CODEWITHSUNDEEP
php
Rickie W
Rickie W

Reputation: 35

How would I tail a dated log file?

I want to tail a dated log file LIVE on the administrator page of my website, but I'm running in to problems. I can get it to work only if I manually input the date in the $filename variable.

Here is the code I found on here too, but would like to know how to get dated files (ex: 2016_04_26.ilog).

<?php
$filename = 'date('Y_m_d').ilog';
$filedir = '/gamelogs/';
$output = shell_exec('exec tail -n250 ' . $filedir$filename');
echo str_replace(PHP_EOL, '<br />', $output);
?>

What is wrong with my code? Thanks in advance!


So after taking a break and coming back to the code, I realized I had the date format wrong and a couple other parts. Here is what I have that works!

<?php
$filename = date('y_m_d').'.ilog'; // misplaced quotes
$filedir = '/home/a3invmgr/destinedtobe/a3ilogs/'; // relative path instead of full path
$output = shell_exec("tail -n100  $filedir$filename"); //exec removed
echo str_replace(PHP_EOL, '<br />', $output); // line OK
?>

Upvotes: 2

Views: 3618

Answers (3)

Pedro Lobito
Pedro Lobito

Reputation: 98921

You've a couple of errors, try this:

<?php
$filename = date('Y_m_d').'.ilog'; // misplaced quotes
$filedir = '/full/path/to/gamelogs/'; // relative path instead of full path
$output = shell_exec("tail -n250  {$filedir}{$filename}"); //exec removed
echo str_replace(PHP_EOL, '<br />', $output); // line OK
?>

Upvotes: 1

Ben
Ben

Reputation: 631

I think your initial $filename might be wrong - I see too many quotes. Assuming that filenames look like 2016_04_27.ilog then you should use:

$filename = date('Y_m_d'). '.ilog';

Also, to further troubleshoot, you might try putting the execution command into a variable and outputting that:

$stringToExecute = ('tail -n250 ' . $filedir$filename);
echo $stringToExecute;
$output = shell_exec(stringToExecute);

Lastly, if you're echoing log files, you might just want to skip the str_replace and just wrap the results in a <pre>. This will tell the browser to keep newlines intact.

echo "<pre>$output</pre>";

Upvotes: 0

Webeng
Webeng

Reputation: 7113

Two problems that need fixing:

  1. $filedir$filename

that is not a valid php variable.

  1. You have an apostrophe attacking your code here

    $filedir$filename');

Take the apostrophe out.

Fix those two issues and let me know if your problem is solved.

Upvotes: 0

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