I am completely new to the shell scripting. Please help me out in below requirement. Thanks
Let us think a,b,c,d are the part of group z.
I want to write a script which runs every day particular time (example 8:00 AM)
The script should look if the user is a part of group z, if he is a part of group z then run particular command (example whoami)
Please help out
Try this:
G='cdrom';
if [ `groups | grep cdrom | wc -l` -eq 1 ];
then whoami;
else echo "Not in group $G";
fi
This should print the user's username with "whoami" if the user is in the group "cdrom". If they're not in the group "cdrom" then it should print "Not in group cdrom". Then you can set it up as a cronjob by editing your crontab like 0 8 * * * /path/to/script with the command crontab -e which will make the program run at 8 am every day.
Please note that this has very basic functionality just so that you can get the idea. You'll have the edit the script to really do anything useful.
Related
I am a beginner in bash scripting and I have created a bash script to add users and remove users on Linux.
But since I am facing some issues with the script not really major issues but would be helpful if anyone could point me how to improve the script and the worst practice I am doing the script would be helpful
however the problem I have noticed is that the script takes -a to add a user -d to remove user and -h to get help the -a flag as 2 optional arguments -p for password and -s for shell so the command would be
./useradd.sh -a user -p password -s shell
this works as expected and the user is added to the system but the problem I am facing is that if I do not enter -a flag and specify the -s and -p flag the script is just exited I want to show a clear idea to the user why it exited and there is so many such errors I am assuming but I have not tested it out so much any help would be appreciated, so here is my script
https://github.com/abhijith7025/bash_scripts/blob/master/useradd.sh
... I have created a bash script to add users and remove users on Linux.
You might want to reconsider this, given that Linux has separate commands for these two operations.
There's a lot more to creating a user than there is is getting rid of one, so you may be asking for trouble trying to conjoin the two.
Anyway:
Your case statement has no "other" branch ("* )").
This may be why you're getting no errors when "misusing" your script.
case $opt in
a )
;;
d )
;;
h )
;;
* )
usage
exit 1
;;
esac
Other things to look out for:
useradd is, perhaps, a poor name for a script that can delete users.
You allow a shell to be specified as an argument, but you don't check to see if that exists.
It's conventional for Linux commands to operate on a "thing" or list of "things" and for that operation to be qualified by options that [usually] precede the thing(s). Thus, your script might be better invoked like this:
./manage_user -a -p password -s shell_exe user1
| | | |
| | | User name
| | Shell for User
| Password for User
Add a User
./manage_user -d user2
| |
| User name
Delete a User
I am trying to run a bash script on start up. The aim of the script is to play an mp3 file with the music player mplayer, at a set time (say at 15 past every hour).
My bash script is the following:
#!/bin/sh
while :
do
S=$(date)
T=${S:14:2} #this gives me the minute column of the current time
if [ $T -eq 15 ]
then
mplayer path_to_mp3_file
fi
done
When I run this bash file from the terminal it works absolutely fine.
However, when I restart my Linux computer with exactly the same script, it fails to work(this is also the case with other music players such as vlc). The script also works when it only contains the mplayer file_path command.
I have tried setting a standard program to open the script, the gnome-terminal.wrapper.
The way in which I was able to set the program as a start up application is in menu>Preferences>Startup applications and add the file to the already existing applications.
There are 2 parts to your problem. 1. play the mp3, 2. (presumably only when the user is logged in). For part 1, use a crontab and have the cron daemon run the script at 15 past the hour. You create your crontab file with crontab -e (to edit your crontab). The format of a crontab entry is:
* * * * * command_to_execute
| | | | |
| | | | +- day of week (0-6) (Sunday = 0)
| | | +--- month (1-12)
| | +----- day of month (1-31)
| +------- hour (0-23)
+--------- minute (0-59)
In your case you would want:
15 * * * * playmymp3.sh
Part 2 In your script you will want to check whether the user is logged in. To only play the music when you are logged in, test that you are logged in with users. Something like this in playmymp3.sh should work:
if grep -q "yourlogin" < <( users ); then
#play mp3 file
fi
Give it a try and let me know.
Try with:
T=$(date +%M)
instead of:
S=$(date)
T=${S:14:2} #this gives me the minute column of the current time
(note: your script enters in a very fast loop when no sound must be played, I suggest you include a sleep command).
Apologies on my part.
When setting the script as a startup file, I had not noticed that I had to provide an actual command, bash script_path, instead of solely the file path.
With this, my problem seems to be solved.
I have a quesiton that puzzles me and I wonder if anyone has attempted to achieve the following:
Let's assume that this is the result of my 'last' command in a Linux environment:
root pts/1 192.168.1.10 Wed Feb 10 07:04 - 07:57 (00:52)
root pts/2 Tue Feb 9 22:00 - 00:13 (02:13)
How can I setup a particular action (say for example a modified MOTD or sending an email) if the the 'root' user has logged in from 192.168.1.10. Is there a way of capturing this information?
