[j0hnbanks]

I'm trying to make a simple text counter for my site, and this is the problem that I am running into, all dirs that this is linked to are set to 777 to remove all problems and the pather are all correct, but for some reason this one line is causing a problem, what am i missing??

This is the error that I am getting:
Warning: fgets(): supplied argument is not a valid stream resource in /storage/bandm/public_html/count.php on line 54


Line 54 says:
$count = fgets("$logfile", 8);

Here is my code, what is the problem with it?:
<?


$datadir = "/storage/bandm/public_html/counters/";

$snoop = 1;

if(phpversion() >= "4.2.0"){
extract($_POST);
extract($_SERVER);
extract($_ENV);
}


function do_snoop($datadir,$REMOTE_ADDR,$HTTP_USER_AGENT,$H TTP_REFERER,$REQUEST_URI)
{
$filename = str_replace("/","_",$REQUEST_URI);
if ( (substr($filename, -1)) == "_" )
$filename .= "index";
$date = date("m/d/Y h:i:s A");
$logfile = fopen("$datadir" . "$filename" . "-info.txt", "a+");
fputs("$logfile", $date . "\t" . $REMOTE_ADDR . "\t" . $HTTP_USER_AGENT . "\t " . $HTTP_REFERER . "\n");
fclose($logfile);
}

$filename = str_replace("/","_",$REQUEST_URI);
if ( (substr($filename, -1)) == "_" )
$filename .= "index";

$filetest = file_exists( "$datadir" . "$filename");
if (!$filetest)
{

$logfile = fopen( "$datadir" . "$filename", "w");
if (!$logfile)
exit;
else
{
$count += 1;
fputs("$logfile", $count);
fclose("$logfile");
exit;
}

}

else

$logfile = fopen("$datadir" . "$filename", "r+");
if (!$logfile)
exit;
else
{
$count = fgets("$logfile", 8);
if (!$count)
exit;
else
{
$count += 1;

echo "<font face=\"verdana\" size=2 color=\"#000000\"> $count </font>";

fseek("$logfile", 0);
fputs("$logfile", $count);
fclose($logfile);
}
}

if ($snoop==1) do_snoop($datadir,$REMOTE_ADDR,$HTTP_USER_AGENT,$H TTP_REFERER,$REQUEST_URI);

exit;

?>

[NeverMind]

when ever you are passing a a resource handle, NEVER put it inside double quotes!
this line was expecting a resource handle:

PHP Code:

$count = fgets("$logfile", 8); 

but instead, you gave him a weird string !!
note that when you put the variable inside double quotes, you are converting it into a string!
so this one should work:

PHP Code:

$count = fgets($logfile, 8); 

I can see that in your code, you are putting all variables inside quotes, it's not a good practice ..
always when you pass arguments to function which are stored into variables, don't duoble quotes them..