Jump to content
Please Use CODE Tags

Getting PHP Errors, can anyone help?

mattonfire
By mattonfire
in Programming
 Share
Go to solution Solved by mattonfire,
40 minutes ago, homeap5 said:

Third error means that you're using new version if php without that function enabled.

Try to use mysqli_real_escape_string this instead

http://php.net/manual/en/mysqli.real-escape-string.php

Changed my usage to mysqli instead of mysqli seems to be working better.

34 minutes ago, vorticalbox said:

  


$username = $_POST['username'];$password = $_POST['password'];

 

 

You are looking for username and password in the POST but their name in the form is user and pass

  


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

 

This worked. Thank you.

  

<?php

$con = mysqli_connect("localhost","root","","");

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

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysqli_real_escape_string($con, $username);
$password = mysqli_real_escape_string($con, $password);

$result = mysqli_query($con,"SELECT * FROM users WHERE username = '$username' and password = '$password'")
			or die("Failed to Query Database. ".mysqli_error($con));
$row = mysqli_fetch_array($result);
if ($row['username'] == $username && $row['password'] == $password ){
	echo "Login Successful. Welcome, ".$row['username'];
} else {
	echo "Login Failed. Please try again.";
}

?>

*EDIT*

Found it. You both get rep.

image.png

Posted 
<html>
<head><title>Login page</title></head>
<link rel="stylesheet" type="text/css" href="style.css">
<body>
	<div id="frm">
		<form action="process.php" method="POST">
			<p>
				<label>Username:</label>
				<input type="text" id="user" name="user"  />
			</p>
			<p>
				<label>Password:</label>
				<input type="password" id="pass" name="pass"  />
			</p>
			<p>
				<input type="submit" id="btn" name="login"  />
			</p>
		</form>
	</div>
</body>
</html>
<?php
$username = $_POST['username'];
$password = $_POST['password'];

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);

mysql_connect("localhost", "root", "");
mysql_select_db("login");

$result = mysql_query("SELECT * FROM users WHERE username = '$username' and password = '$password'")
			or die("Failed to Query Database.".mysql_error());
$row = mysql_fetch_array($result);
if ($row['username'] == $username && $row['password'] == $password ){
	echo "Login Successful. Welcome, ".$row['username'];
} else {
	echo "Login Failed. Please try again.";
}
?>

image.png.0e886369100184d5e41157ab7fec187a.png

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

What are you using is that Notepad++

ASTRAY NOIR

CPU

Ryzen 5 3600

Cooler

Corsair H100I PRO

RAM

PNY XLR-8 EPIC-X DDR4-3200 CL16 32GB

GPU

ASUS ROG RTX 2070

STORAGE

SAMSUNG 850 EVO, 2xSEAGATE BARRACUDA PRO 1TB,A2000 1TB

Case

NZXT H500 

PSU

Enermax Revolution DF 650W

MONITOR

MSI MAG272,LG 22MP68BQ,DELL D2015H

PERIPHIRALS

Ducky One 2 (HORIZON) || Glorious Model O- Pink || G305 || Razer Nari Essential || HyperX Cloud II

 

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author
1 minute ago, truecrafting said:

What are you using is that Notepad++

Yeah, what would that change?

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
2 minutes ago, mattonfire said:

Yeah, what would that change?

Well sorry mate i can't help you i forgot how to code HTML it's almost been 2 years since i last touched notepad++ really sorry

ASTRAY NOIR

CPU

Ryzen 5 3600

Cooler

Corsair H100I PRO

RAM

PNY XLR-8 EPIC-X DDR4-3200 CL16 32GB

GPU

ASUS ROG RTX 2070

STORAGE

SAMSUNG 850 EVO, 2xSEAGATE BARRACUDA PRO 1TB,A2000 1TB

Case

NZXT H500 

PSU

Enermax Revolution DF 650W

MONITOR

MSI MAG272,LG 22MP68BQ,DELL D2015H

PERIPHIRALS

Ducky One 2 (HORIZON) || Glorious Model O- Pink || G305 || Razer Nari Essential || HyperX Cloud II

 

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
1 hour ago, mattonfire said:

image.png.0e886369100184d5e41157ab7fec187a.png

  

<html>
<head><title>Login page</title></head>
<link rel="stylesheet" type="text/css" href="style.css">
<body>
	<div id="frm">
		<form action="process.php" method="POST">
			<p>
				<label>Username:</label>
				<input type="text" id="user" name="user"  />
			</p>
			<p>
				<label>Password:</label>
				<input type="password" id="pass" name="pass"  />
			</p>
			<p>
				<input type="submit" id="btn" name="login"  />
			</p>
		</form>
	</div>
</body>
</html>

<?php
if (isset($_POST['username']) and isset($_POST['password'])){
$username = $_POST['username'];
$password = $_POST['password'];

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);

mysql_connect("localhost", "root", "");
mysql_select_db("login");

$result = mysql_query("SELECT * FROM users WHERE username = '$username' and password = '$password'")
			or die("Failed to Query Database.".mysql_error());
$row = mysql_fetch_array($result);
if ($row['username'] == $username && $row['password'] == $password ){
	echo "Login Successful. Welcome, ".$row['username'];
} else {
	echo "Login Failed. Please try again.";
}
}
?>

 

First two errors are easy. In fact you can prevent them by configuring php different. Or use one "if" like in code above (should help).

