Passing a variable from PHP to ubuntu - php

I want to run a script on my ubuntu server with a variable from php in it.
Here are some of the things i've tried to pass a variable to ubuntu...
shell_exec('vpsName=HI3');
vpsName=`mysql -D jake_db -h 127.0.0.1 -u jake -pXXXXXXX -se "SELECT vpsName FROM reinstalls WHERE status = 'pending'"`;
The final way in which i thought I could fix it was to avoid running the script through ubuntu all together and run it from shell_exec(); but it fails on running the guestfish commands.
Here is my entire reinstalls.sh script.
sudo rm /var/lib/libvirt/images/"$vpsName".qcow2 && sudo wget -O /var/lib/libvi$
guestfish -a /var/lib/libvirt/images/"$vpsName".qcow2 <<'EOF'
run
mount /dev/ubuntu-vg/root /
rm /etc/network/interfaces
EOF
sudo fusermount -u /mnt && virsh start "$vpsName" && echo "IT WORKED!"
I am open to any way of getting this to work, as long as its secure,
Thanks in advance,
Jake
EDIT:
If I run the script with a VPS name instead of a variable, it works. I just can't find a way to pass the variables from the website to the ubuntu16.04 OS.

Well one way would be to run each command from PHP:
shell_exec("sudo rm /var/lib/libvirt/images/" . $vpsName . ".qcow2");
shell_exec("sudo wget -O /var/lib/libvi ... etc");
The other would be to invoke your shell script from PHP, passing the vps name as a parameter:
shell_exec("reinstalls.sh " . $vpsName)
But then you'd have to rewrite the shell script to pick up the command line parameter and apply it as necessary. In the case of bash, this explains how to go about that.

Related

Incorrect Webpage Output for PHP exec running bash script for audtool (part of Audacious)

On a local linux server (Rapsberry Pi debian stretch with desktop), I am working on sending "audtool" commands to a running Audacious media player using php, exec and bash scripts. Audacious is autostarted when the server starts up with user "pi". I have apache2 and php set up and working on the server, and I can ssh to the server and run all the commands from the cli. I believe I have resolved the issues with running audtool (dbus and setting the right environment variables) and running the php on the command line works successfully. However when running the php on a webpage I get back a long string of information about apache2
I have spent several hours (getting on for a whole day) researching this on the web in order to get to this stage, so close I can almost touch it, but stuck on this last element. The example is to display the current song from a running instance of Audacious. Audtool requires a running dbus (looks for a display). Using exec or shell_exec I have no problems running bash commands such as whoami or ls.
The php page (cursong.php):
<?php
echo exec('/var/www/html/cursong.sh');
?>
The bash script (cursong.sh):
#!/bin/bash
##call current song
pid=`pidof audacious`
user=`ps -p $pid -o user=`
export `strings /proc/$pid/environ | grep DBUS_SESSION_BUS_ADDRESS`
sudo -E -su $user /usr/bin/audtool --current-song
(from here: https://redmine.audacious-media-player.org/boards/1/topics/1058?r=1059)
Output from command line:
php -f cursong.php
Artist - Song Title (for example - so this works)
Output on webpage:
declare -x APACHE_LOCK_DIR="/var/lock/apache2" declare -x
APACHE_LOG_DIR="/var/log/apache2" declare -x
APACHE_PID_FILE="/var/run/apache2/apache2.pid" declare -x
APACHE_RUN_DIR="/var/run/apache2" declare -x APACHE_RUN_GROUP="www-
data" declare -x APACHE_RUN_USER="www-data" declare -x
INVOCATION_ID="4ce76136ca8842bd9108d6b1b9a5b9ed" declare -x
JOURNAL_STREAM="8:23896" declare -x LANG="C" declare -x OLDPWD
declare -x
PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin"
declare -x PWD="/var/www/html" declare -x SHLVL="1"
I have set www-data, the apache2 user with the following in
/etc/sudoers:
www-data ALL=NOPASSWD: ALL
and /var/www/html is rwx for anyone
Obviously, I am expecting to see "Artist - Song Title" on the webpage, but instead I get back all the apache2 info. What am i missing, or where have I gone wrong?
I hate answering my own question, makes it look like I wasn't trying hard enough! After a further five hours or so of searching around and attempting fixes, I happened upon this post on SO:
Running command-line application from PHP as specific user
which suggested putting a "sudo -u user" in the exec of the php file. I tried this with the "pi" user and it still didn't work, then I simply tried it with "sudo" and hey presto!!
The php file now looks like this:
<?php
echo shell_exec('sudo /var/www/html/cursong.sh 2>&1');
?>
Now to do some testing on how it works with the other audtool commands that don't ask for a response but require action from audacious, and to see how I can reduce scripting php files by passing a parameter to the bash script!
Just for completeness, the php and bash scripts for both a request and an action, using a parameter fed to the php url and then on to the bash script:
PHP File with Parameter
<?php
$request = $_GET["request"];
echo shell_exec("sudo /var/www/html/cursong.sh \"${request}\" 2>&1");
?>
url example:
http://192.168.1.92/cursong.php?request="--playlist-shuffle-status"
Bash Script with parameter
#!/bin/bash
##call request
pid=`pidof audacious`
user=`ps -p $pid -o user=`
export `strings /proc/$pid/environ | grep DBUS_SESSION_BUS_ADDRESS`
sudo -E -su $user /usr/bin/audtool $1
PHP file for an action
<?php
$action = $_GET["action"];
shell_exec('sudo /var/www/html/playsong.sh \"${request}\" ');
?>
url example:
http://192.168.1.92/cursong.php?action="--playback-play"
Bash script for an action
#!/bin/bash
##call action
pid=`pidof audacious`
user=`ps -p $pid -o user=`
export `strings /proc/$pid/environ | grep DBUS_SESSION_BUS_ADDRESS`
sudo -E -su $user /usr/bin/audtool $1

