PHP IF statement checks age

Question

SQLpower 0 Light Poster

12 Years Ago

Hello, folks.

I got another problem.

Basically, I wanna check if someone is old enough to watch a movie.

I got 2 sessions - one for the certificate( 12, 15, 18) and 1 for the year of birth.

My question is can i use somtehing similiar to this or? Cause now is not working...

<?php
session_start();


if  (!(isset($_SESSION["gatekeeper"]) && $_SESSION['gatekeeper'] !=''))
{
	header("Location:mainpage.php");
   echo "Not an user, leave";
}
else
		
{
		echo "<center><h3>You are logged in as " . $_SESSION['gatekeeper'] . "</h3></center>";


// Read the quantity from the query string on the searchresult.php
$old = 2012 - $_SESSION['agez'];

if ($old < $_SESSION['certificate'])
{

echo "<script type ='text/javascript'> window.alert('Sorry, you are not old enough')</script>";

exit;
}
else
{
// Echo the form, with hidden fields containing the quanti

echo "<form method='post' action='buyfilm2.php' >";
echo "The quantity of tickets you want to buy?: <input name='maxtickets' value='1' />";
echo "<input type='hidden' name='id' value='$_GET[filmID]' /> ";
echo "<input type='submit' value='Go!' />";
}
}

?>

php

Edited 12 Years Ago by SQLpower because: n/a

6 Contributors
17 Replies
190 Views
3 Days Discussion Span
Latest Post 12 Years Ago Latest Post by nidayaqoob1230

broj1 356 Humble servant

12 Years Ago

Line 8 will never get executed, but that is not the problem anyway. The code seems OK. What actually is not working? Please describe (error messages, what is supposed to happen but it does not...).

diafol

12 Years Ago

I'd use a DOB not an age, as you can be different ages in asingle year, depending when you have your birthday.

$MyDob = '10/10/1968'; ..this comes from a form etc (many formats allowed e.g. US / European)

$myDobF = strftime('%Y-%m-%d', strtotime($MyDob));
$dob = new DateTime($myDobF);
$current = new DateTime(date('Y-m-d'));
$interval = $dob->diff($current);
$age = $interval->format('%y');

Then check - but actually needs to exist

if($age < $_SESSION['certificate']){
  header(...);
  exit;
}else{
  ...
}

diafol

12 Years Ago

$myDobF = strftime('%y-%m-%d', strtotime("$dayofbirth/$monthofbirth/$yearofbirth"));
//and
$age = $interval->format('%y');

you can store what you like in a session. Just remember to place session_start(); as the very first thing at the top of EVERY page. Then:

$_SESSION['age'] = $age;

GliderPilot 33 Posting Whiz

12 Years Ago

Usually that happens when MySQL returns an empty record set and you try to minipulate it. Place your mysql_num_rows() function into the if statement before it or add an or die to your mysql_query

$result = mysql_query ($sql);

if ($result)
{
  if(mysql_num_rows($result)==0)
  {
    echo "ERROR - No mactching rows from the database!<br/>";
  }else{
    $_SESSION['gatekeeper'] = $_POST["username"];
    $_SESSION['agez']       = $_POST["yearofbirth"];
    $_SESSION['admin'] 	 = $_POST["isadmin"];
    $_SESSION['age'] 	= $age;

    // Redirect to the main menu
    header ("Location: mainpage.php");
    $errorMessage = "logged on";
  }

}
else
{
  $errorMessage = "Error logging on";
}

Biiim 182 Junior Poster

12 Years Ago

that var isn't done correctly

string(16) "row[certificate]"

it sounds like you've put in

echo "<form action='formprocess.php?certificate=row[certificate]>";

instead of

echo "<form action='formprocess.php?certificate=$row[certificate]>";

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SQLpower 0 Light Poster · Answer 1 · 2012-03-07T04:14:03+00:00

I am aware of the 8th line, however the script works but it doesn't do what it has to do. For instance, it always load the page even if the user is not old enough.