 

Third error means that you're using new version if php without that function enabled.

Try to use mysqli_real_escape_string this instead

http://php.net/manual/en/mysqli.real-escape-string.php

 

or use your own str_replace code.

Link to comment
Share on other sites
Link to post
Share on other sites
Posted 


$username = $_POST['username'];$password = $_POST['password'];

 

You are looking for username and password in the POST but their name in the form is user and pass 


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

                     ¸„»°'´¸„»°'´ Vorticalbox `'°«„¸`'°«„¸
`'°«„¸¸„»°'´¸„»°'´`'°«„¸Scientia Potentia est  ¸„»°'´`'°«„¸`'°«„¸¸„»°'´

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
20 minutes ago, vorticalbox said:

You are looking for username and password in the POST but their name in the form is user and pass.

Damn, you're right. :)

Anyway, these errors are not related to it, so both solutions should be used for this to work.

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author
  • Solution
40 minutes ago, homeap5 said:

Third error means that you're using new version if php without that function enabled.

Try to use mysqli_real_escape_string this instead

http://php.net/manual/en/mysqli.real-escape-string.php

Changed my usage to mysqli instead of mysqli seems to be working better.

34 minutes ago, vorticalbox said:

  


$username = $_POST['username'];$password = $_POST['password'];

 

 

You are looking for username and password in the POST but their name in the form is user and pass

  


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

 

This worked. Thank you.

  

<?php

$con = mysqli_connect("localhost","root","","");

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

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysqli_real_escape_string($con, $username);
$password = mysqli_real_escape_string($con, $password);

$result = mysqli_query($con,"SELECT * FROM users WHERE username = '$username' and password = '$password'")
			or die("Failed to Query Database. ".mysqli_error($con));
$row = mysqli_fetch_array($result);
if ($row['username'] == $username && $row['password'] == $password ){
	echo "Login Successful. Welcome, ".$row['username'];
} else {
	echo "Login Failed. Please try again.";
}

?>

*EDIT*

Found it. You both get rep.

image.png

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
On 9/19/2018 at 6:15 PM, mattonfire said:

Changed my usage to mysqli instead of mysqli seems to be working better.

This worked. Thank you.

  


<?php

$con = mysqli_connect("localhost","root","","");

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

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysqli_real_escape_string($con, $username);
$password = mysqli_real_escape_string($con, $password);

$result = mysqli_query($con,"SELECT * FROM users WHERE username = '$username' and password = '$password'")
			or die("Failed to Query Database. ".mysqli_error($con));
$row = mysqli_fetch_array($result);
if ($row['username'] == $username && $row['password'] == $password ){
	echo "Login Successful. Welcome, ".$row['username'];
} else {
	echo "Login Failed. Please try again.";
}

?>

*EDIT*

Found it. You both get rep.

image.png

I assume you use PHP 7 right? Mysql has been removed as of PHP 7.0, use mysqli or PDO ;) 

Business Management Student @ University St. Gallen (Switzerland)

HomeServer: i7 4930k - GTX 1070ti - ASUS Rampage IV Gene - 32Gb Ram

Laptop: MacBook Pro Retina 15" 2018

Operating Systems (Virtualised using VMware): Windows Pro 10, Cent OS 7

Occupation: Software Engineer

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

the mysql_  functions are deprecated , mysqli "module" is supposed to be used, as it has more functionality and it's safer.

 

You can use mysqli as functions like mysqli_real_escape_string  or mysqli::real_escape_string ( string $escapestr ) 

See  http://php.net/manual/ro/mysqli.real-escape-string.php

 

Note that you need to have a connection created to the mysql server before you escape the variables, as escaping the strings depends on connection parameters to the database.

 

Also you get those warnings about username and passwords ....  if user doesn't enter name or password then browser won't send those with the request (instead of an empty string) so php won't great the entry in the _POST array

You can avoid logging errors doing something like this

 

$username =  (isset($_POST['username'])==TRUE) ? $_POST['username'] :  '';

 

it's a simple if then else  .. if the post entry exists then username becomes that entry else the variable becomes '' instead of  being NULL or something undefined (because _post entry doesn't exist ... and the warning isn't triggered because isset doesn't activate the warning text.

 

Oh, and there's $_GET and $_POST arrays ,  $_GET array gets filled with stuff sent through the url  like index.php?page=1 will get you a $_GET['page'] = 1;  See http://php.net/manual/en/reserved.variables.get.php

 

But note that by default, php will be configured to put both the GET and POST parameters and the cookies  in a single array called _REQUEST, so it may be better to get used to using _REQUEST in some situations, see http://php.net/manual/en/reserved.variables.request.php 

Link to comment
Share on other sites
Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share
Go to topic listing Programming

×