No output??

[Ali_Baba2]

Anyone have an idea why I dont have any output on this on?

Maybe there are some flaws in the script, since I dont have output I cant control it.

so...any help would be appreciated.

include ("db_connect.inc");

if (!mysql_select_db('counter')) die(mysql_error);

$REMOTE_ADDR = $HTTP_SERVER_VARS['REMOTE_ADDR'];

$sql = "SELECT * FROM IP_STATS WHERE IP_ADDRESS = '$REMOTE_ADDR'"; 
$result = mysql_query($sql, $link_id) or die("Problem with select query: ".mysql_error());
$row = mysql_fetch_assoc($result);

if ($row['IP_ADDRESS'] !== '$REMOTE_ADDR')
{
$PERCENTAGE=$row['HITS_IP']/100*100;
$REMOTE_ADDR = $HTTP_SERVER_VARS['REMOTE_ADDR'];
mysql_query("Update IP_STATS 
SET IP_ADDRESS=$REMOTE_ADDR,
HITS_IP=HITS_IP+1,
PERCENTAGE=$PERCENTAGE,
WHERE IP_ADDRESS='$REMOTE_ADDR'"); 

}

else if ($row['IP_ADDRESS'] == '$REMOTE_ADDR')
	{
	$PERCENTAGE=$row['HITS_IP']/$row['TOTAL_VISITS']*100;
	$HITS_IP = 1;


$sql = "INSERT into IP_STATS (HITS_IP, PERCENTAGE) VALUES ('$HITS_IP', '$PERCENTAGE')";
$result = mysql_query($sql, $link_id) or die("Problem with query: ".mysql_error());
}

mysql_query("Update IP_STATS 
SET TODAY_HITS=TODAY_HITS+1, 
MONTH_HITS=MONTH_HITS+1,
TOTAL_VISITS=TOTAL_VISITS+1,
UNIQUE_VISITS=UNIQUE_VISITS+1"); 


$sql = "SELECT * FROM IP_STATS ORDER BY HITS_IP"; 
$result = mysql_query($sql, $link_id) or die("Problem with select query: ".mysql_error());
while ( $row = mysql_fetch_array($result) ) {
echo $row['IP_ADDRESS'];
}

maybe my table has some errors, so i put it here.

CREATE TABLE IP_STATS (
  ID INT(20) NOT NULL auto_increment,
  IP_ADDRESS varchar(60) NOT NULL default '',
  HITS_IP int(11), 
  PERCENTAGE INT  CHECK ((Percentage > 0) AND (Percentage <= 100)),
  TODAY_HITS int(11),
  MONTH_HITS int(11),
  TOTAL_VISITS int(11),
  UNIQUE_VISITS int(11),
  PRIMARY KEY  (id))
  TYPE=MyISAM";

This is just a counter scipt.
check if IP is in DB if true update some counters
and if false update also update some counter.

Now would I like to have an output ordered on HITS_IP, so I can see what IP_adres has the most hits.

Like this

IPaddress__most hits ip____highest percentage

and a table with todays hits, this month hits, total hits and unique hits.

There is no timeframe yet in the script. need to search for it.
my goal with this question this get to work.
IPaddress__most hits ip____highest percentage

[rachel2004]

I only looked at the first few lines of the code, but I already spotted some problems:

if (!mysql_select_db('counter')) die(mysql_error);

Try using mysql_error() so you can get error messages from MySQL:

if (!mysql_select_db('counter')) die(mysql_error());

You need to remove the comma after your variable $PERCENTAGE:

PERCENTAGE=$PERCENTAGE 
    WHERE IP_ADDRESS='$REMOTE_ADDR'");

Use mysql_error() after your mysql_query() statements. You are doing it on some of your queries, but you are not in others.