SQLpower 0 Light Poster · Answer 2 · 2012-03-07T05:57:15+00:00

Still doesn't work, I can still see the page even if I am underaged. :(

SQLpower 0 Light Poster · Answer 3 · 2012-03-07T07:04:33+00:00

Can I actually get from a form $dayofbirth, $yearofbirth, $monthofbirth from my database and put it into

$myDobF = strftime('%yearofbirth-%monthofbirth-%dayofbirth', strtotime($MyDob));
    $dob = new DateTime($myDobF);
    $current = new DateTime(date('Y-m-d'));
    $interval = $dob->diff($current);
    $age = $interval->format('%yearofbirth');

?
And can I also store $age in to a $_SESSION variable after that?

SQLpower 0 Light Poster · Answer 4 · 2012-03-07T18:32:37+00:00

I have no idea how to fix this, the code seems alright, I have no idea where the problem is. It is always displaying the page even if the user is 1 years old and there is a minimum age of 18....I will make it properly after I fix that bloody thing... :S

<?php
if(!isset($_SESSION))
{
session_start();
}  



	if  (!(isset($_SESSION["gatekeeper"]) && $_SESSION['gatekeeper'] !=''))
{
	echo "Not an user, leave";
	header("Location:mainpage.php");
	exit;
}
	else
{

	echo "It's displaying the age" . $_SESSION['old'] . "WTF";

	if ($_SESSION['certificate'] > $_SESSION['old'])
		{
	header("Location:mainpage.php");
	exit;
		}
	else
{

echo "<center><h3>You are logged in as " . $_SESSION['gatekeeper'] . "</h3></center>";


// Read the quantity from the query string on the searchresult.php



// Echo the form, with hidden fields containing the quanti

echo "<form method='post' action='buyfilm2.php' >";
echo "The quantity of tickets you want to buy?: <input name='maxtickets' value='1' />";
echo "<input type='hidden' name='id' value='$_GET[filmID]' /> ";
echo "<input type='submit' value='Go!' />";

}
}

?>

SQLpower 0 Light Poster · Answer 5 · 2012-03-07T18:52:45+00:00

Ah need help here, got stuck. Anyway

Thisi s my form:

<!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">
<?php


include('functions.php');
write_head("Solent Login Page", "newCSS");

?>
<body>

<div id    ="body"    >
<div id    ='sidebar' > <form method='post' action='login.php'>
<label for = 'username'> Username:</label>
<input name='username' id='username'/><br/>  
<label for = 'password'>Password:</label>
<input name='password' id='password'/><br/>
<label for ='yearofbirth'> Year of Birth:</label>
<input name='yearofbirth' id='yearofbirth' /><br/>
<label for ='monthofbirth'> Month of Birth:</label>
<input name='monthofbirth' id='monthofbirth' /><br/>
<label for ='dayofbirth'> Day of Birth:</label>
<input name='dayofbirth' id='dayofbirth' /><br/>
<td>Admin?</td> 
<td>
<input type = "checkbox" name = "isadmin" value = "1" /> Yes
<input type = "checkbox" name = "isadmin" value = "0" /> No
</td>
</br>
<input type='submit'  value='Log in' /></form>
<!-- / end of login page -->
</div>

</body>
</html>

This is my PHP script

<?php
//prevent sql injesctions
session_start();
error_reporting (E_ALL ^ E_NOTICE);
// for error handling
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);

$uname 	= $_POST["username"];
$pword 	= $_POST["password"];
$ybirth = $_POST["yearofbirth"];
$ybirth = $_POST["monthofbirth"];
$ybirth = $_POST["dayofbirth"];
$admin  = $_POST["isadmin"];
$num_rows = 0;
$errorMessage = "";

$uname  = htmlspecialchars($uname);
$pword  = htmlspecialchars($pword);
$ybirth = htmlspecialchars($ybirth);
$mbirth = htmlspecialchars($mbirth);
$dbirth = htmlspecialchars($dbirth);
$admin  = htmlspecialchars($admin);


