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PHP Error on line

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Join Date: Oct 2009

Posts: 9

Reputation: foxwizzy is an unknown quantity at this point

Solved Threads: 1

foxwizzy

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Newbie Poster

PHP Error on line

Oct 31st, 2009

hello all i am new to this forum and need some help whit php.
I am trying to create a script that shows the number of users online.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
 
	$host = "xx.xxx.xxx.xx";
	$user = "xxxx";
	$pass = "xxxx";
	$db = "users"
 
	$conn = mysql_connect("$host", "$user", "$pass") or die ("Unable to connect to database."); 
mysql_select_db("$db", $conn);
 
 
    include_once ("usersOnline.class.php");
$visitors_online = new usersOnline();
 
if (count($visitors_online->error) == 0) {
 
	if ($visitors_online->count_users() == 1) {
		echo "There is " . $visitors_online->count_users() . " visitor online";
	}
	else {
		echo "There are " . $visitors_online->count_users() . " visitors online";
	}
}
else {
	echo "<b>Users online class errors:</b><br /><ul>\r\n";
	for ($i = 0; $i < count($visitors_online->error); $i ++ ) {
		echo "<li>" . $visitors_online->error[$i] . "</li>\r\n";
	}
	echo "</ul>\r\n";
 
}
?>
<?php
	
	$host = "xx.xxx.xxx.xx";
	$user = "xxxx";
	$pass = "xxxx";
	$db = "users"
	
	$conn = mysql_connect("$host", "$user", "$pass") or die ("Unable to connect to database."); 
mysql_select_db("$db", $conn);

	
    include_once ("usersOnline.class.php");
$visitors_online = new usersOnline();

if (count($visitors_online->error) == 0) {

	if ($visitors_online->count_users() == 1) {
		echo "There is " . $visitors_online->count_users() . " visitor online";
	}
	else {
		echo "There are " . $visitors_online->count_users() . " visitors online";
	}
}
else {
	echo "<b>Users online class errors:</b><br /><ul>\r\n";
	for ($i = 0; $i < count($visitors_online->error); $i ++ ) {
		echo "<li>" . $visitors_online->error[$i] . "</li>\r\n";
	}
	echo "</ul>\r\n";

}
?>

My problem is : Parse error: parse error in C:\wamp\www\Untitled-2.php on line 24

My line 24 is this

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$conn = mysql_connect("$host", "$user", "$pass") or die ("Unable to connect to database."); 
mysql_select_db("$db", $conn);
$conn = mysql_connect("$host", "$user", "$pass") or die ("Unable to connect to database."); 
mysql_select_db("$db", $conn);

This is my usersOnline.class.php

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 PHP Syntax (Toggle Plain Text) 
class usersOnline {
 
	var $timeout = 600;
	var $count = 0;
	var $error;
	var $i = 0;
 
	function usersOnline () {
		$this->timestamp = time();
		$this->ip = $this->ipCheck();
		$this->new_user();
		$this->delete_user();
		$this->count_users();
	}
 
	function ipCheck() {
		if (getenv('HTTP_CLIENT_IP')) {
			$ip = getenv('HTTP_CLIENT_IP');
		}
		elseif (getenv('HTTP_X_FORWARDED_FOR')) {
			$ip = getenv('HTTP_X_FORWARDED_FOR');
		}
		elseif (getenv('HTTP_X_FORWARDED')) {
			$ip = getenv('HTTP_X_FORWARDED');
		}
		elseif (getenv('HTTP_FORWARDED_FOR')) {
			$ip = getenv('HTTP_FORWARDED_FOR');
		}
		elseif (getenv('HTTP_FORWARDED')) {
			$ip = getenv('HTTP_FORWARDED');
		}
		else {
			$ip = $_SERVER['REMOTE_ADDR'];
		}
		return $ip;
	}
 
	function new_user() {
		$insert = mysql_query ("INSERT INTO useronline(timestamp, ip) VALUES ('$this->timestamp', '$this->ip')");
		if (!$insert) {
			$this->error[$this->i] = "Unable to record new visitor\r\n";			
			$this->i ++;
		}
	}
 
	function delete_user() {
		$delete = mysql_query ("DELETE FROM useronline WHERE timestamp < ($this->timestamp - $this->timeout)");
		if (!$delete) {
			$this->error[$this->i] = "Unable to delete visitors";
			$this->i ++;
		}
	}
 
	function count_users() {
		if (count($this->error) == 0) {
			$count = mysql_num_rows ( mysql_query("SELECT DISTINCT ip FROM useronline"));
			return $count;
		}
	}
 
}
 
?>
class usersOnline {

	var $timeout = 600;
	var $count = 0;
	var $error;
	var $i = 0;
	
	function usersOnline () {
		$this->timestamp = time();
		$this->ip = $this->ipCheck();
		$this->new_user();
		$this->delete_user();
		$this->count_users();
	}
	
	function ipCheck() {
		if (getenv('HTTP_CLIENT_IP')) {
			$ip = getenv('HTTP_CLIENT_IP');
		}
		elseif (getenv('HTTP_X_FORWARDED_FOR')) {
			$ip = getenv('HTTP_X_FORWARDED_FOR');
		}
		elseif (getenv('HTTP_X_FORWARDED')) {
			$ip = getenv('HTTP_X_FORWARDED');
		}
		elseif (getenv('HTTP_FORWARDED_FOR')) {
			$ip = getenv('HTTP_FORWARDED_FOR');
		}
		elseif (getenv('HTTP_FORWARDED')) {
			$ip = getenv('HTTP_FORWARDED');
		}
		else {
			$ip = $_SERVER['REMOTE_ADDR'];
		}
		return $ip;
	}
	
