[Idolize]

So this is the error I am getting basically.



Here is the code:

PHP Code


<?php
error_reporting(E_STRICT);
$name = trim(strtolower($_GET['name']));


$name = str_replace(".", "\.", $name);
$name = str_replace("'", "\'", $name);
$name = str_replace("_", "\_", $name);

$log = array_pop(glob('logs/debug-*-*.log'));
$contents = file_get_contents($log);

if(preg_match('/' . $name . '.*?-r(\d+\.\d+\.\d+\.\d+)/', $contents, $results))
{
        echo $results[1];
}
else
{
        
        if($error)
        {    
            echo 'That user is not in our DB, please resovle them first!';
        }
}

?>

Basically, I have a folder full of logs, I want it to print out every IP that is connected with this username, but removing the duplicates and looking through all the .log files.

[Yck]

What is the $log, value or an array?
array_pop expects an array variable (by reference), not a value.

[Novalty]

Here is your problem...

PHP Code


<?php 
$log = array_pop(glob('logs/debug-*-*.log')); 


array_pop takes an argument that is passed by reference. That means you must give it a variable whereas you are just passing in an array which is returned from glob.

Changing it to the following should fix the issue:

PHP Code


<?php 
$filesArray = glob('logs/debug-*-*.log');
$log = array_pop($filesArray); 


[Idolize]

Now it wont print out anything :/

[c00perzz]

Either im confused, or :

a) there is no array to remove from
b) $log = array_pop(glob('logs/debug-*-*.log')); will end up in an empty array since its removing all log files.

Shouldnt it be like this? :

PHP Code


<?php 
$logs = glob('logs/debug-*-*.log');
foreach ($logs as $log) {
$contents = file_get_contents($log);

if(preg_match('/' . $name . '.*?-r(\d+\.\d+\.\d+\.\d+)/', $contents, $results))
{
        echo $results[1];
        array_pop($log); // not even sure if this is needed
}
else
{
        
        if($error)
        {    
            echo 'That user is not in our DB, please resovle them first!';
        }
}
}