$myDobF = strftime('%y-%m-%d', strtotime("$dbirth/$mbirth/$ybirth"));
$dob = new DateTime($myDobF);
$current = new DateTime(date('Y-m-d'));
$interval = $dob->diff($current);
$age = $interval->format('%y');



if(!ctype_alnum($uname))
{
    echo "ERROR: Input contains characters other than letters and numbers.";
}



$conn = mysql_connect("localhost", "yvrachev", "Eer3num6") or die("cannot connect"); 
mysql_select_db("yvrachev") or die (mysql_error());




$sql = ("SELECT * FROM sc_users WHERE username = '$uname' and password = '$pword' and yearofbirth = '$ybirth' and monthofbirth='$mbirth' and dayofbirth='$dbirth' isadmin='$admin'");


$result = mysql_query ($sql);


 if ($result) 
{
}
else 
{
$errorMessage = "Error logging on";
}


 
 if(mysql_num_rows($result)==0)
{
    echo "ERROR - No mactching rows from the database!<br/>";
}
else
{
	$_SESSION['gatekeeper'] = $_POST["username"];
	$_SESSION['agez']       = $_POST["yearofbirth"];
	$_SESSION['admin'] 		= $_POST["isadmin"];
	$_SESSION['age'] 		= $age;




   
    // Redirect to the main menu
    header ("Location: mainpage.php");
	$errorMessage = "logged on";

}

?>

I am getting Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /home/yvrachev/public_html/project/login.php on line 63 ERROR - No mactching rows from the database! and it is not working :( :s

Biiim 182 Junior Poster · Answer 6 · 2012-03-08T00:08:11+00:00

put this in, my guess is $_SESSION isn't what you expect or even just a string

var_dump($old);
var_dump($_SESSION['certificate']);
if ($old < $_SESSION['certificate'])
{

SQLpower 0 Light Poster · Answer 7 · 2012-03-08T02:03:09+00:00

int(7) string(16) "row[certificate]" This is what i get....

<?php
if(!isset($_SESSION))
{
session_start();
}  


$certificate = $_GET["certificate"];
$old = 2012 - $_SESSION['agez'];

var_dump($old);
 
var_dump($certificate);
   


   

	if  (!(isset($_SESSION["gatekeeper"]) && $_SESSION['gatekeeper'] !=''))
{
	echo "Not an user, leave";
	header("Location:mainpage.php");
	exit; 
}
	else
{

	

if ($old < $certificate)
{
	header("Location:mainpage.php");
	exit;
}
	else
{

echo "<center><h3>You are logged in as " . $_SESSION['gatekeeper'] . "</h3></center>";


// Read the quantity from the query string on the searchresult.php



// Echo the form, with hidden fields containing the quanti

echo "<form method='post' action='buyfilm2.php' >";
echo "The quantity of tickets you want to buy?: <input name='maxtickets' value='1' />";
echo "<input type='hidden' name='id' value='$_GET[filmID]' /> ";
echo "<input type='submit' value='Go!' />";

}
}

?>

Biiim 182 Junior Poster · Answer 8 · 2012-03-08T16:13:42+00:00

oh and also when values are passed through post and get vars they are returned as strings

Such as $_GET might equal string(2) "18".

php will usually evaluate it fine but remember for comparisons it isn't an integer and will fail is_int($_GET).

you can convert by using intval()

$certificate = inval($_GET['certificate']);

It's always best whatever function you have to make sure the right type of values are being passed to it and to error/correct if they arn't, can save many headaches trying to trace it back later on.

if(isset($_GET['certificate']) && (ctype_digit($_GET['certificate']) || is_int($_GET['certificate']))){
	$certificate = intval($_GET['certificate']);
}else{
	$certificate = 21;//Don't know, worst case?
}

Now certificate is always an integer, if its set wrong it guesses worst case 21

SQLpower 0 Light Poster · Answer 9 · 2012-03-08T17:33:12+00:00

if(!isset($_SESSION))
{
session_start();
}  

#var_dump($old);
#var_dump($certificate);
   