The second part of this question is that how can I make the above check a bit more robust - i.e. if I have the following:
mary pts/1 192.168.1.10 Wed Feb 10 07:04 - 07:57 (00:52)
bob pts/2 Tue Feb 9 22:00 - 00:13 (02:13)
Now I'd like to perform an action if the username is equal to 'mary' and the host is 192.168.1.10.
Any suggestions are welcomed.
Thank you in advance.
There's a special file /etc/ssh/sshrc where you can put some commands that will runs each time someone connect by ssh. I wrote that for you :
#!/bin/bash
mail=user#domain.tld
monitored_user=root
monitored_ip=x.x.x.x
hostname=$(hostname)
# add a welcome message:
printf >&2 "\nWelcome on $hostname $USER\n"
read -d " " ip <<< $SSH_CONNECTION
[[ $ip == $monitored_ip && $USER == $monitored_user ]] || exit 0
date=$(date "+%d.%m.%Y %Hh%M")
reverse=$(dig -x $ip +short)
mail -s "Connexion of $USER on $hostname" $mail <<EOF
IP: $ip
Reverse: $reverse
Date: $date
EOF
Put this script in a file, then put the full path of the script in /etc/ssh/sshrc
In man ssh :
/etc/ssh/sshrc :
Commands in this file are executed by ssh when the user
logs in, just before the user's shell (or command) is started. See the
sshd(8) manual page for more information.
Thanks for all your replies. Eventually I managed to find a solution which does work for the time being but it does have one flaw which I'll point out in a minute.
I have added the following to my /etc/bashrc file (or /etc/bash.bashrc whatever environment you're using):
HOST="192.168.0.1"
RHOST=`who am i | sed -n 's/.*(\([^) ]*\).*/\1/p; 1q'`
if [ "$RHOST" == "$HOST" ]; then
echo "SAY WHAT!"
#add further actions here if needed
fi
The flaw that I was talking about before may actually not be a flaw. If you're already SSH-ed into the system, and you want to SSH to a host which lives on the same IP, say ssh root#your-host who am i would then print 'your-host' but I think that's the way it should be.
Needless to say that the above sed statement can be modified so you can capture the username as well, and you can extend the if/else statement to suite your needs.
Thank you again for all your replies.
You can add something to /etc/profile or equivalent that does something depending on the value of $SSH_CLIENT.
It looks like you are using last because it reads /var/log/wtmp by default which is a record of logins. The who command also allows you to read the same file but with an interface more to your needs.
For example:
$ who --ips /var/log/wtmp | grep '^msw.*127.0.0.1'
msw pts/2 2012-10-07 15:52 127.0.0.1
msw pts/3 2012-10-07 15:55 127.0.0.1
where neither of those sessions were active, but rather historic and logged.
In ubuntu i put a script in
/etc/profile.d
and when someone(user ssh) log in, it send an email to my mail
#/etc/profile.d/run_on_loggin.sh
echo $(who i am) | mail -s 'SSH Login Notification' mymail#hotmail.com
I want to create a php file with smtp, to send email with my mail to me...
some times hotmail saved in spam...
if i have the php file i will run like this...
if i want to pass var to file php run like this...
excuse my english :3
note: i think this command run from user, be carefully if the user doen't has permission to use some command or send email.
One way would be to run a simple script periodically:
#!/bin/bash
users=$(last | sed -ne '/192\.168\.1\.10/ s/\([^ ]*\).*/\1/p')
for user in $users; do
sendmail "$user" < email.txt
done
This would pipe the last command into sed to extract a user list and save it into the variable $users. The sed command uses the -n flag so it only prints what we tell it to. First, we select lines that contain the specified IP, with the /192\.168\.1\.10/ "address". On those lines, we attempt to extract the characters before a space, and if we succeed we print the result.
Then, we can loop through the $users variable and act accordingly.
One way to call this repeatedly would be through cron, and a simpler way would be to do while true; do ./my_script.bash; sleep 60; done.
I want to create a script that, after knowing that my machine has been turned on for at least 7h, it does something.
Is this possible? Is there a system variable or something like that that shows me the last time the machine was turned on?
The following command placed in /etc/rc.local:
echo 'touch /tmp/test' | at -t $(date -d "+7 hours" +%m%d%H%M)
will create a job that will run a touch /tmp/test in seven hours.
To protect against frequent reboots and prevent adding multiple jobs you could use one at queue exclusively for this type of jobs (e.g. c queue). Adding -q c to the list of at parameters will place the job in the c queue. Before adding new job you can delete all jobs from c queue:
for job in $(atq -q c | sed 's/[ \t].*//'); do atrm $job; done
You can parse the output of uptime I suppose.
