
 <tr>
     <td></td>
     <td><input type='submit' name='loginbtn' value='Login'  /</td>
 </tr>
 </table>
 </form>";

 if ($_POST['loginbtn']){
     $user = $_POST['user'];
     $password = $_POST['password'];

     if ($user){
         if ($password){
             require("connect.php");

             $password = md5(md5("sjkdhfgh".$password."asdfhh"));
             // make sure login info correct
    $query = mysql_query("SELECT * FROM users WHERE username='$user'");
             $numrows = mysql_num_rows($query);
             if ($numrows == 1){
                 die(mysql_error());
                 $row = mysql_fetch_essoc($query);
                 $dbid = $row ['id'];
                 $dbuser = $row ['username'];
                 $dbpass = $row ['password'];
                 $dbactive = $row ['active'];

                 if ($password == $dbpass){
                     if ($dbactive == 1){

                         // set seccion info
                         $_SESSION['userid'] = $dbid;
                         $_SESSION['username'] = $dbuser;
            echo "You Have Been Logged In As <b>$dbuser</b>. <a
  href='index.html'>Click Here</a> To Go Home";

                     }
                     else
        echo "You Must Activate You Accout To Login. $form";
                 }
                 else
    echo "You Did Not Enter The Right Password.       $form";

             }
             else
                 echo "The Username You Entered Was Not Found. $form";

             mysql_close();

         }
         else
             echo "You Must Enter Your Password. $form";
     }
     else  
         echo "You Must Enter Your UserName. $form";
}        
else
    echo $form;
 ?>
 <body>
 </html>

我正在尝试创建一个登录网页，但我不断收到此错误

连接到 MySQL

警告：mysql_num_rows() 期望参数 1 是资源，布尔值在 D:\Hosting\11742398\html\Test\Login.php 第 41 行给出的用户名未找到。

4

2 回答 2

1

用这个 - 

$user = mysql_real_escape_string($_POST['user']);

使用调试你的 sql 查询mysql_error()

mysql_query("SELECT * FROM users WHERE username='$user'") or die(mysql_error());

你不应该使用 mysql_* 函数。

于 2013-09-13T10:35:23.217 回答
0

修改行：

$query = mysql_query("SELECT * FROM users WHERE username='$user'");

至

$query = mysql_query("SELECT * FROM users WHERE username='$user'") or die(mysql_error());

你会在那里得到确切的错误。

于 2013-09-13T10:35:40.793 回答