PHP Login Form Problem

CripDawg · February 27, 2015

does anyone know what is wrong with my code i think it may have something to do with my brackets my login screen is no longer appearing i just get a blank page thanks in advance

<?phperror_reporting (E_ALL ^ E_NOTICE);session_start ();?> <html xmlns="http://www.w3.org/1999/xhtml"><head>     <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />     <title>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"); $password = md5(md5("yjtfjtfjj".$password."xjtfjtfjj")); echo "$password";  $query = mysql_query("SELECT * FROM users WHERE username='$user'");$numrows = msql_num_rows($query);if ($numrows == 1){$row = msql_fetch_assoc($query); $dbid = $row['id'];$dbuser = $row['username'];$dbpass = $row['password'];$dbactive = $row['active']; if ($password == $dbpass){if dbactive == 1) {$_SESSION['userid'] = $dbid;$_SESSION['username'] = $dbuser; echo "you have been logged in as <b>$dbuser</b> <a href 'members.php'> Click here</a> to go to the Members page";}else echo "you must activate your account to login $form";}else echo "incorrect password $form";  }else echo "Incorrect username. $form";      mysql_close();     }     else      echo "You must enter your password. $form";               }           else                echo "You must enter your username. $form";       }    else       echo $form;    ?></html>

EvilWiffles · February 27, 2015

does anyone know what is wrong with my code i think it may have something to do with my brackets my login screen is no longer appearing i just get a blank page thanks in advance

<?phperror_reporting (E_ALL ^ E_NOTICE);session_start ();?> <html xmlns="http://www.w3.org/1999/xhtml"><head>     <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />     <title>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"); $password = md5(md5("yjtfjtfjj".$password."xjtfjtfjj")); echo "$password";  $query = mysql_query("SELECT * FROM users WHERE username='$user'");$numrows = msql_num_rows($query);if ($numrows == 1){$row = msql_fetch_assoc($query); $dbid = $row['id'];$dbuser = $row['username'];$dbpass = $row['password'];$dbactive = $row['active']; if ($password == $dbpass){if dbactive == 1) {$_SESSION['userid'] = $dbid;$_SESSION['username'] = $dbuser; echo "you have been logged in as <b>$dbuser</b> <a href 'members.php'> Click here</a> to go to the Members page";}else echo "you must activate your account to login $form";}else echo "incorrect password $form";  }else echo "Incorrect username. $form";      mysql_close();     }     else      echo "You must enter your password. $form";               }           else                echo "You must enter your username. $form";       }    else       echo $form;    ?></html>

Try this. I edited and fixed a mistake you made with the form and not properly(?) ending the form tag.

SSL · February 27, 2015

gg my nested if statement experience.

This is really hard to read. I would try to find a way to combine some of those conditions, and try to indent things consistently. It will make it easier to debug your own code.

if (!empty($user) && !empty($password) {  // bejf}

Lots of security issues here as well; you should sanitize all database input and limit the number of results from the user table query to 1, to avoid an attacker being able to get the entire contents of that table.

CripDawg · February 27, 2015

Try this. I edited and fixed a mistake you made with the form and not properly(?) ending the form tag.

i tried using this but it didnt work if it helps im using this youtube tutorial

Prashank · February 28, 2015

i tried using this but it didnt work if it helps im using this youtube tutorial

There are bazallion issues, including syntax. You need to post more info other than "its not working", post the exact error you get.

P.S: My first post here

Mojo-Jojo · February 28, 2015

$user = $_POST['user'];/* etc */ $query = mysql_query("SELECT * FROM users WHERE username='$user'");

Dat gaping SQL injection hole

Prashank · February 28, 2015

$user = $_POST['user'];/* etc */ $query = mysql_query("SELECT * FROM users WHERE username='$user'");

Dat gaping SQL injection hole

Haha yea, working on fixing/refactoring up his code a little.

