How to set the process name of a shell script? - shell

Is there any way to set the process name of a shell script? This is needed for killing this script with the killall command.

Here's a way to do it, it is a hack/workaround but it works pretty good. Feel free to tweak it to your needs, it certainly needs some checks on the symbolic link creation or using a tmp folder to avoid possible race conditions (if they are problematic in your case).
Demonstration
wrapper
#!/bin/bash
script="./dummy"
newname="./killme"
rm -iv "$newname"
ln -s "$script" "$newname"
exec "$newname" "$#"
dummy
#!/bin/bash
echo "I am $0"
echo "my params: $#"
ps aux | grep bash
echo "sleeping 10s... Kill me!"
sleep 10
Test it using:
chmod +x dummy wrapper
./wrapper some params
In another terminal, kill it using:
killall killme
Notes
Make sure you can write in your current folder (current working directory).
If your current command is:
/path/to/file -q --params somefile1 somefile2
Set the script variable in wrapper to /path/to/file (instead of ./dummy) and call wrapper like this:
./wrapper -q --params somefile1 somefile2

You can use the kill command on a PID so what you can do is run something in the background, get its ID and kill it
PID of last job run in background can be obtained using $!.
echo test & echo $!

You cannot do this reliably and portably, as far as I know. On some flavors of Unix, changing what's in argv[0] will do the job. I don't believe there's a way to do that in most shells, though.
Here are some references on the topic.
Howto change a UNIX process and child process name by modifying argv0
Is there a way to change the effective process name in Python?

This is an extremely old post. Pretty sure the original poster got his/her answer long ago. But for newcomers, thought I'd explain my own experience (after playing with bash for a half hour). If you start a script by script name w/ something like:
./script.sh
the process name listed by ps will be "bash" (on my system). However if you start a script by calling bash directly:
/bin/bash script.sh
/bin/sh script.sh
bash script.sh
you will end up with a process name that contains the name of the script. e.g.:
/bin/bash script.sh
results in a process name of the same name. This can be used to mark pids with a specific script name. And, this can be useful to (for example) use the kill command to stop all processes (by pid) that have a process name containing said script name.

You can all use the -f flag to pgrep/pkill which will search the entire command line rather than just the process name. E.g.
./script &
pkill -f script

Include
#![path to shell]
Example for path to shell -
/usr/bin/bash
/bin/bash
/bin/sh
Full example
#!/usr/bin/bash

On Linux at least, killall dvb works even though dvb is a shell script labelled with #!. The only trick is to make the script executable and invoke it by name, e.g.,
dvb watch abc write game7 from 9pm for 3:30
Running ps shows a process named
/usr/bin/lua5.1 dvb watch ...
but killall dvb takes it down.

%1, %2... also do an adequate job:
#!/bin/bash
# set -ex
sleep 101 &
FIRSTPID=$!
sleep 102 &
SECONDPID=$!
echo $(ps ax|grep "^\(${FIRSTPID}\|${SECONDPID}\) ")
kill %2
echo $(ps ax|grep "^\(${FIRSTPID}\|${SECONDPID}\) ")
sleep 1
kill %1
echo $(ps ax|grep "^\(${FIRSTPID}\|${SECONDPID}\) ")

I put these two lines at the start of my scripts so I do not have to retype the script name each time I revise the script. It won't take $0 of you put it after the first shebang. Maybe someone who actually knows can correct me but I believe this is because the script hasn't started until the second line so $0 doesn't exist until then:
#!/bin/bash
#!/bin/bash ./$0
This should do it.

My solution uses a trivial python script, and the setproctitle package. For what it's worth:
#!/usr/bin/env python3
from sys import argv
from setproctitle import setproctitle
from subprocess import run
setproctitle(argv[1])
run(argv[2:])
Call it e.g. run-with-title and stick it in your path somewhere. Then use via
run-with-title <desired-title> <script-name> [<arg>...]

Run bash script with explicit call to bash (not just like ./test.sh). Process name will contain script in this case and can be found by script name. Or by explicit call to bash with full path as
suggested in display_name_11011's answer:
bash test.sh # explicit bash mentioning
/bin/bash test.sh # or with full path to bash
ps aux | grep test.sh | grep -v grep # searching PID by script name
If the first line in script (test.sh) explicitly specifies interpreter:
#!/bin/bash
echo 'test script'
then it can be called without explicit bash mentioning to create process with name '/bin/bash test.sh':
./test.sh
ps aux | grep test.sh | grep -v grep
Also as dirty workaround it is possible to copy and use bash with custom name:
sudo cp /usr/bin/bash /usr/bin/bash_with_other_name
/usr/bin/bash_with_other_name test.sh
ps aux | grep bash_with_other_name | grep -v grep

Erm... unless I'm misunderstanding the question, the name of a shell script is whatever you've named the file. If your script is named foo then killall foo will kill it.

