[SOLVED] Functions and Include

Akenatehm · February 28, 2009

Hey Guys,

I thought I'd give creating my own functions a go. I thought I'd try and clean up my code so it doesn't look so messy so I put most of my common functions in a file called functions.php and using

include 'functions.php';

I would call on them as needed. But there seems to be a problem with MySQL resources and interchanging them between different functions.

I get the following error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /Library/WebServer/Documents/game/functions.php on line 13

On a registration page.

Here's functions.php:

<?php

function mysqlcheck($query)
{
if (!$query)
{
	die('MySQL Check Error with ' . mysql_error());
}	
}

function userexists()
{
if(mysql_num_rows($checkuser) > 0)
{
$error = '<font color="red"> User Already Exists </font>';
}
}

function activationkey()
{
$activationKey =  mt_rand() . mt_rand() . mt_rand() . mt_rand() . mt_rand();	
}

function getplayerid($username)
{
$findplayerid = mysql_query("SELECT playerid FROM accounts WHERE username = '$username'");	
}

function createresourceuser($playerid)
{
$createresourcesuser = mysql_query("INSERT INTO resources (playerid,gold,wood,stone) VALUES('$playerid','200','200','200')");	
}
?>

and register.php:

<?php
session_start();

if (isset($_POST['submit']))
{
if($_POST['password'] == $_POST['confirmpassword'])
{
		include 'connect.php';
		include 'functions.php';
		connect();
		$username = $_POST['username'];

		$checkuser = mysql_query("SELECT * FROM accounts WHERE username = '$username'");

		mysqlcheck("$checkuser");
		userexists();
		$activationkey = activationkey();	

		$username = $_POST['username'];
		$password = $_POST['password'];
		$email = $_POST['email'];
		$password = md5($password);

		$adduser = mysql_query("INSERT INTO accounts (username, password, email, activationkey, last_online) VALUES ('$username' , '$password' , '$email', '$activationkey', '0')");

		mysqlcheck($adduser);
		getplayerid($username);					
		mysqlcheck($findplayerid);									

		$transferplayerid = mysql_fetch_assoc($findplayerid);
		$playerid = $transferplayerid['playerid'];	
		createresourceuser($playerid);
		mysqlcheck($createresourceuser);

		$result = '<font color="green">User Added Succesfully </font>';
}

else
{				
$result = '<font color="red"> Passwords Do Not Match </font>';		
}
}
?>
<html>
<head>
<title>Registration - Fall of The Empire</title>
<link rel="stylesheet" href="styles.css" type="text/css"/>
</head>
<body bgcolor="272d31">
	<div align="center" class="header">
		<img src="images/logo.png" width="473" height="227" alt="Logo">
</div>
<br>
<center>
<div class="main">
<div class="navbar">
	<img src="images/nav_top.png" width="232" height="46" alt="Nav Top"><br>
	<a href="index.php"><img src="images/nav_home.png" width="232" height="33" alt="Nav Home" border="0"></a><br>
	<a href="news.php"><img src="images/nav_news.png" width="232" height="32" alt="Nav News" border="0"></a><br>
	<a href="about.php"><img src="images/nav_about.png" width="232" height="23" alt="Nav About" border="0"></a><br>
	<a href="login.php"><img src="images/nav_login.png" width="232" height="32" alt="Nav Login" border="0"></a><br>
	<a href="register.php"><img src="images/nav_register.png" width="232" height="25" alt="Nav Register" border="0"></a><br>
	<a href="contactus.php"><img src="images/nav_bottom.png" width="232" height="179" alt="Nav Bottom" border="0"><br></a>
	</div>	

	<div class="cbtop" align="center">
	<img src="images/register.png" width="171" height="25" alt="Register" class="welcome" align="right">	
	</div>
	<br><br>
	<div class="middle"><hr><br>
	<br>
	<table width="300px" align="right">
		<form action="<?php echo $PHP_SELF; ?>" method="POST">
			<tr>
				<td>
					<font color="white">
						Username:
					</font>
				</td>

				<td>
					<input type="text" name="username">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Email
					</font>			
				</td>

				<td>
					<input type="text" name="email">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Password:
					</font>
				</td>

				<td>
					<input type="password" name="password">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Confirm Password:
					</font>
				</td>

				<td>
					<input type="password" name="confirmpassword">
				</td>
			</tr>

			<tr>
				<td>
					<br>
				</td>
				<td>
				</td>
			<tr>
				<td>
				</td>
		<td align="right" style="padding-right:8px">
		<input type="submit" name="submit" value="Register"align="right">
		</td>
		</tr>
		</form>
	</table>
	<p><?php echo $result; ?></p>
	<p class="heading" align="right"><br><br><br><br><br><br><br><br><br><br></p>	
	<div class="cbbottom" align="center">
<img src="images/bottom.png" width="486" height="25" alt="Bottom" align="center">
	</div>
	</div>		
	<br>
	</div>
	</div>
	</center>
</body>
</html>

Any help, would be greatly appreciated.

dt192 · February 28, 2009

function userexists($in)

{

if(mysql_num_rows($in) > 0)

{

$error = '<font color="red"> User Already Exists </font>';

}

$checkuser = mysql_query("SELECT * FROM accounts WHERE username = '$username'");

userexists($checkuser);

Akenatehm · February 28, 2009

Ok Sweet. I fixed that. But now I am getting this:

MySQL Check Error with Duplicate entry 'Cody' for key 'username'

Any idea how i can get rid of it and make it display the error I already have ?

dt192 · March 1, 2009

i think you have username set as a key which meeans its uniqe and your trying to add to rows to the database both with the username cody, but i could be wrong lol

Akenatehm · March 1, 2009

Ok, i've fixed those.