Prashank · February 28, 2015

<?php // login.phperror_reporting (E_ALL ^ E_NOTICE);session_start();if(isset($_POST['loginbtn'])) {	$message  = $success = '';	$user     = mysqli_real_escape_string(trim($_POST['user']));	$password = mysqli_real_escape_string($_POST['password']);	if(!$user)	{		$message  .= "You must enter your username.<br />";	}	if(!$password)	{		$message  .= "You must enter your password.<br />";	}	require("connect.php"); // use http://php.net/manual/en/book.mysqli.php instead of mysql	$password = md5(md5("yjtfjtfjj".$password."xjtfjtfjj"));	//echo "$password"; // debugging like a champ	$query   = mysqli_query("SELECT * FROM users WHERE username='$user'");	$numrows = mysqli_num_rows($query);	if($numrows != 1)	{		$message .= "Incorrect username."; // Its better not to give hint about what was wrong, just show them generic message that the login is invalid	}	else	{		$row      = mysqli_fetch_assoc($query); 		$dbid     = $row['id'];		$dbuser   = $row['username'];		$dbpass   = $row['password'];		$dbactive = (int)$row['active'];		if($password === $dbpass)		{			if($dbactive === 1)			{				$_SESSION['userid']   = $dbid;				$_SESSION['username'] = $dbuser;				$success = "you have been logged in as <b>$dbuser</b> <a href 'members.php'> Click here</a> to go to the Members page";			}			else			{				$message .= "you must activate your account to login";			}		}		else 		{			$message .= "incorrect password";		}	}	mysqli_close();}?><html xmlns="http://www.w3.org/1999/xhtml"><head>	<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />	<title>Login</title></head><body>	<?php	if(!empty($success))	{		echo $success;	}	else	{		echo $message	?>	<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>	<?php	}	?></body></html>

I have refactored this code and added some security tho I have not tested this, look through it and google anything you don't know about because you should.

Happy Learning

SSL · March 1, 2015

I'd just find a decent tutorial and start over from scratch.

CripDawg · March 5, 2015

<?php // login.phperror_reporting (E_ALL ^ E_NOTICE);session_start();if(isset($_POST['loginbtn'])) {	$message  = $success = '';	$user     = mysqli_real_escape_string(trim($_POST['user']));	$password = mysqli_real_escape_string($_POST['password']);	if(!$user)	{		$message  .= "You must enter your username.<br />";	}	if(!$password)	{		$message  .= "You must enter your password.<br />";	}	require("connect.php"); // use http://php.net/manual/en/book.mysqli.php instead of mysql	$password = md5(md5("yjtfjtfjj".$password."xjtfjtfjj"));	//echo "$password"; // debugging like a champ	$query   = mysqli_query("SELECT * FROM users WHERE username='$user'");	$numrows = mysqli_num_rows($query);	if($numrows != 1)	{		$message .= "Incorrect username."; // Its better not to give hint about what was wrong, just show them generic message that the login is invalid	}	else	{		$row      = mysqli_fetch_assoc($query); 		$dbid     = $row['id'];		$dbuser   = $row['username'];		$dbpass   = $row['password'];		$dbactive = (int)$row['active'];		if($password === $dbpass)		{			if($dbactive === 1)			{				$_SESSION['userid']   = $dbid;				$_SESSION['username'] = $dbuser;				$success = "you have been logged in as <b>$dbuser</b> <a href 'members.php'> Click here</a> to go to the Members page";			}			else			{				$message .= "you must activate your account to login";			}		}		else 		{			$message .= "incorrect password";		}	}	mysqli_close();}?><html xmlns="http://www.w3.org/1999/xhtml"><head>	<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />	<title>Login</title></head><body>	<?php	if(!empty($success))	{		echo $success;	}	else	{		echo $message	?>	<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>	<?php	}	?></body></html>

I have refactored this code and added some security tho I have not tested this, look through it and google anything you don't know about because you should.

Happy Learning

thanks dude that was heaps of help but its sill not showing anything when i press the login button could it be that there is something wrong with my connect.php file hers whats currently in it as it is what the tutorial specifies or should i change it to some sqli comand now that youve changed it. thaks for your help agin

mysql_connect("localhost", "root", "toblerone")mysql_select_db("users");

Prashank · March 5, 2015

thanks dude that was heaps of help but its sill not showing anything when i press the login button could it be that there is something wrong with my connect.php file hers whats currently in it as it is what the tutorial specifies or should i change it to some sqli comand now that youve changed it. thaks for your help agin
mysql_connect("localhost", "root", "toblerone")mysql_select_db("users");

yea it will be replaced with mysqli, don't use mysql its deprecated and will be removed.

http://php.net/manual/en/mysqli.construct.php

mysqli_connect("localhost", "root", "toblerone", "users");

CripDawg · March 6, 2015

yea it will be replaced with mysqli, don't use mysql its deprecated and will be removed.

http://php.net/manual/en/mysqli.construct.php
mysqli_connect("localhost", "root", "toblerone", "users");

I'm getting these errors PS i can see them now i have Xampp instead of using mowes portable

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 27

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 64

Prashank · March 6, 2015

I'm getting these errors PS i can see them now i have Xampp instead of using mowes portable

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 27

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 64

Read the documentation: http://php.net/manual/en/mysqli.query.php

You're using the procedural method so you will have to pass the link identifier as first parameter.

assign the mysqli_connect to a variable and use that as first parameter.