[Ali_Baba2]

I was looking with more logic to some lines like these

if ($row['IP_ADDRESS'] !== '$REMOTE_ADDR') 
    { 
    $PERCENTAGE=$row['HITS_IP']/100*100; 
    $REMOTE_ADDR = $HTTP_SERVER_VARS['REMOTE_ADDR']; 
    mysql_query("Update IP_STATS  

    SET IP_ADDRESS=$REMOTE_ADDR, 
    HITS_IP=HITS_IP+1, 
    PERCENTAGE=$PERCENTAGE, 
    WHERE IP_ADDRESS='$REMOTE_ADDR'");  

} 

I try to translate in my opinion of view.

if visitors ipadress is not in the database then update the row where ipadress is that of visitors one. that isnt logical because he couldnt find it in the first place, so how can he update it??

is this the right way of looking to these lines???

[rachel2004]

It looks like you have your IF and ELSE IF actions mixed up. I think you want this:

if ($row['IP_ADDRESS'] !== '$REMOTE_ADDR')
{
    // do INSERT of new info
}
else if ($row['IP_ADDRESS'] == '$REMOTE_ADDR') 
{
    // do UPDATE of existing record
}

However, it's possible that your IF check might never be TRUE if I am a new visitor with an IP address that isn't already in your database (I can't say for sure since I haven't tested it myself), and thus, my IP address will never be added to your database. It might be better to write it like this:

$result = mysql_query($sql, $link_id) 
    or die("Problem with select query: ".mysql_error());
$num = mysql_num_rows($result);
$row = mysql_fetch_assoc($result); 
if ($num == 0) // the IP address is not in database
{
    // do INSERT for this new IP address
}
else if ($row['IP_ADDRESS'] == '$REMOTE_ADDR')
{
    //  do UPDATE for this existing IP address
}
else 
{
    echo "<p>Problem encountered: ".$mysql_error()."</p>";
}

[Ali_Baba2]

Thanks rachel.

I got it finally working now.
I have one last question, why is there a @ in this line??
what is it purpose??

$result = @mysql_query ($query); 

final code

$db_host = "localhost"; 
$db_user = "xxx"; 
$db_pass = "xxx"; 

//connect to database
mysql_connect($db_host, $db_user, $db_pass); 
$link_id = mysql_connect ($db_host, $db_user, $db_pass);

if (!mysql_select_db('counter')) die(mysql_error());
echo ("database connected...<br>");

$get_ip = $_SERVER['REMOTE_ADDR']; 

/* create query to select IP addresses  and related info*/ 
$sql = "SELECT u.ip_address, u.hits_ip FROM IP_STATS as u WHERE u.ip_address = '$get_ip'";
$result = mysql_query($sql, $link_id) or die("<p>Couldn't execute query :".mysql_error()); 
$row = mysql_fetch_assoc($result);  

$num = mysql_num_rows($result);

if ($num == 0) 
{ 
    echo ("new visitor <br>");

$newuser = "INSERT INTO IP_STATS (ip_address,ip_hits) VALUES('$get_ip',1)";
$row = mysql_fetch_array($result); 
$remote_ip = $row['ip_address'];  

}			


else if ($num !== 0)

{	
	echo ("old visitor updating hits-ip <br>");
	$query = "update ip_stats set hits_ip = hits_ip + 1 where ip_address='$get_ip'" OR die ('Could not run query!'); 

$result = @mysql_query ($query); 

}
else  

{ 
    echo "Problem encountered: "; 
}


$sql = "SELECT * FROM IP_STATS";
$result = mysql_query($sql, $link_id) or die("<p>Couldn't execute query :".mysql_error()); 

while ($row = mysql_fetch_row ($result))
{ 
    echo "ID: " . $row[0] . " Ip-address: " . $row[1] . " Ip-hits" . $row[2] . "<br>"; 
}


[\php]

once again thanks!!

[rachel2004]

The "@" symbol, in the way it is being used, suppresses error messages from the PHP server. So if this line:

$result = @ ($query);

had a problem, you (and visitors to your site) would not see the error message get outputted to the browser. If you are developing your PHP code, it is better not to use it, but if you are at the point where you code is going to go "live" to the Web, it is better to use it.

Review this link for options regarding error handling:

http://www.php.net/manual/en/ref.errorfunc.php