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使用MySQL登录PHP

[英]PHP Login using MySQL

 原文  2016-08-05 04:42:03  0   7    php/ mysql/ login

问题描述

我有3个文件，分别是connection.php，work.php和login.php，目前正在使用登录功能。

我的数据库名称是野生生物，它的字段为wrd_username（用于用户名）和wrd_password（用于密码）

我遇到了困难，因为SQL无法检测到用户名和密码。

此PHP需要MySQL。 请帮助我。

connection.php 

<?php
$conn = mysql_connect('localhost', 'root', '', 'wildlife');
if (!$conn) 
{
    die('Connect Error: ' . mysql_errno());
}
else
{
    echo ("connected from connection.php");
}?>

work.php 

  <?php include('login.php'); ?> <html> <head><title>howww</title> </head> <body> <form action="login.php" method="post"> <!-- Sign In Process --> Username: <input type="text" name="user" id="emp_username"style="width:150"> <br /> Password: <input type="password" name="pass" id="emp_password"style="width:153"> <br /> <br /> <input type="submit" value="submit"> </form> </body> </html>

login.php 

enter code here
    <?php
    // Try and connect to the database
    include('connection.php');
    $selected = mysql_select_db("wildlife",$conn)
      or die("Could not select ");

       //$myusername = mysql_real_escape_string($conn,$_POST['username']);
       //$mypassword = mysql_real_escape_string($conn,$_POST['password']); 
        if(isset($_POST['user']) && isset($_POST['pass'])){
                $emp_username = $_POST['emp_username'];
                $emp_password = $_POST['emp_username'];

                $query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");
                if(mysql_num_rows($query) > 0 ) 
                { 
                    //check if there is already an entry for that username
                    echo "DETECTED Username AND PASS already exists!";
                }
                else
                {

                    //header("location:index.php");
                    echo(" okay");
                    //header('work.php');
                }
        }
        mysql_close();
    ?>

每当我尝试输入用户名和密码时，总是从login.php以ok结尾

7 个解决方案

解决方案1
 1  2016-08-05 04:47:21

您的错误似乎是为密码变量分配了错误的$_POST变量： 

$emp_password = $_POST['emp_username'];

应该是这样： 

$emp_password = $_POST['emp_password'];

另外，您的代码容易受到SQL注入的攻击。 尝试学习使用准备好的陈述 。

解决方案2
 0  2016-08-05 04:50:19

您需要对login.php文件进行这些更改 

<?php
// Try and connect to the database
include('connection.php');
$selected = mysql_select_db("wildlife",$conn)
  or die("Could not select ");

   //$myusername = mysql_real_escape_string($conn,$_POST['username']);
   //$mypassword = mysql_real_escape_string($conn,$_POST['password']); 
    if(isset($_POST['user']) && isset($_POST['pass'])){
            $emp_username = $_POST['user']; //name as of html form
            $emp_password = $_POST['pass']; //name as of html form

            $query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");
            if(mysql_num_rows($query) > 0 ) 
            { 
                //check if there is already an entry for that username
                echo "DETECTED Username AND PASS already exists!";
            }
            else
            {

                //header("location:index.php");
                echo(" okay");
                //header('work.php');
            }
    }
    mysql_close();
?>

解决方案3
 0  2016-08-05 04:51:52

您的$user和$pass变量始终为空。我已经更新了下面的login.php页面。 使用下面的代码 

<?php
        // Try and connect to the database
        include('connection.php');
        $selected = mysql_select_db("wildlife",$conn)
          or die("Could not select ");

           //$myusername = mysql_real_escape_string($conn,$_POST['username']);
           //$mypassword = mysql_real_escape_string($conn,$_POST['password']); 
            if(isset($_POST['user']) && isset($_POST['pass'])){
                    $user = $_POST['user'];
                    $pass = $_POST['pass'];

                    $query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");
                    if(mysql_num_rows($query) > 0 ) 
                    { 
                        //check if there is already an entry for that username
                        echo "DETECTED Username AND PASS already exists!";
                    }
                    else
                    {

                        //header("location:index.php");
                        echo(" okay");
                        //header('work.php');
                    }
            }
            mysql_close();
        ?>

解决方案4
 0  2016-08-05 04:52:38

我相信问题是：

$query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and emp_password='$pass'");

应该：

$query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$emp_username' and emp_password='$emp_password'");

看起来您在对MySQL的查询中只是有一些未使用的变量（$ user和$ pass并未真正分配给任何东西）。 除此之外，一切看起来都不错。

解决方案5
 0  2016-08-05 04:53:33

仅用3行，它应该是这样的： 

$user = $_POST['emp_username'];
$pass = $_POST['emp_password'];

$query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");

解决方案6
 -1  2016-08-05 04:59:00

在connection.php页面中，无需在login.php页面mysql_select_db（“ wildlife”，$ conn）和查询中使用mysql_connect（'localhost'，'root'，``，'wildlife'）添加数据库只需在条件emp_username ='$ emp_username'和emp_password ='$ emp_password'“添加适当的变量

connection.php 

<?php
$conn = mysql_connect('localhost', 'root', '');
if (!$conn) 
{
    die('Connect Error: ' . mysql_errno());
}
else
{
    echo ("connected from connection.php");
}?>

work.php 

<?php include('login.php'); ?>

<html>
<head><title>PLEASE GUMANA KA NA</title>
</head>
<body>
<form action="login.php" method="post"> <!-- Sign In Process -->
Username: <input type="text" name="user" id="emp_username"style="width:150">
<br />
Password: <input type="password" name="pass" id="emp_password"style="width:153">
<br />
<br />
<input type="submit" value="submit">
</form>
</body>
</html>

login.php 

enter code here
    <?php
    // Try and connect to the database
    include('connection.php');
    $selected = mysql_select_db("wildlife",$conn)
      or die("Could not select ");

         if(isset($_POST['user']) && isset($_POST['pass'])){
                $emp_username = $_POST['user'];
                $emp_password = $_POST['pass'];

                $query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$emp_username' and  emp_password='$emp_password'");
                if(mysql_num_rows($query) > 0 ) 
                { 
                    //check if there is already an entry for that username
                    echo "DETECTED Username AND PASS already exists!";
                }
                else
                {

                    //header("location:index.php");
                    echo(" okay");
                    //header('work.php');
                }
        }
        mysql_close();
    ?>

解决方案7
 -1  2016-08-05 05:01:20

$emp_username = $_POST['emp_username'];
$emp_password = $_POST['emp_username'];

$query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");

该代码段应更改为： 

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

$query = mysql_query("SELECT * FROM wrd_users WHERE emp_username='$user' and  emp_password='$pass'");

因为表单具有user和pass参数，所以没有emp_username或emp_password

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