We won't be able to find pid of the shell script using "ps -ef | grep {scriptName}" unless the name of script is overridden using shebang. Although all the running shell scripts come in response of "ps -ef | grep bash". But this will become trickier to identify the running process as there will be multiple bash processing running simultaneously.
So a better approach is to give an appropriate name to the shell script.
Edit the shell script file and use shebang (the very first line) to name the process e.g. #!/bin/bash /scriptName.sh
In this way we would be able to grep the process id of scriptName using
"ps -ef | grep {scriptName}"

Related

How can i find the process name by the shell script that invokes it?

Is there a way, I can find the process name of bash script by the shell script that was used to invoke it? or can I set process name of bash script to something such as
-myprocess
(I have looked into argv[0], but I am not clear about it)
so when I use
ps -ef | grep -c '[M]yprocess'
I get only all the instances of myprocess?
To obtain the command name of the current script that is running, use:
ps -q $$ -o comm=
To get process information on all running scripts that have the same name as the current script, use:
ps -C "$(ps -q $$ -o comm=)"
To find just the process IDs of all scripts currently being run that have the same name as the current script, use:
pgrep "$(ps -q $$ -o comm=)"
How it works
$$ is the process ID of the script that is being run.
The option -q $$ tells ps to report on only process ID $$.
The option -o comm= tells ps to omit headers and to skip the usual output and instead print just the command name.
The parent process id can be obtained from $PPID on bash and ksh. We can read the fields from ps into an array.
For bash you can try this. The problem with ps is that many options are non-standard, so I have kept that as generic as possible:
#!/bin/bash
while read -a fields
do
if [[ ${fields[0]} == $PPID ]]
then
echo "Shell: ${fields[3]}"
echo "Command: ${fields[4]}"
fi
done < <(ps -p $PPID)
You have tagged bash and ksh, but they have different syntax rules. To read into an array bash uses -a but ksh uses -A, So for korn shell you would need to change the read line (and the #! line):
while read -A fields
Not sure what you mean by that, but let's go with an example script b.sh.
#!/usr/local/bin/bash
echo "My name is $0, and I am running under $SHELL as the shell."
Running the script will give you:
$ bash b.sh
My name is b.sh, and I am running under /usr/local/bin/bash as the shell.
For more check this answer: HOWTO: Detect bash from shell script

pgrep command doesn't show bash script

When I run a simple bash script, say myscript.sh
#!/bin/bash
sleep 30
from the terminal, and then do pgrep myscript.sh I don't get any result. Why?
You're probably doing this:
pgrep myscript.sh
This won't show the process you're running because it is /bin/bash that is running your script.
You should be doing:
pgrep -fl myscript.sh
To list your process.
As per man pgrep:
-f Match the pattern anywhere in the full argument string of the process instead of just the executable name.
Your just running your bash script, u need to use -f flag, please check man page
Check pgrep is installed in your machine. Just do man pgrep, if you get command not found that install the utility.
pgrep looks through the currently running processes and lists the process IDs which matches the selection criteria to stdout. All the criteria have to match.
Example usage:
pgrep name | xargs kill
If you use pgrep name | kill, the ouput of pgrep name is feed to stdin of kill. Because kill does not read arguments from stdin, so this will not work.
Using xargs, it will build arguments for kill from stdin. Example:
$ pgrep bash | xargs echo
5514 22298 23079

