[system(), passthru(), shell_exec(), exec() hangs]

Thanks Thorpe ... it looks like I'm going to have to look to another solution to the problem.

[system(), passthru(), shell_exec(), exec() hangs]

Hi Thorpe,

The purpose of the pHP code is to actually start the program and not bother with anything that the script outputs.  The job of the PHP code is to start the script in event that the script stops.

I've got it so that it knows when the script needs to be run again, but it just hangs on issuing the exec command.

Michelle.

[system(), passthru(), shell_exec(), exec() hangs]

Same problem as a post a few months ago, but a different cause.

I write tools to support people running L2J servers, and I have a problem ...

The script detects whether the gameserver is down, and if it is, then it it tasked with restarting it.

The program correctly executes the script, which contains a command ...
#! /bin/sh
./GameServer_loop.sh &.
... which then calls a looping .sh file.

The issue of this is that, of course, the whole thing never ends and even if I shut down the gameserver, the program hangs ... presumably waiting for some return from the command so that the script can continue executing.

This, obviously, won't happen.

What I need is a way of running a system command, and telling the PHP script not to bother waiting for a response ... just continue running.

Any ideas please?

[creat directory]

Sorry if I'm off the mark.  The second variable is the permissions, I believe.

Here's some code I use in case it helps ...

$file_loc = $server_dir . 'log' . $svr_dir_delimit . 'knightchat' . $svr_dir_delimit . 'chat.log';
$dir_loc = $server_dir . 'log' . $svr_dir_delimit . 'knightchat';
if (!is_dir($dir_loc))
{ mkdir($dir_loc, 0700); }

I did have issues with writing directories that the script didn't have permissions to create.  It is a matter of working out what credentials the PHP script is running with.  I did this by setting a directory to 777, running the script to create the directory, then I could see what user name and group it created the directory with, by using the ls -la.  Then I tightened the security on the parent directory again.

Re: the safe mode, I think there is no way other than to switch the php.ini file to not operate in safe mode.  I don't think it can be done by a session variable, but I'm no expert.

[Newbie with a pass by reference problem]

Thanks very much for the help.

I was right at the begining of my PHP project when I needed to talk with the gameserver via telnet, so I had to copy someone elses telnet code and adjust it.  The last two months has seen me learn a lot about PHP, but still not enough to get to this kind of detail yet.

[Newbie with a pass by reference problem]

Hi All,

Sorry for being a complete newbie  ;D

A nice simple one for you.  I dabble in different languages, and am working on a PHP project to help people running L2J servers to administer the game.  My work is free to the community.

I'm having a problem, however, with code that is passing two variables by reference.  I understand what I am doing, and I understand the error, which is that passing variables by reference is depreciated in the later versions of PHP.  I just don't know enough to be confident of my solution.

Although setting the variable in the php.ini file will get around it, and I do set a temporary session variable, this isn't an ideal solution.  Also, although fumbling around in the dark is my prefered way of learning, when other people are using my code I sort of owe it to them to say, "I need help!" and then go get it ... hence I'm here.

[code]$usetelnet = fsockopen($telnet_host, $telnet_port, &$errno, &$errstr, $telnet_timeout);[/code]This is what I'm doing wrong ... $errno and $errstr are being passed by reference, but if I don't pass them by reference I'm sort of stuffed.

To my amateure mind, the other side is probably working on the reference and if I don't pass the right numbers, then it may well write in to an unknown place in memory.  That is why I'm concerned.

If I do something like ...
[code]$errnoref = &$errno;
$errstrref = &$errstr;
$usetelnet = fsockopen($telnet_host, $telnet_port, $errnoref, $errstrref, $telnet_timeout);[/code]... will that solve my problem; or will I just crash someones server!

Any help gratefully appreciated.