help with login page coding

Question

digital_plunder 0 Newbie Poster

13 Years Ago

when i try to access this login page..an empty page open...please help with this coding

<!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>Latest</title>
</head>

<body>

<?php
include("db.php");

if (isset($_POST["submit"]))
{
$username=$_POST["username"];
$password=$_POST["password"];
$query="select * from latest where username='$username' && password='$password'";
$query1=mysql_query($query);
if(mysql_num_rows($query1)>0)

{?>

<font color="#FF0000" size="+4"><b>
<?php  

 echo "welcome";
 echo ($username);

}
else 
{

?>
 

<center>
<table align="center" border="2">
<form enctype="multipart/form-data" action="<?php $_SERVER['PHP_SELF']  ?>" method="get">
<tr><td></td></tr>
<tr><td>Enter Username</td>
<td><input type="text" name="username" />
</td></tr>
<tr><td>Enter Password</td>
<td><input type="password" name="password" /></td>
</tr>
<tr><td>Submit</td><td><input type="submit" value="submit" name="submit"/></td></tr>
</form>
<?php } } ?>
</table>

</center>
</body>
</html>

php

5 Contributors
11 Replies
149 Views
1 Day Discussion Span
Latest Post 13 Years Ago Latest Post by digital_plunder

diafol

13 Years Ago

Your mix of html and php is a bit difficult to follow as your indenting is all to hell.

broj1 356 Humble servant

13 Years Ago

Move the curly brace form line 49 to a line 29. Your second if block should be a part of your first if block which means if form has been submitted and if password is correct then display the greeting else display the table with the form. See the code below.

There are other issues like security. Clean your input to prevent nasty guys and girls to inject nasty code.

<!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>Latest</title>
</head>

<body>

<?php
include("db.php");

if (isset($_POST["submit"]))
{
$username=$_POST["username"];
$password=$_POST["password"];
$query="select * from latest where username='$username' && password='$password'";
$query1=mysql_query($query);
if(mysql_num_rows($query1)>0)

{?>

<font color="#FF0000" size="+4"><b>
<?php  

 echo "welcome";
 echo ($username);

}
}
else 
{

?>
 

<center>
<table align="center" border="2">
<form enctype="multipart/form-data" action="<?php $_SERVER['PHP_SELF']  ?>" method="get">
<tr><td></td></tr>
<tr><td>Enter Username</td>
<td><input type="text" name="username" />
</td></tr>
<tr><td>Enter Password</td>
<td><input type="password" name="password" /></td>
</tr>
<tr><td>Submit</td><td><input type="submit" value="submit" name="submit"/></td></tr>
</form>
<?php } ?>
</table>

</center>
</body>
</html>

effu 13 Junior Poster in Training

13 Years Ago

It could be your SQL on line 17 of your code. Try using the word AND instead of the logical operator &&. Like this:

$query="select * from latest where username='$username' AND password='$password'";

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digital_plunder 0 Newbie Poster · Answer 1 · 2011-10-17T21:51:04+00:00

thx for the help bruh....i have improved it little bit...but m getting error...can you please resolve it...

<!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>Latest</title>
    </head>
     
    <body>
     
    <?php
    include("db.php");
     
    if(isset($_POST["submit"]))
    {
    $username=$_POST["username"];
    $password=$_POST["password"];
    $query="select * from latest where username='$username' && password='$password'";
    $query1=mysql_query($query);
    if(mysql_num_rows($query1)>0)
     
    {?>
     
    <font color="#FF0000" size="+4"><b>
    <?php
     
    echo "welcome "."$username";
      exit;  
    	}
		else
			{
				$msg="Invalid Username or Password";		
			}
			
	}
     
     ?>
    <table align="center" border="2">
    <form enctype="multipart/form-data" action="login.php" method="post">
    <tr><td> <?php echo($msg);?></td></tr>
    <tr><td>Enter Username</td>
    <td><input type="text" name="username" />
    </td></tr>
    <tr><td>Enter Password</td>
    <td><input type="password" name="password" /></td>
    </tr>
    <tr><td>Submit</td><td><input type="submit" value="submit" name="submit"/></td></tr>
    </form>
  
    </table>
     
    </center>
    </body>
    </html>

stoopkid 6 Junior Poster in Training · Answer 2 · 2011-10-18T02:43:48+00:00

There are a lot of things you need to change with this code.

Since this is a login code, you have 2 main goals: security & security. First, when accepting $_POST data, make sure you verify it.

if(isset($_POST['username'])) {
  $username = mysql_real_escape_string($_POST['username']);
}

You need to do the same for the password variable as well. Also, when dealing with passwords, always encrypt them...don't leave themas their original strings. Use PHPs md5 encryption so that passwords aren't retrieved/stored as their original values.

if(isset($_POST['password'])) {
 $password = mysql_real_escape_string($_POST['password']);
  $password = md5($password);
}

Lastly, how are you storing the login details? The proper way to do it is through session data.

digital_plunder 0 Newbie Poster · Answer 3 · 2011-10-18T07:03:12+00:00

That error comes that is not sql error,the error is on 39 line,when someone try to login with wrong username....can smeone help me out.

diafol · Answer 4 · 2011-10-18T12:43:52+00:00

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">

try that if the error is coming from line 39. You don't need enctype as you're not sending a file. Also, post is better as you don't see the details in the url.

broj1 356 Humble servant Featured Poster · Answer 5 · 2011-10-18T12:50:28+00:00

Also line 19

if(mysql_num_rows($query1)>0)

should be

if($query1 and mysql_num_rows($query1)>0)

since query1 is false when wrong username or password. And also initialize $msg first. Put the following code say on line 12:

$msg = '';

.

digital_plunder 0 Newbie Poster · Answer 6 · 2011-10-18T14:43:59+00:00

thanks broj1...my problem has been resolved...thx a lots....

broj1 356 Humble servant Featured Poster · Answer 7 · 2011-10-18T15:00:17+00:00

Thats why we are here for:-). But do not forget to improve the code with security measures.

digital_plunder 0 Newbie Poster · Answer 8 · 2011-10-18T15:36:08+00:00

digital_plunder 0 Newbie Poster

13 Years Ago

Ok ok bro,i will improve it..thx again.....