Linux crontab doesnt launch a script

I have this user crontab (accessed via the command crontab -e):
# m h dom mon dow command
*/3 * * * * sh /home/FRAPS/Desktop/cronCheck.sh
The script cronCheck.sh looks like that:
#!/bin/sh
SERVICE='Script'
if ps ax | grep -v grep | grep -i "$SERVICE" > /dev/null
then
echo "######## $SERVICE service running, everything is fine ##################\n" >> CronReport.txt
else
echo "$SERVICE is not running. Launching it now\n" >> CronReport.txt
perl Script.pl
fi
When I launch the script (cronCheck.sh) from its own directory, it works like a charm, but when cron launches it, it always "# $SERVICE service running, everything is fine ###"
despite 'Script' is not running.
Thanks,
Here's an even better way to write that conditional:
services=$(ps -e -o comm | grep -cFi "$SERVICE")
case "$services" in
(0)
# restart service
;;
(1)
# everything is fine
;;
(*)
# more than one copy is running
;;
esac
By using ps -e -o comm you avoid having to do the silly grep -v grep thing, because only the actual process name appears in the ps output, not the arguments. And grep -cFi counts up the matches and gives you a number, so you don't have to deal with the exit status of a pipeline.
Also, as other posters have implied, you should lead off this script by setting the PATH variable.
PATH=/bin:/usr/bin:/sbin:/usr/sbin
export PATH
You might or might not want to put /usr/local/bin at the beginning of that list, depending on your system. Don't do it if you don't need anything from there.
Final piece of advice: When writing scripts that will execute without user supervision (such as cron jobs), it's a good idea to put set -e at the beginning. That makes them exit unsuccessfully if any command fails.
You need to put the grep -v grep after the grep -i "$SERVICE". The way you have it now it's guaranteed to be true.
Checking the return status of a pipe like that could be problematic. You should either check the $PIPESTATUS array, or you can pipe the final grep into wc -l to count the number of lines.
cron typically does not set up a lot of the environment like a user account does. You may need to modify your script to get things setup properly.
Cron jobs don't get the same environment settings that you get at a shell prompt - those are generally set up by your shell on login - so you want to use absolute rather than relative paths throughout. (i.e. don't assume the PATH environment variable will exist or be set up the same as it is for you at a shell prompt, and don't assume the script will run with PWD set to your home directory, etc.) So:
in your crontab entry replace sh with /bin/sh (or remove it if cronCheck.sh is executable, the shebang line will do).
in cronCheck.sh add paths to the log file and the perl script.
cronCheck.sh should end up looking something like:
#!/bin/sh
SERVICE='Script'
if ps ax | grep -v grep | grep -i "$SERVICE" > /dev/null
then
echo "######## $SERVICE service running, everything is fine ##################\n" >> CronReport.txt
else
# Specify absolute path to a log file that's writeable for the user the
# cron runs as (probably you). Example: /tmp/CronReport.txt
echo "$SERVICE is not running. Launching it now\n" >> /tmp/CronReport.txt
# Specify absolute path to both perl and the script. Example: /usr/bin/perl
# and /home/FRAPS/scripts/Script.pl
/usr/bin/perl /home/FRAPS/scripts/Script.pl
fi
(Again you can get rid of the /usr/bin/perl bit if Script.pl is executable and has the path to the right perl in the shebang line.)