Lumi · March 6, 2015

Here's one i made quite a while ago. This is procedural code, you should be able to get a good understanding of it.

<?phpinclude 'inc/config.inc.php';echo $includes;session_start("dfhbuijdfhiufdshikfdshjku");$email = (isset($_POST['email']) ? htmlspecialchars($_POST['email']) : "");$password = (isset($_POST['password']) ? htmlspecialchars($_POST['password']) : "");function handleLogin($email, $password, $db){	global $return;	global $handle;	$email = strtolower(mysqli_real_escape_string($db, $email));	$password = hash('sha512', mysqli_real_escape_string($db, $password));	$query = mysqli_query($db, "SELECT * FROM `users` WHERE email='$email'");	$row = mysqli_fetch_assoc($query);	if($password == $row['password']):		$ip = @$_SERVER['HTTP_X_FORWARDED_FOR'];		$updateip = mysqli_query($db, "UPDATE `users` SET ip='$ip' WHERE email='$email'");		$_SESSION['email'] = $email;		$_SESSION['name'] = $row['name'];		$handle = '<div class="alert alert-dismissible alert-success"><b>Alright!</b><p>You have successfully logged in! Redirecting in 3 seconds.</p></div>		<meta http-equiv="refresh" content="3; url=index.php" />';	else:		$return = '<div class="alert alert-dismissible alert-warning"><b>Uh oh!</b><p>Invalid email/password combination.</p></div>';	endif;}$handle = '		<div class="alert alert-dismissible alert-warning"><b>Can\'t login?</b><p>We have upgraded encryption, everyone must do a password reset. Click <a href="reset.php">here</a> to do so.</p></div>			<form method="POST" action="login.php">				<input class="form-control" type="email" placeholder="example@[member='Example'].com" name="email"/>				<br/>				<input class="form-control" type="password" placeholder="password" name="password"/>				<br/>				<input style="width: 100%;" class="btn btn-success" type="submit" value="Login"/>			</form>';if(@isset($_SESSION['email'])){	$handle = '<div class="alert alert-dismissible alert-danger"><b>Uh oh!</b><p>You are already logged in and may not use this page! Redirecting.</p></div><meta http-equiv="refresh" content="3; url=index.php" />';}elseif($email && $password){	if($email !== "" && $password !== ""):		handleLogin($email, $password, $db);	else:		//impossible error but yolo		$return = '<div class="alert alert-dismissible alert-warning"><b>What the fucking shit!</b><p>An error fucking occured somehow!</p></div>';	endif;}?><html><head><title>GouldingKV's - Login</title></head><body>	<div class="well" style="max-width: 35%; margin: auto auto; margin-top: 10%;">		<center><a href="index.php"><img src="img/logo4.png"/></a></center>		<br/>			<?php				echo $handle;				if(@$return):					echo '<br/>' . $return;				endif;			?>			<p style="float: left;">Don't have an account? <a href="register.php">Register here.</a></p> <p style="float: right;">Forgot your password? <a href="reset.php">Reset it here.</a></p>			<br/>			<br>			<a href="https://fakewebsite.io/"><div class="alert alert-dismissible alert-info"><p>Need a VPS? Get one from fakewebsite.io! (Click this ad anywhere)</p></div></a>	</div></body></html>

CripDawg · March 6, 2015

Read the documentation: http://php.net/manual/en/mysqli.query.php

You're using the procedural method so you will have to pass the link identifier as first parameter.

assign the mysqli_connect to a variable and use that as first parameter.

so at the line "require (connect.php)" i replace it with $link = mysqli_connect("localhost", "joseph.foley", "toblerone", "users"); is that correct and also dose this remove the need for connect.php altogether?

sorry I'm really amateur at php i not long switched from visual basic

Lumi · March 7, 2015

so at the line "require (connect.php)" i replace it with $link = mysqli_connect("localhost", "joseph.foley", "toblerone", "users"); is that correct and also dose this remove the need for connect.php altogether?

sorry I'm really amateur at php i not long switched from visual basic

the point of "connect.php" is basically the same point as "config.php".

You set it in that file so you can include it in multiple files later without having to retype >mysqli_connect("localhost" etc) in every file.