As Pavel and thkala point out below, this is not a robust solution. See their comments!
The uptime command shows you how long the system has been running.
To accomplish your task, you can make a script that first does sleep 25200 (25200 seconds = 7 hours), and then does something useful. Have this script run at startup, for example by adding it to /etc/rc.local. This is a better idea than polling the uptime command to see if the machine has been up for 7 hours (which is comparable to a kid in the backseat of a car asking "are we there yet?" :-))
Just wait for uptime to equal seven hours.
http://linux.die.net/man/1/uptime
I don't know if this is what you are looking for, but uptime command will give you for how many computer was running since last reboot.
$ cut -d ' ' -f 1 </proc/uptime
This will give you the current system uptime in seconds, in floating point format.
The following could be used in a bash script:
if [[ "$(cut -d . -f 1 </proc/uptime)" -gt "$(($HOURS * 3600))" ]]; then
...
fi
Add the following to your crontab:
#reboot sleep 7h; /path/to/job
Either /etc/crontab, /etc/cron.d/, or your users crontab, depending on whether you want to run it as root or the user -- don't forget to put "root" after "#reboot" if you put it in /etc/crontab or cron.d
This has the benefit that if you reboot multiple times, the jobs get cancelled at shut down, so you won't get a bunch of them stacking up if you reboot several times within 7 hours. The "#reboot" time specification triggers the job to be run once when the system is rebooted. "sleep 7h;" waits for 7 hours before running "/path/to/job".
I'm using a basic shell script to log the results of top, netstat, ps and free every minute.
This is the script:
/scripts/logtop:
TERM=vt100
export TERM
time=$(date)
min=${time:14:2}
top -b -n 1 > /var/log/systemCheckLogs/$min
netstat -an >> /var/log/systemCheckLogs/$min
ps aux >> /var/log/systemCheckLogs/$min
free >> /var/log/systemCheckLogs/$min
echo "Message Content: $min" | mail -s "Ran System Check script" email#domain.com
exit 0
When I run this script directly it works fine. It creates the files and puts them in /var/log/systemCheckLogs/ and then sends me an email.
I can't, however, get it to work when trying to get cron to do it every minute.
I tried putting it in /var/spool/cron/root like so:
* * * * * /scripts/logtop > /dev/null 2>&1
and it never executes
I also tried putting it in /var/spool/cron/myservername and also like so:
* * * * * /scripts/logtop > /dev/null 2>&1
it'll run every minute, but nothing gets created in systemCheckLogs.
Is there a reason it works when I run it but not when cron runs it?
Also, here's what the permissions look like:
-rwxrwxrwx 1 root root 326 Jul 21 01:53 logtop
drwxr-xr-x 2 root root 4096 Jul 21 01:51 systemCheckLogs
Normally crontabs are kept in "/var/spool/cron/crontabs/". Also, normally, you update it with the crontab command as this HUPs crond after you're done and it'll make sure the file gets in the correct place.
Are you using the crontab command to create the cron entry? crontab to import a file directly. crontab -e to edit the current crontab with $EDITOR.
All jobs run by cron need the interpreter listed at the top, so cron knows how to run them.
I can't tell if you just omitted that line or if it is not in your script.
For example,
#!/bin/bash
echo "Test cron jon"
When running from /var/spool/cron/root, it may be failing because cron is not configured to run for root. On linux, root cron jobs are typically run from /etc/crontab rather than from /var/spool/cron.
When running from /var/spool/cron/myservername, you probably have a permissions problem. Don't redirect the error to /dev/null -- capture them and examine.
Something else to be aware of, cron doesn't initialize the full run environment, which can sometimes mean you can run it just fine from a fully logged-in shell, but it doesn't behave the same from cron.
In the case of above, you don't have a "#!/bin/shell" up top in your script. If root is configured to use something like a regular bourne shell or cshell, the syntax you use to populate your variables will not work. This would explain why it would run, but not populate your files. So if you need it to be ksh, "#!/bin/ksh". It's generally best not to trust the environment to keep these things sane. If you need your profile run the do a ". ~/.profile" up front as well. Or a quick and dirty way to get your relatively full env is to do it from su as such "* * * * * su - root -c "/path/to/script" > /dev/null 2>&1
Just some things I've picked up over the years. You're definitely expecting a ksh based on your syntax, so you might want to be sure it's using it.
Thanks for the tips... used a little bit of each answer to get to the bottom of this.
I did have the interpreter at the top (wasn't shown here), but may have been wrong.
Am using #!/bin/bash now and that works.
Also had to tinker with the permissions of the directory the log files are being dumped in to get things working.