I am trying to execute the following command through PHP using shell_exec() function:
sudo -u USERNAME -pPASSWORD echo "SOME-STRING" > /dev/tcp/127.0.0.1/64444
but nothing happens (if I copy/paste it manually to the terminal it works). There is also nothing in the $output of the shell_exec() function.
So with that command I am trying to send a string over TCP through a different user with sudo privileges.
EDIT #1: The full PHP command looks like this:
shell_exec('sudo -u USERNAME -pPASSWORD echo "RLSET|79-192.168.1.33-0-1" > /dev/tcp/127.0.0.1/64444');
EDIT #2: I've also checked the phpinfo() function and safe_mode is off also the shell_exec() function is not on the disallowed functions list.
EDIT #4: My PHP PATH variable contains the following locations:
/usr/local/bin:/usr/bin:/bin
I'm lost here, any and all help would be appreciated.
Scape your quotes put your full command path:
shell_exec("sudo -u USERNAME -pPASSWORD echo \"FULL-PATH-COMMAND\" > /dev/tcp/127.0.0.1/64444")
maybe your binary (sudo, echo, ...) is not in the PATH env variable :
putenv('PATH', '/bin:/sbin:/usr/bin:/usr/sbin:') // etc...
Have you tried putting a space after -p?
So that should be
shell_exec('sudo -u USERNAME -p PASSWORD echo "RLSET|79-192.168.1.33-0-1" > /dev/tcp/127.0.0.1/64444');
I just tried some commands here on my own machine.
I tried
sudo -u USERNAME -p PASSWORD ls
and it works but when I tried
sudo -u USERNAME -pPASSWORD ls
It give me the usage information.
Related
Referencing this post as well(How to create database using php shell_exec and sql command line)
I've followed the accepted answer in that post, but I am curious as to why the shell_exec is unable to run mysql code. It merely returns NULL.
The user already has sudo access, when i copy the following command line into the console log manually it works.
$cmd = escapeshellcmd('sudo mysql -u root -e "create database somedb"');
$test = shell_exec($cmd);
var_dump($test);
Edit 1:
Updated [username], if root may cause some issues
You will need to use the full path to mysql
Type whereis mysql to find it.
It will most likely be /usr/bin/mysql
Change your command to:
$cmd = escapeshellcmd('sudo /usr/bin/mysql -u [username] -e "create database somedb"');
You may need full path to sudo .. same principles will apply.
I am trying to execute this command with php:
system ('mysqldump -u myUser myDbname | mysql -u myUser -A myDbBackupName');
This does not return a error, but does nothing.
The same command executed in server by command line works perfectly.
I am using .my.cnf and i configured the user to mysql, mysqldump and client.
I don't know what is happening. Can somebody help me?
I solved this issue. Put this 2>&1 in the end of the command to force return the output of method exec() or system() to facilit the debug.
The correct mysql command is without space between -p[password]. In fact the password is necessery, independent if you using the archive .my.cnf
Like this:
mysqldump -u user -ppassword myDbName | mysql -u user -ppassword myDbBackupName
I am trying to execute
exec('mysqldump -u user -pPASSWORD db1 tbl1 > /etc/table1.sql' ,$output, $return_var);
and also I have tried,
exec('mysqldump -u user -pPASSWORD db1 tbl1 > /etc/table1.sql');
but the $output is empty, and $return_var is 2
I have change apache user's permissions in file /etc/sudoers
#User privilege specification
www-data ALL=(ALL:ALL) NOPASSWD:ALL
Also the following is running on cmd
mysqldump -u user -pPASSWORD db1 tbl1 > /etc/table1.sql
Can someone help me?
I would rather post this as a comment rather than an answer, but I suggest using popen as it might provide more helpful information why your command fails.
Full function signature:
popen (string $command, string $mode)
It works sorta like fopen, including the modes. In your case you will probably want to use 'r' as your mode. The function returns a handle like fopen would do, so you can fread it for the data. Don't forget to pclose the hande after you're done!
To capture any error output (the function returns the handle even if command fails), add 2>&1 (leading space included) after your command. You can probably remove that afterwards, all it does is redirecting errors to the output.
Source:
http://php.net/manual/en/function.popen.php
I'm trying to run a mysqldump from my php file using exec(). File is located in my website's www directory and I'm accessing it from browser. But, no matter what I try, I'm simply not getting ANY results and no sql file is created.
Here's my command:
$export = exec("/usr/bin/mysqldump -u MY_USERNAME -pMY_PASS DATABASE_NAME products > /path/to/www/directory/sync/products.sql");
exec() is enabled. sync directory permissions are 777.
If I use the same command, but add $output, $return_var like this:
$export = exec("/usr/bin/mysqldump -u MY_USERNAME -pMY_PASS DATABASE_NAME products > /path/to/www/directory/sync/products.sql", $output, $return_var);
$output is an empty array and $return_var echoes 127.
Path to mysqldump is correct:
root#server [~]# which mysqldump
/usr/bin/mysqldump
root#server [~]#
If I run the command from console, it works just fine and the new file gets created in my sync directory.
Can someone please tell me what I'm doing wrong and what I'm missing here? The script used to work just fine until we moved to a new server...
I've been searching all over the place and have stumbled upon something that turned out to solve my problem.
Instead of using
$export = exec("/usr/bin/mysqldump -u MY_USERNAME -pMY_PASS DATABASE_NAME products > /path/to/www/directory/sync/products.sql");
I am now using this and THIS WORKS:
$export = exec('mysqldump -u MY_USERNAME -pMY_PASS DATABASE_NAME products -r "/path/to/www/directory/sync/products.sql"');
Using path to mysqldump doesn't make a difference. Using options (in this case -r) solved it for me, although I don't really understand why the original solution didn't work.
Maybe worth mentioning - I noticed that I can't import a database using the standard mysql -u -p db < table.sql - running that command just shows me help instructions, as if I ran mysql -?. I'm not sure if this is specific to me and my server, but maybe the solution above helps someone in a similar situation.
I have a string of script which working in terminal but does not work when I use it in PHP with shell_exec().I know a lot of questions similar to this question has been asked already but in my case the problem I am facing is that I have already tried the proposed solutions I found. Below is my simple code.
<?php
$output = shell_exec('mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql');
echo "<pre>$output</pre>";
?>
So far this is the best solution I have found.
Does anyone knows what might be wrong?
Your shell_exec is this:
$output = shell_exec('mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql');
And your command is this:
mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql
The reason that command works when you are in the shell is the binary path to mysql is part of your user login profile.
To see what I mean, login to the shell as yourself and then type echo $PATH and what you will see is a list of search paths the shell uses to figure out where binaries you are attempting to run are located.
But when you attempt to run a script via shell_exec() the Apache server user running PHP is making the sell call. And that user typically does not have binary paths set. So you need to provide the full path to mysql which might be:
/usr/bin/mysql
Or:
/usr/local/bin/mysql
The best solution is from the shell use the which command like so:
which mysql
And then take the full path provided and adjust your shell_exec() command as follows; using /usr/bin/ for example:
$output = shell_exec('/usr/bin/mysql -u root -pmypass -h 127.0.0.1 mydatabase < db.sql');
Also, where is db.sql actually located? You would have to prepend the full path to that MySQL script like this as well; using /full/path/to/this/ for example:
$output = shell_exec('/usr/bin/mysql -u root -pmypass -h 127.0.0.1 my database < /full/path/to/this/db.sql');