Any var you set in a file that you're including/requiring you can use in your script later.

example:

in connect.php we have

$link = mysqli_connect("localhost", "joseph.foley", "toblerone", "users");

if we open up login.php and put

include 'connect.php';

later down the line we can use $link

so if you wanted to escape strings for mysql you'd do

mysqli_real_escape_string($link, $var)

CripDawg · March 7, 2015

the point of "connect.php" is basically the same point as "config.php".

You set it in that file so you can include it in multiple files later without having to retype >mysqli_connect("localhost" etc) in every file.

Any var you set in a file that you're including/requiring you can use in your script later.

example:

in connect.php we have

$link = mysqli_connect("localhost", "joseph.foley", "toblerone", "users");

if we open up login.php and put

include 'connect.php';

later down the line we can use $link

so if you wanted to escape strings for mysql you'd do

mysqli_real_escape_string($link, $var)

so how would i fix these two lines?

$user     = mysqli_real_escape_string(trim($_POST['user']));$password = mysqli_real_escape_string($_POST['password']);

Lumi · March 7, 2015

so how would i fix these two lines?

$user     = mysqli_real_escape_string(trim($_POST['user']));$password = mysqli_real_escape_string($_POST['password']);

$user = mysqli_real_escape_string($link, $_POST['user']);

$password = mysqli_real_escape_string($link, $_POST['password']);

CripDawg · March 7, 2015

$user = mysqli_real_escape_string($link, $_POST['user']);

$password = mysqli_real_escape_string($link, $_POST['password']);

if i change those lines to what you said the errors say it received no parameters instead of 1/2? I'm very confused sorry

Lumi · March 7, 2015

if i change those lines to what you said the errors say it received no parameters instead of 1/2? I'm very confused sorry

paste both errors

CripDawg · March 7, 2015

paste both errors

$user = mysqli_real_escape_string($link, $_POST['user']);$password = mysqli_real_escape_string($link, $_POST['password']);

Errors

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 29

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 65

$user     = mysqli_real_escape_string(trim($_POST['user']));$password = mysqli_real_escape_string($_POST['password']);

Errors

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 29

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 65

Lumi · March 7, 2015

Errors

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 29

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 65

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

You will only get this error when the post data isn't set. Don't worry about it.

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

as goes, $query = mysqli_query($link, "YOUR QUERY STRING HERE");

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 29

This will error because mysqli_query failed.

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 65

mysqli_close($link)

Hope this helps.

CripDawg · March 7, 2015

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9
Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

You will only get this error when the post data isn't set. Don't worry about it.

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 28

as goes, $query = mysqli_query($link, "YOUR QUERY STRING HERE");

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 29

This will error because mysqli_query failed.

Warning: mysqli_close() expects exactly 1 parameter, 0 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 65

mysqli_close($link)

Hope this helps.

that fixed most of it but it sill says the top two and it still won't log me in with the details i entered in the database

outputs:

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 9

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/xampp/login.php on line 10

You must enter your username.

You must enter your password.

Incorrect username.

  <?php // login.phperror_reporting (E_ALL ^ E_NOTICE);session_start(); if(isset($_POST['loginbtn'])) {$message  = $success = ''; $user     = mysqli_real_escape_string(trim($_POST['user']));$password = mysqli_real_escape_string($_POST['password']); if(!$user){$message  .= "You must enter your username.<br />";}if(!$password){$message  .= "You must enter your password.<br />";} include("connect.php");  $password = md5(md5("yjtfjtfjj".$password."xjtfjtfjj")); //echo "$password"; // debugging like a champ $query   = mysqli_query($link, "SELECT * FROM users WHERE username='$user'");$numrows = mysqli_num_rows($query); if($numrows != 1){$message .= "Incorrect username."; // Its better not to give hint about what was wrong, just show them generic message that the login is invalid}else{$row      = mysqli_fetch_assoc($query); $dbid     = $row['id'];$dbuser   = $row['username'];$dbpass   = $row['password'];$dbactive = (int)$row['active']; if($password === $dbpass){if($dbactive === 1){$_SESSION['userid']   = $dbid;$_SESSION['username'] = $dbuser; $success = "you have been logged in as <b>$dbuser</b> <a href 'members.php'> Click here</a> to go to the Members page";}else{$message .= "you must activate your account to login";}}else {$message .= "incorrect password";}  } mysqli_close($link);}?> <html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>Login</title></head><body><?phpif(!empty($success)){echo $success;}else{echo $message?><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><?php}?></body></html>

Prashank · March 8, 2015

move

include("connect.php");

to the very top of the file, just after <?php line

and pass $link to all mysqli functions first parameter.

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PHP Login Form Problem

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