  
$certificate = $_GET["certificate"];
$old = date("Y") - $_SESSION['agez'];




   

	if  (!(isset($_SESSION["gatekeeper"]) && $_SESSION['gatekeeper'] !=''))
{
	echo "Not an user, leave";
	header("Location:mainpage.php");
	exit; 
}
	else
{

	

if ($old < $certificate)
{
	header("Location:mainpage.php");
	exit;
}
	else
{
}

Works now, I changed the form as well. Thanks. I do have another question tho. So when I enter my Login and Password, how can I create a session from the SQL query about my Date of birth, month of birth, year of birth and then use it to check if the user is old enough to see a movie
Login.php

<?php
//prevent sql injesctions
session_start();
error_reporting (E_ALL ^ E_NOTICE);
// for error handling
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);

$uname 	= $_POST["username"];
$pword 	= $_POST["password"];
$num_rows = 0;
$errorMessage = "";
$uname  = htmlspecialchars($uname);
$pword  = htmlspecialchars($pword);




if(!ctype_alnum($uname))
{
    echo "ERROR: Input contains characters other than letters and numbers.";
}



$conn = mysql_connect("localhost", "yvrachev", "Eer3num6") or die("cannot connect"); 
mysql_select_db("yvrachev") or die (mysql_error());





$sql = ("SELECT * FROM sc_users WHERE username = '$uname' and password = '$pword'");
$result = mysql_query ($sql);



if($result) 
{
}
else 
{
$errorMessage = "Error logging on";
}



 
 if(mysql_num_rows($result)==0)
{
    echo "ERROR - No mactching rows from the database!<br/>";
}
else
{
	$sql_check = "SELECT * FROM sc_users WHERE (username = '$uname' and password='$pword' AND isadmin= 1)";
	$sql_result = mysql_query($sql_check) or die(mysql_error());
	
	if(mysql_num_rows($sql_result)==0)
	{	
		$_SESSION['gatekeeper'] = $_POST["username"];
		$_SESSION['agez'] = $_GET["yearofbirth"];
		header("Location: mainpage.php");
	}
	else
	{
	$_SESSION['gatekeeper'] = $_POST["username"];
	$_SESSION['admin']      = $_POST["username"];
	$_SESSION['agez'] = $_GET["yearofbirth"];
		header("Location: newfilm.php");
	}
	

   
    // Redirect to the main menu
    
	$errorMessage = "logged on";
	
}


?>

Biiim 182 Junior Poster · Answer 10 · 2012-03-08T18:19:05+00:00

how i do logins is i keep it all in a database eg,

userid,username,hash,salt,token,access_level,active,dob
1,bim,3Fsd4sdf32dsfsd54r,4sdf34,sdfrgsgfgfsgergsdfgrtgdfgfg,10,1,1998-03-06

then on every page of the project i include a function that checks the session token and pull the users logged in data

app.php

<?
//connect db
$Q = "select `username`,`access_level`,`active`,`dob` from `users` where `token` = '{$_SESSION['token']} LIMIT 1";
$R = mysqli_query($con,$Q);
if($R !== false){
$A = mysql_fetch_assoc($R);
$A['loggedin'] = true;
}else{
//header('Location: login.php');
}

<?
require('app.php');
//$A['dob'] == users dob in mysql datetime format
//user is 14 years old, cant view anything cert 15 or above
//rest of somepage.php...
?>

broj1 356 Humble servant Featured Poster · Answer 11 · 2012-03-09T00:48:33+00:00

The problem is probably in line 50

$result = mysql_query($sql);

. Function mysql_query returns either a resource if everything is OK or FALSE if there is an error. In your case it returned FALSE as the error message says

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given...

Probably it could not read data from the database. Check whether your query is OK. Put this on line 48 so you get the query displayed and check it in phpMyAdmin (or mysql client).

die($sql);

nidayaqoob1230 0 Newbie Poster · Answer 12 · 2012-03-10T13:44:18+00:00

thank you ardav its really great and use full reply, hope that it will also solve my problem, keep it up