Allow PHP/Apache to shell_execute commands on Ubuntu

I'm trying to execute a command through PHP with shell_exec. The PHP file is hosted by Apache on my Ubuntu server.
When I run this:
echo shell_exec("ps ax | grep nginx");
Then I get to see data. But when I run another command, for example:
echo shell_exec("cat /usr/local/nginx/config/nginx.config");
Then it's not showing anything at all. But when I copy that command and paste it in my terminal, then it executes fine.
My Apache server is running as user www-data. So I edited sudoers and added this line:
www-data ALL=(ALL:ALL) ALL
I know this is a security risk, but I wanted to make sure (for now) that www-data is able to execute all commands. But, for some reason I'm still not able to execute all commands with my PHP script.
Anyone any idea what to do?
have you read http://php.net/manual/en/function.shell-exec.php
There is quite a discussion in comments section. Top comment is:
If you're trying to run a command such as "gunzip -t" in shell_exec and getting an empty result, you might need to add 2>&1 to the end of the command, eg:
Won't always work:
echo shell_exec("gunzip -c -t $path_to_backup_file");
Should work:
echo shell_exec("gunzip -c -t $path_to_backup_file 2>&1");
In the above example, a line break at the beginning of the gunzip output seemed to prevent shell_exec printing anything else. Hope this saves someone else an hour or two.
echo shell_exec("sudo cat /usr/local/nginx/config/nginx.config");
Try that.

