[Resolved] Time and Counter

[NZ_Kiwi]

Hi,

I'm making a counter code and i'm getting a timp stamp added to my database along with a visitors IP, now when the visit again i'm searching for there IP and i want to either exit the code or add another line deoending on when they last visited.

how would i run a query if the old time stamp next to the last IP match in my database and against the new time stamp of the online now to see if is more than 10 minutes

[devinemke]

i assume your data structure looks something like this:
counter
count_id (INT)
timestamp (INT)
ip (VARCHAR)

$result = mysql_query("SELECT * FROM counter WHERE ip = '" . $SERVER['REMOTE_ADDR'] . "' AND timestamp >= " time() - 600);
if (mysql_num_row($result))
{
	echo 'The IP address ' . $SERVER['REMOTE_ADDR'] . ' has visited in the last 10 minutes.  No record added.';
}
else
{
	$result = mysql_query("INSERT INTO counter SET ip = '" . $SERVER['REMOTE_ADDR'] . "', timestamp = " time());
	if ($result) {echo 'New record added.';}
}

[NZ_Kiwi]

Originally posted by devinemke
i assume your data structure looks something like this:
counter
count_id (INT)
timestamp (INT)
ip (VARCHAR)

$result = mysql_query("SELECT * FROM counter WHERE ip = '" . $SERVER['REMOTE_ADDR'] . "' AND timestamp >= " time() - 600);
if (mysql_num_row($result))
{
	echo 'The IP address ' . $SERVER['REMOTE_ADDR'] . ' has visited in the last 10 minutes.  No record added.';
}
else
{
	$result = mysql_query("INSERT INTO counter SET ip = '" . $SERVER['REMOTE_ADDR'] . "', timestamp = " time());
	if ($result) {echo 'New record added.';}
}

[/B]

okay i can see how that works but i get a parse error on this line

$SERVER['REMOTE_ADDR'] . "' AND timestamp >= " time() - 600);

what is wrong?

[bodzan]

It should look like:

$result = mysql_query("INSERT INTO counter SET ip='".$SERVER['REMOTE_ADDR']."', timestamp=".time());

a "." sign before time()...

[NZ_Kiwi]

Originally posted by bodzan
It should look like:

$result = mysql_query("INSERT INTO counter SET ip='".$SERVER['REMOTE_ADDR']."', timestamp=".time());

a "." sign before time()...

Okay, i've fixed it to what you said now it looks like this

$result = mysql_query("INSERT INTO counter SET IP = '".$SERVER['REMOTE_ADDR']."', timestamp="."time()); 

but i still get a parse error on line 12, this is line 12,

i thought it would be more like

$result = mysql_query("SELECT * FROM counter WHERE IP = '".$SERVER['REMOTE_ADDR']."'); 

Then go on and do a if statement about how the timestamp is similar to the new one if it its -600 then exit if not insert a new row???

[bodzan]

in my post I meant just to put a period or . before time()...

Just delete the " before time()...

Sory for my unclearness...

[devinemke]

sorry. those were typos from me, here's an update:

$result = mysql_query("SELECT * FROM counter WHERE ip = '" . $SERVER['REMOTE_ADDR'] . "' AND timestamp >= " . time() - 600); 
if (mysql_num_row($result)) 
{ 
    echo 'The IP address ' . $SERVER['REMOTE_ADDR'] . ' has visited in the last 10 minutes.  No record added.'; 
} 
else 
{ 
    $result = mysql_query("INSERT INTO counter SET ip = '" . $SERVER['REMOTE_ADDR'] . "', timestamp = "  . time()); 
    if ($result) {echo 'New record added.';} 
}

[bodzan]

$result = mysql_query("INSERT INTO counter SET ip = \'" . $SERVER['REMOTE_ADDR'] . "\', timestamp = " . time());

In both lines there is this tiny problem (well I'm guessing it is a problem) of no escaping the ' sign and thus the code might be misread other than that, I don't know, it looks fine to me...

I think this one could also solve this issue:
$result = mysql_query("INSERT INTO counter SET ip = '$SERVER[REMOTE_ADDR]', timestamp = " . time());

well, I use this kind of writting...

[NZ_Kiwi]

Everything you have givin me to try does not work, they all give me the same error 😕

I've made this code,

$newIP = $_SERVER['REMOTE_ADDR'];
$self = $_SERVER['PHP_SELF'];
$time = time();

$result = mysql_query ("SELECT * FROM counter WHERE IP = $newIP"); 
$row = mysql_fetch_array($result);

$IP = $row["IP"];
$timestamp = $row["timestamp"];


if ($timestamp  <= 600)
{
echo 'sorry';
} else {
$result = mysql_query ("INSERT INTO counter (`id`,`newIP`,`self`,`time`) Values ('','$newIP','$self',$time')") or die("INSERT error: ".mysql_error()); 
}

problem with this code it that if the person is new they will not be in the database.... then you get a Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/site.com/httpdocs/counter2.php on line 11

any ideas?

[bodzan]

And for the latest set of code I can see where the problem is so you get the error that you get:

$result = mysql_query ("SELECT * FROM counter WHERE IP = $newIP");
$row = mysql_fetch_array($result);

If there is no record of somebody visiting before from that IP (i.e. visits for the first time) the $result will return an empty array and thus will be the MySQL result resource...

I would check if there is a result at all like

if(mysql_num_rows($result)!=0){
$row = mysql_fetch_array($result);

$IP = $row['IP'];
$timestamp = $row['timestamp'];

}

this way you avoid this little error.

Will it perform like you need it I don't know (I haven't seen your entire code)...

[devinemke]

let's all backup a sec. my original code was was based purley on a proposed data structure (just guessing at field names). what is the actual data structure of your table (table name, field names, field types, etc...)

[NZ_Kiwi]

okay my table is the same

<?php

mysql_connect (localhost, root, y4worl);
mysql_select_db (bevan);
$result = mysql_query("SELECT * FROM counter WHERE IP = '" . $SERVER['REMOTE_ADDR'] . "' AND timestamp >= " . time() - 600);  


if(mysql_num_rows($result)!=0){ 
$row = mysql_fetch_array($result); 

} else {


if (mysql_num_row($result))  

{  

    echo 'The IP address ' . $SERVER['REMOTE_ADDR'] . ' has visited in the last 10 minutes.  No record added.';  

}  

else  

{  

    $result = mysql_query("INSERT INTO counter SET ip = '" . $SERVER['REMOTE_ADDR'] . "', timestamp = "  . time());  

    if ($result) {echo 'New record added.';}  

} 

}

?>

The error is: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/site.com/httpdocs/counter.php on line 8

Fatal error: Call to undefined function: mysql_num_row() in /home/httpd/vhosts/site.com/httpdocs/counter.php on line 14