Determining whether shell script was executed "sourcing" it

Is it possible for a shell script to test whether it was executed through source? That is, for example,
$ source myscript.sh
$ ./myscript.sh
Can myscript.sh distinguish from these different shell environments?
I think, what Sam wants to do may be not possible.
To what degree a half-baken workaround is possible, depends on...
...the default shell of users, and
...which alternative shells they are allowed to use.
If I understand Sam's requirement correctly, he wants to have a 'script',
myscript, that is...
...not directly executable via invoking it by its name myscript
(i.e. that has chmod a-x);
...not indirectly executable for users by invoking sh myscript or
invoking bash myscript
...only running its contained functions and commands if invoked by
sourcing it: . myscript
The first things to consider are these
Invoking a script directly by its name (myscript) requires a first line in
the script like #!/bin/bash or similar. This will directly determine which
installed instance of the bash executable (or symlink) will be invoked to run
the script's content. This will be a new shell process. It requires the
scriptfile itself to have the executable flag set.
Running a script by invoking a shell binary with the script's (path+)name as
an argument (sh myscript), is the same as '1.' -- except that the
executable flag does not need to be set, and said first line with the
hashbang isn't required either. The only thing needed is that the invoking
user needs read access to the scriptfile.
Invoking a script by sourcing its filename (. myscript) is very much the
same as '1.' -- exept that it isn't a new shell that is invoked. All the
script's commands are executed in the current shell, using its environment
(and also "polluting" its environment with any (new) variables it may set or
change. (Usually this is a very dangerous thing to do: but here it could be
used to execute exit $RETURNVALUE under certain conditions....)
For '1.':
Easy to achieve: chmod a-x myscript will prevent myscript from being
directly executable. But this will not fullfill requirements '2.' and '3.'.
For '2.' and '3.':
Much harder to achieve. Invokations by sh myscript require reading
privileges for the file. So an obvious way out would seem to chmod a-r
myscript. However, this will also dis-allow '3.': you will not be able to
source the script either.
So what about writting the script in a way that uses a Bashism? A Bashism is a
specific way to do something which other shells do not understand: using
specific variables, commands etc. This could be used inside the script to
discover this condition and "do something" about it (like "display warning.txt",
"mailto admin" etc.). But there is no way in hell that this will prevent sh or
bash or any other shell from reading and trying to execute all the following
commands/lines written into the script unless you kill the shell by invoking
exit.
Examples: in Bash, the environment seen by the script knows of $BASH,
$BASH_ARGV, $BASH_COMMAND, $BASH_SUBSHELL, BASH_EXECUTION_STRING... . If
invoked by sh (also if sourced inside a sh), the executing shell will see
all these $BASH_* as empty environment variables. Again, this could be used
inside the script to discover this condition and "do something"... but not
prevent the following commands from being invoked!
I'm now assuming that...
...the script is using #!/bin/bash as its first line,
...users have set Bash as their shell and are invoking commands in the
following table from Bash and it is their login shell,
...sh is available and it is a symlink to bash or dash.
This will mean the following invokations are possible, with the listed values
for environment variables
vars+invok's | ./scriptname | sh scriptname | bash scriptname | . scriptname
---------------+--------------+---------------+-----------------+-------------
$0 | ./scriptname | ./scriptname | ./scriptname | -bash
$SHLVL | 2 | 1 | 2 | 1
$SHELLOPTS | braceexpand: | (empty) | braceexpand:.. | braceexpand:
$BASH | /bin/bash | (empty) | /bin/bash | /bin/bash
$BASH_ARGV | (empty) | (empty) | (empty) | scriptname
$BASH_SUBSHELL | 0 | (empty) | 0 | 0
$SHELL | /bin/bash | /bin/bash | /bin/bash | /bin/bash
$OPTARG | (empty) | (empty) | (emtpy) | (emtpy)
Now you could put a logic into your text script:
If $0 is not equal to -bash, then do an exit $SOMERETURNVALUE.
In case the script was called via sh myscript or bash myscript, then it will
exit the calling shell. In case it was run in the current shell, it will
continue to run. (Warning: in case the script has any other exit statements,
your current shell will be 'killed'...)
So put into your non-executable myscript.txt near its beginning something like
this may do something close to your goal:
echo BASH=$BASH
test x${BASH} = x/bin/bash && echo "$? : FINE.... You're using 'bash ...'"
test x${BASH} = x/bin/bash || echo "$? : RATS !!! -- You're not using BASH and I will kick you out!"
test x${BASH} = x/bin/bash || exit 42
test x"${0}" = x"-bash" && echo "$? : FINE.... You've sourced me, and I'm your login shell."
test x"${0}" = x"-bash" || echo "$? : RATS !!! -- You've not sourced me (or I'm not your bash login shell) and I will kick you out!"
test x"${0}" = x"-bash" || exit 33
This may or may not be what the asker wanted but, on a similar situation, I wanted a script to indicate that it is meant to be sourced and not directly run.
To achieve this effect my script reads:
#!/bin/echo Should be run as: source
export SOMEPATH="/some/path/on/my/system"
echo "Your environment has been set up"
So when I run it either as a command or sourced I get:
$ ./myscript.sh
Should be run as: source ./myscript.sh
$ source ./myscript.sh
Your environment has been set up
You can of course fool the script by running it as sh ./myscript.sh, but at least it gives the correct expected behaviour on 2 out of 3 cases.
This is what I was looking for:
[[ ${BASH_SOURCE[0]} = $0 ]] && main "$#"
I cannot add comment yet (stackexchange policies) so I add my own answer:
This one may works regardless if we do:
bash scriptname
scriptname
./scriptname.
on both bash and mksh.
if [ "${0##/*}" == scriptname ] # if the current name is our script
then
echo run
else
echo sourced
fi
If you have a non-altering file path for regular users, then:
if [ "$(/bin/readlink -f "$0")" = "$KNOWN_PATH_OF_THIS_FILE" ]; then
# the file was executed
else
# the file was sourced
fi
(it can also easily be loosened to only check for the filename or whatever).
But your users need to have read permission to be able to source the file, so absolutely nothing can stop them from doing what they want with the file. But it might help them out to not use it in the wrong way.
This solution is not dependent on Bashisms.
Yes it is possible. In general you can do the following:
#! /bin/bash
sourced () {
echo Sourced
}
executed () {
echo Executed
}
if [[ ${0##*/} == -* ]]; then
sourced
else
executed $#
fi
Giving the following output:
$ ./myscript
Executed
$ . ./myscript
Sourced
Based on Kurt Pfeifle’s answer, this works for me
if [ $SHLVL = 1 ]
then
echo 'script was sourced'
fi
Example
Since all of our machines have history, I did this:
check_script_call=$(history |tail -1|grep myscript.sh )
if [ -z "$check_script_call" ];then
echo "This file should be called as a source."
echo "Please, try again this way:"
echo "$ source /path/to/myscript.sh"
exit 1
fi
Everytime you run a script (without source), your shell creates a new env without history.
If you want to care about performance you can try this:
if ! history |tail -1|grep set_vars ;then
echo -e "This file should be called as a source.\n"
echo "Please, try again this way:"
echo -e "$ source /path/to/set_vars\n"
exit 1
fi
PS: I think Kurt's answer is much more complete but I think this could help.
In the first case, $0 will be "myscript.sh". In the second case, it will be "./myscript". But, in general, there's no way to tell source was used.
If you tell us what you're trying to do, instead of how you want to do it, a better answer might be forthcoming.