Now I get this stupid thing:

MySQL Check Error with

Saying that there is no error and that it is just syaing it... anyone know how to get rid of this?

trq · March 1, 2009

You pass the variable $createresourceuser to mysqlcheck() but never seem to assign anything to this variable.

Akenatehm · March 1, 2009

$createresourceuser is a variable in the createresourcesuser() function in functions.php . Is it not possible to check a function with another?

trq · March 1, 2009

You really need to read up on the use of functions and variable scope. Variables created within functions only exist within that function. That is part of the point of functions, to encapsulate code. Code within them should not mess up the global namespace.

Akenatehm · March 1, 2009

I see. I've fixed that. Is there any way to stop a certain amount of code from being executed if something happens?

trq · March 1, 2009

I see. I've fixed that. Is there any way to stop a certain amount of code from being executed if something happens?

I'm not sure what you meen. die will stop execution.

Akenatehm · March 1, 2009

I used. and If ...Else function.

Here's my code:

functions.php

<?php

function mysqlcheck($query)
{
if (!$query)
{
	die('MySQL Check Error with ' . mysql_error());
}	
}

?>

register.php

<?php
session_start();

if (isset($_POST['submit']))
{
if($_POST['password'] == $_POST['confirmpassword'])
{
		include 'connect.php';
		include 'functions.php';
		connect();
		$username = $_POST['username'];

		$checkuser = mysql_query("SELECT * FROM accounts WHERE username = '$username'");

		mysqlcheck($checkuser);
		if(mysql_num_rows($checkuser) > 0)
		{
		$error = '<font color="red"> User Already Exists </font>';
		}
		else
		{

		$activationkey = mt_rand() . mt_rand() . mt_rand() . mt_rand() . mt_rand();	

		$username = $_POST['username'];
		$password = $_POST['password'];
		$email = $_POST['email'];
		$password = md5($password);

		$adduser = mysql_query("INSERT INTO accounts (username, password, email, activationkey, last_online) VALUES ('$username' , '$password' , '$email', '$activationkey', '0')");

		mysqlcheck($adduser);

		$findplayerid = mysql_query("SELECT playerid FROM accounts WHERE username = '$username'");

		mysqlcheck($findplayerid);									

		$transferplayerid = mysql_fetch_assoc($findplayerid);
		$playerid = $transferplayerid['playerid'];	

		$createresourcesuser = mysql_query("INSERT INTO resources (playerid,gold,wood,stone) VALUES('$playerid','200','200','200')");	

		mysqlcheck($createresourceuser);

		$result = '<font color="green">User Added Succesfully </font>';
}
}
else
{				
$result = '<font color="red"> Passwords Do Not Match </font>';		
}
}
?>

<html>
<head>
<title>Registration - Fall of The Empire</title>

<?php 
include 'page_elements.php';
abovenav();
navbar();
?> 
<div class="cbtop" align="center">
	<img src="images/register.png" width="171" height="25" alt="Register" class="welcome" align="right">	
	</div>
	<br><br>
	<div class="middle"><hr><br>
	<br>
	<table width="300px" align="right">
		<form action="<?php echo "$PHP_SELF";?>" method="POST">
			<tr>
				<td>
					<font color="white">
						Username:
					</font>
				</td>

				<td>
					<input type="text" name="username">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Email
					</font>			
				</td>

				<td>
					<input type="text" name="email">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Password:
					</font>
				</td>

				<td>
					<input type="password" name="password">
				</td>
			</tr>

			<tr>
				<td>
					<font color="white">
						Confirm Password:
					</font>
				</td>

				<td>
					<input type="password" name="confirmpassword">
				</td>
			</tr>

			<tr>
				<td>
					<br>
				</td>
				<td>
				</td>
			<tr>
				<td>
				</td>
		<td align="right" style="padding-right:8px">
		<input type="submit" name="submit" value="Register"align="right">
		</td>
		</tr>
		<tr>
			<td>
				<br>
			</td>
			<td>
			</td>
		<tr>
			<td>
			</td>
			<td><?php echo $result; ?>
			</td>
			</tr>
		</form>
	</table>
<?php
footer();
?>


</body>
</html>

and it is still showing up with:

MySQL Check Error with

Do you have any idea why its doing that?

Akenatehm · March 1, 2009

Does Anyone??

trq · March 1, 2009

Its hard to tell which piece of code is throwing that error, but if your looking for last inserted players id take a look at mysql_insert_id.

Akenatehm · March 1, 2009

Ok. mysql_insert_id is working. But it is still showing that same error.

trq · March 1, 2009

You need to debug your code and find out which call to mysqlcheck() is causing the error.

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[SOLVED] Functions and Include

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