2014-10-06 144 views
-1

我想知道这个PHP编码的错误,只是想知道在这个编码中的错误在哪里我跳过页面输入设计的HTML部分,因为我只想知道PHP部分, ,我试图使用该用户输入用户名和密码登录到网站和该特定网站应该是密码保护密码php mysql用户名密码验证

<?php> 
if isset($_POST=['submit'])); 
{ 
$inputuser = $_POST['user']; 
$inputpass = $_POST['pass']; 

$user = "root"; 
$password = ""; 
$database = "Tutorial"; 

$connect = mysql_connect("localhost",$user,$password); 
@mysql_select_db($database) or ("database not found"); 

$query = " SELECT * FROM 'users' WHERE 'user' =        i $inputuser";//for query specific data 
$querypass = "SELECT * FROM 'users' WHERE 'user' =   $  i $inputpass'"; 

$result = mysql_query($query); 
$resultpass = mysql_query($querypass); 

$row = mysql_fetch_array($result); 
$rowpass = mysql_fetch_array($resultpass); 

$serveruser = $row["user"]; 
$serverpass = $row["password"]; 

if($serveruser&&$serverpass){ 
if (!$result) { 
die("username and password is invalid"); 
} 

echo "<br> <center>database output</b></center><br><br>"; 
mysql_close(); 
echo $inputpass; 
echo $serverpass; 
if ($inputpass == $serverpass) { 
header('location: Home.php'); 
} else { 
header('location: fail.php'); 
} 
} 
+1

什么是你所得到的错误?你能对这个问题更具体一点吗? – addicted20015 2014-10-06 10:00:39

+3

请停止使用'mysql_ *'函数,因为这些函数已被弃用 – Azrael 2014-10-06 10:04:55

+1

您应该在E_ALL上设置错误报告并删除与数据库连接的错误抑制。你想要打印错误,我们希望看到你的错误:) – 2014-10-06 10:06:51

回答

0

试试这个方法:

$query="SELECT * FROM users WHERE user='" . mysql_real_escape_string($inputuser) . "' AND psw='" . mysql_real_escape_string($inputpsw) . "'"; 
$qr=mysql_query($query) or die (mysql_error()); 
if(mysql_num_rows($qr)>0) //admitting that usernames and psw are unique 
{ 
    //success 
    header('location:home.php'); 
} 
else //no rows=no username responding to $inputusername 
{ 
    header('location:fail.php'); 
} 
+0

感谢您的信息 – 2014-10-08 08:00:01