	function new_user() {
		$insert = mysql_query ("INSERT INTO useronline(timestamp, ip) VALUES ('$this->timestamp', '$this->ip')");
		if (!$insert) {
			$this->error[$this->i] = "Unable to record new visitor\r\n";			
			$this->i ++;
		}
	}
	
	function delete_user() {
		$delete = mysql_query ("DELETE FROM useronline WHERE timestamp < ($this->timestamp - $this->timeout)");
		if (!$delete) {
			$this->error[$this->i] = "Unable to delete visitors";
			$this->i ++;
		}
	}
	
	function count_users() {
		if (count($this->error) == 0) {
			$count = mysql_num_rows ( mysql_query("SELECT DISTINCT ip FROM useronline"));
			return $count;
		}
	}

}

?>

Can any one tell me where my prob is?

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Join Date: Oct 2006

Posts: 1,102

Reputation: ardav will become famous soon enough

Solved Threads: 139

ardav

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Veteran Poster

-1

Oct 31st, 2009

I haven't looked closely at your code, but it looks as though you've left out a semicolon at the end of:

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$db = "users"

If you don't reply to your own thread or you can't find the solved link - you're off my Christmas list - permanently! Bah humbug!

•

Join Date: Oct 2009

Posts: 9

Reputation: foxwizzy is an unknown quantity at this point

Solved Threads: 1

foxwizzy

Offline

Newbie Poster

Oct 31st, 2009

What wandres a small semicolon can do

the script is ok now thenx

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Join Date: Aug 2007

Posts: 130

Reputation: hemgoyal_1990 is an unknown quantity at this point

Solved Threads: 8

hemgoyal_1990 hemgoyal_1990 is offline

Offline

Junior Poster

-1

Oct 31st, 2009

@above.
if your script is work and ur problem is solved then please mark this thread to solve.

http://www.kuchamancity.com
Hem Web Solution..
Behind Every Successful Man, There is an Untold Pain in His Heart.

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Join Date: Oct 2009

Posts: 9

Reputation: foxwizzy is an unknown quantity at this point

Solved Threads: 1

foxwizzy

Offline

Newbie Poster

Nov 1st, 2009

Now that i got the number of users on work i have another prob...
I am going to show the total number of registerd users(userID,user name,E-Mail)...and i dont now what this is :
"Notice: Undefined index: id in C:\wamp\www\Untitled-1.php on line 3"

The script works but thet line is buging me

this is line 3

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$myID = $_GET['id'];
$myID = $_GET['id'];

This is the script :

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<?php
 
$myID = $_GET['id'];
 
$server   = 'host';
$username = 'dbuser';
$password = 'dbpass';
$database = 'db';
 
$connect = mysql_connect($server, $username, $password)
	or die("Error connecting: " . mysql_error());
$select = mysql_select_db($database, $connect)
	or die("Error selecting DB: " . mysql_error());
 
// Select a specific entry
if(strlen($myID) > 0) {
	$query = "SELECT * FROM users WHERE id='$myID'"; }
// Select EVERY entry
else {
	$query = "SELECT * FROM users"; }
?>
 
<center>
<table border='1' cellspacing='0' cellpadding='2'>
	<tr>
		<th>ID</th>
		<th>Nume Utilizator</th>
		<th>E-Mail</th>
	</tr>
 
<?php
 
$run = mysql_query($query)
	or die("Error running query: " . mysql_error());
while($row = mysql_fetch_array($run)) {
	print("\n<tr><td>{$row['ID']}</td>");
	print("\n<td>{$row['username']}</td>");
	print("\n<td>{$row['email']}</td></tr>"); }
 
?>
 
</table>
</center>
<?php

$myID = $_GET['id'];

$server   = 'host';
$username = 'dbuser';
$password = 'dbpass';
$database = 'db';

$connect = mysql_connect($server, $username, $password)
	or die("Error connecting: " . mysql_error());
$select = mysql_select_db($database, $connect)
	or die("Error selecting DB: " . mysql_error());

// Select a specific entry
if(strlen($myID) > 0) {
	$query = "SELECT * FROM users WHERE id='$myID'"; }
// Select EVERY entry
else {
	$query = "SELECT * FROM users"; }
?>

<center>
<table border='1' cellspacing='0' cellpadding='2'>
	<tr>
		<th>ID</th>
		<th>Nume Utilizator</th>
		<th>E-Mail</th>
	</tr>
		
<?php

$run = mysql_query($query)
	or die("Error running query: " . mysql_error());
while($row = mysql_fetch_array($run)) {
	print("\n<tr><td>{$row['ID']}</td>");
	print("\n<td>{$row['username']}</td>");
	print("\n<td>{$row['email']}</td></tr>"); }

?>

</table>
</center>

Last edited by foxwizzy; Nov 1st, 2009 at 5:43 am.

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Join Date: Oct 2006

Posts: 1,102

Reputation: ardav will become famous soon enough

Solved Threads: 139

ardav

Offline

Veteran Poster

-1

Nov 1st, 2009

Why is it bugging you?

You don't have any validation:
If expecting an integer, you should check to see that you are getting an integer. What happens if an id is not passed? You script will fail. What should you do in this case? Select a random id from the DB or show an error message before redirecting to another page?
Look up:

is_int() and isset()

Last edited by ardav; Nov 1st, 2009 at 6:09 am.

If you don't reply to your own thread or you can't find the solved link - you're off my Christmas list - permanently! Bah humbug!