launching php script running on server (and opening sockets) from a php page

I'd need you precious help on a matter I am spending hours on.
Scope: Apache2 and PHP running on a raspberry pi;
Premise: my little knowledge of Linux environment!
The objective: launching a long-run php script, that opens sockets, from another php script running as webpage. In other terms, the application is a chat and I need to start the php server script form a web page, for my convenience.
The issue: if I run it from the the console, logged as "pi", with the following command
php -q /var/www/chatSocket.php > /var/www/tmp/socketProcessOutput.txt 2>&1 & echo $!
it works like a charm, but if it try to do so from a script, with the following (don't mind the concatenated strings and assignment of output to variables - it made no difference removing them):
$result .= "Result of pkill (killed process): " .shell_exec('sudo pkill -f SongWebSocket.php') ."\n";
$result .= "Launching new process: id returned:". shell_exec('php -q /var/www/chatSocket.php > /var/www/tmp/socketProcessOutput.txt 2>&1 & echo $!') ."\n";
$result .= "Checking running SongWebSocket.php process:" ."\n";
$result .= shell_exec('ps -A aux| grep -e SongWebSocket.php -e USER') ."\n";
.. it does not work (it seems like it launch the script but the sockets ar not open).
Any clue why this happens?
Also, and this can be for my little knowledge of Linux, why i get a dioffrent aoutput from the command
ps aux| grep -e SongWebSocket.php -e USER
if I launch it from the shell, as user pi, or from the sript, as www-data user.
I Look forward for your help. Thanks in advance!
Marco.
www-data user doesn't have the permisson I guess. why not add "sudo" for every shell_exec line? (sudo starts the programm with root permission). it's not pretty and not secure but it might work on you local home-network. sudo php ... sudo ps -A aux etc. In addition you should make sure that the php safe_mode is off. you can see that by adding phpinfo(); to you php code

Run a command on a SSH screen through SSH with phpseclib

I have a PHP script that starts a detached screen through SSH:
$ssh->exec("screen -m -d -S ".$user);
I now need to execute a command in that screen without being in that screen. I have the code that does that, which I have tested through a SSH client, but when I try to use it with the phpseclib exec command, it does not work. This is the code that works:
screen -S ".$user." -X stuff "cd minecraft/servers/".$user."/;sh start.sh $(printf '\r')"
And this is it in the PHP script:
$ssh->exec("screen -S ".$user." -X stuff \"cd minecraft/servers/".$user."/;sh start.sh $(printf '\r')\"");
I attempted to escape the extra double quotes in the code.
Is there anything I can do to make this work through PHP? Thanks
Hmmm...
create please two bash script, first: create screen with user parameter with name f.e. run_screen, second: tester for SSH client with user parameter with name f.e. run_test.
Run first script:
$ssh->exec('[full_path]/run_screen ' . $user);
and second:
$ssh->exec('[full_path]/run_test ' . $user);
bash syntax is here bash syntax
Sure that the user of server (f.e. Apache) has permissions to run scripts.

Command not found: PHP exec()

This is driving me crazy. I need to have php execute a command to restart a script running in node. I'm using a node app called forever to run said script. The code is as follows:
<?php
echo '<pre>';
echo exec('sudo -u dalton forever restart botti.js 2>&1');
echo '</pre>';
?>
However, when I run that, I get sudo: forever: command not found
Next, I try which forever and type forever, both which give me:
forever: /usr/local/bin/forever
I edit my code to:
echo exec('sudo -u dalton /usr/local/bin/forever restart botti.js 2>&1');
Edit: After a typo, the error is now:
/usr/bin/env: node: No such file or directory
I'm at my wit's end. Any ideas?
As the forever command only runs, when you give the full path, I suspect, that /usr/local/bin is not in your PATH environment variable, which contains all directories, that are searched for executable commands by default, separated by : (I suspect you're on Linux, may differ for other OS)
I suspect forever calls /usr/bin/env node. The error from env is probably caused by node being outside your PATH too.
To set your PATH in php, use putenv('PATH=<your path here>');
e.g. to append /usr/local/bin:
putenv('PATH=' . getenv('PATH') . ':/usr/local/bin')
This may also be a sudo issue, try the -E (preserve environment) switch.
Figured it out, I needed to define node as well:
$asdf = system('sudo -E -u dalton /usr/local/bin/node /usr/local/bin/forever restart botti.js 2>&1');
Create a symbolic link for forever
ln -s /usr/local/bin/forever /usr/bin/env/forever
And also for nodejs if incase it's still called "nodejs". Make it call as "node"
ln -s /usr/bin/nodejs /usr/bin/node
I will solve the forever execution problem.
For php side, try with this
echo shell_exec("your command sh");

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