PHP shell_exec() - having problems running command - php

Im new to php shell commands so please bear with me. I am trying to run the shell_exec() command on my server. I am trying the below php code:
$output = shell_exec('tesseract picture.tif text_file -l eng');
echo "done";
I have the picture.tif in the same directory as the php file. In my shell I can run this without a problem.
It takes a while to run the code, then it doesnt make the text_file like it does when I run it in command prompt.

Per your comment:
Should I write a loop in shell
instead?
You can write a very simple shell script to run the command in a loop. Create the script file first:
touch myscript.sh
Make the script executable:
chmod 700 myscript.sh
Then open it with a text editor such as vim and add this:
for (( i = 0 ; i <= 5; i++ ))
do
tesseract picture.tif text_file -l eng
done
Thats the very basics of it (not sure what else you need), but that syntax should help get you started. To run the script, do this if you're in the same directory as the script:
./myscript.sh
Or specify the full path to run it from anywhere:
/path/to/mydir/myscript.sh

Could this be a permissions issue? My guess is that PHP isn't running with the same permissions that you do when you execute the command directly from the command prompt. What OS are you running on?

Related

Issues With PHP exec & shell_exec - Shell Script Execution

I've been unable to run php scripts that I need to use to start and stop webcam services that run on the local machine with the scripts. I can find nothing in the logs to indicate why the script doesn't' work.
I confess to being severely handicapped regarding PHP, especially server-side scripting.
The environment is Debian Jesse running Nginx with all required SSH and PHP modules installed
I have added www-data to the sudoers file with:
www-data ALL=(ALL) NOPASSWD: /var/www/html/start_webcam.sh
Enabled the $PATH environment for www-data at:
/etc/php5/fpm/pool.d/www.conf
The shell script resides in the .../html directory and runs from the terminal with no issues.
This is the code for both the php and shell scripts:
start_webcam.php:
<?php
echo exec('sudo bash /var/www/html/aspirebox/start_webcam.sh 2>&1, $output');
print_r($output);
?>
The $output and print_r stuff is there because it was the last thing I tried based on a post I found out here somewhere.
start_webcam.sh
#!/bin/bash
service motion start
Thanks in advance to anyone out here that has a clue. After 2 days of wrestling with this, I am sure that I do not.
according to Passing Variables to shell_exec()? you should change your code like this:
<?php
$output = exec('/var/www/html/aspirebox/start_webcam.sh 2>&1 ');
print_r($output);
?>
and let your bash script execute as all (no need to sudo bash):
chmod a+x /var/www/html/aspirebox/start_webcam.sh
Thank you very much - that worked.
I worked through getting the path straight for the directory the shell script runs in, and the correct path to run "service".
All I have now is to figure out why I'm getting "Failed to start motion.service: Access denied"
I've given www-data permission to run the script without a password on sudoers, have to keep digging.
Thanks again!

How to plan job in php script via exec and 'at'

I try to plan one-time job with 'at' command. There is next code in script:
$cmd = 'echo "/usr/bin/php '.$script_dir.$script_name.' '.$args.'"|/usr/bin/at "'.$time.'" 2>&1';
exec($cmd, $output , $exit_code);
When I run this command from script it adds the job to the schelude. This I see by the line in logs job 103 at Thu Sep 3 15:08:00 2015 (same text contains $output). But then nothing happens in specified time like at ignores the job. And there are no error messages in logs.
When I run same command with same args from command line on server it scheludes the job and than runs it at specified time.
I found out that when I try to plan a job via php script it runs under apache user. I tried to run next in command line on server:
sudo -u apache echo "/usr/bin/php /var/www/pant/data/www/pant.com/scripts/Run.php firstarg secondarg "|/usr/bin/at "16:00 03.09.2015"
It works correct too. I checked sudoers and have added apache user with NOPASSWD privileges. Script Run.php has execute rights.
at.deny is empty. at.allow does not exist.
So question is: why 'at' does not run command given via php script (exec) but runs same command in command line? How to run it?
Thanks to all.
I found by chance answer at stackexchange.com:
The "problem" is typically PHP is intended to run as module in a webserver. You may need to install the commandline version of php before you can run php scripts from the commandline

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.

Can't open gnome-terminal with php

I actually try to laucnh a gnome-term with a php script, seems i have some problems with the users www-data;
my script make only a ls -l command in a directory (is just for a test) and i run it with a php page in my local-web site.
here the gnome-terminal command in my bash script (he run perfectly when i double-click on him) :
gnome-terminal --working-directory=/opt/cuckoo -x bash -c "ls -l"
and here is the call on the php-page :
system("/my/path/to/the/script/script.sh");
i have some echo in my script and i see them in the php page after i try to run the script with the php.page.
i think www-data don't have the right to do so i give the ownership of the script with the chown command, and at last a try the sudo visudo command and make the script execute like the user www-data is root (with NO PASSWD arg)
But i can't open the terminal and make a ls at last, i try with exec too, and show the result with $ouput butthe result is the same as well.
At last my question is : Php can really run a terminal or maybe a fool myself^^? Thanks for taking time to rescure me ;)
PHP can run everything, but depends who spawns it. Forget just running X apps from a web server - you'll need more than just executing them (permissions, DISPLAY and Xauth settings). Read more about the X clients and architecture.
Probably the right place to ask this is at SuperUser, since the problem is not in the coding itself.

php command line exec() multiple execution and directories?

I am trying to Execute a multiple commands in php using exec() and shell_exec but i am getting a null value back which i shouldn't and nothing is happening (if i copy and paste the strings below in the command line it will work fine and accomplish the job needed) this is the commands i am using:
$command = "cd /../Desktop/FolderName;";
$command .= 'export PATH=$PATH:`pwd`;';
$command .= 'Here i execute a compiler;';
and then i use the escapeshellcmd()
$escaped_command = escapeshellcmd($command);
then
shell_exec($escaped_command);
any ideas what i am doing wrong and i also tried escapeshellarg() instead of escapeshellcmd()?
Solution: the Problem was the permission of the execution compiler for other owners is non and this was the problem.
because when you are using exec() function in php the owner of the file will be www-data so you need to give permission for the www-data either from the ACL of ubuntu or whatever linux based operating system(you can know the owner by doing this exec('whoami')), or by the files you need to execute.
(Sorry my bad English)
On Linux you can add your Commands in a Shell Script.
You can put this in any file:
#!/bin/bash
cd /../Desktop/FolderName
export PATH=$PATH:`pwd`
EXECUTE COMPILER
And save this as fille.sh
Then, add execution permissions:
chmod +x path/to/file.sh
From PHP, you can call this Script executing:
shell_exec('sh path/to/file.sh');
Hope this helps!

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