How to change argv0 in bash so command shows up with different name in ps?

In a C program I can write argv[0] and the new name shows up in a ps listing.
How can I do this in bash?
You can do it when running a new program via exec -a <newname>.
Just for the record, even though it does not exactly answer the original poster's question, this is something trivial to do with zsh:
ARGV0=emacs nethack
I've had a chance to go through the source for bash and it does not look like there is any support for writing to argv[0].
I'm assuming you've got a shell script that you wish to execute such that the script process itself has a new argv[0]. For example (I've only tested this in bash, so i'm using that, but this may work elsewhere).
#!/bin/bash
echo "process $$ here, first arg was $1"
ps -p $$
The output will be something like this:
$ ./script arg1
process 70637 here, first arg was arg1
PID TTY TIME CMD
70637 ttys003 0:00.00 /bin/bash ./script arg1
So ps shows the shell, /bin/bash in this case. Now try your interactive shell's exec -a, but in a subshell so you don't blow away the interactive shell:
$ (exec -a MyScript ./script arg1)
process 70936 here, first arg was arg1
PID TTY TIME CMD
70936 ttys008 0:00.00 /bin/bash /path/to/script arg1
Woops, still showing /bin/bash. what happened? The exec -a probably did set argv[0], but then a new instance of bash started because the operating system read #!/bin/bash at the top of your script. Ok, what if we perform the exec'ing inside the script somehow? First, we need some way of detecting whether this is the "first" execution of the script, or the second, execed instance, otherwise the second instance will exec again, and on and on in an infinite loop. Next, we need the executable to not be a file with a #!/bin/bash line at the top, to prevent the OS from changing our desired argv[0]. Here's my attempt:
$ cat ./script
#!/bin/bash
__second_instance="__second_instance_$$"
[[ -z ${!__second_instance} ]] && {
declare -x "__second_instance_$$=true"
exec -a MyScript "$SHELL" "$0" "$#"
}
echo "process $$ here, first arg was $1"
ps -p $$
Thanks to this answer, I first test for the environment variable __second_instance_$$, based on the PID (which does not change through exec) so that it won't collide with other scripts using this technique. If it's empty, I assume this is the first instance, and I export that environment variable, then exec. But, importantly, I do not exec this script, but I exec the shell binary directly, with this script ($0) as an argument, passing along all the other arguments as well ($#). The environment variable is a bit of a hack.
Now the output is this:
$ ./script arg1
process 71143 here, first arg was arg1
PID TTY TIME CMD
71143 ttys008 0:00.01 MyScript ./script arg1
That's almost there. The argv[0] is MyScript like I want, but there's that extra arg ./script in there which is a consequence of executing the shell directly (rather than via the OS's #! processing). Unfortunately, I don't know how to get any better than this.
Update for Bash 5.0
Looks like Bash 5.0 adds support for writing to special variable BASH_ARGV0, so this should become far simpler to accomplish.
(see release announcement)
( exec -a foo bash -c 'echo $0' )
ps and others inspect two things, none of which is argv0: /proc/PID/comm (for the "process name") and /proc/PID/cmdline (for the command-line). Assigning to argv0 will not change what ps shows in the CMD column, but it will change what the process usually sees as its own name (in output messages, for example).
To change the CMD column, write to /proc/PID/comm:
echo -n mynewname >/proc/$$/comm; ps
You cannot write to or modify /proc/PID/cmdline in any way.
Process can set their own "title" by writing to the memory area in which argv & envp are located (note that this is different than setting BASH_ARGV0). This has the side effect of changing /proc/PID/cmdline as well, which is what some daemons do in order to prettify (hide?) their command lines. libbsd's setproctitle() does exactly that, but you cannot do that in Bash without support of external tools.
I will just add that this must be possible at runtime, at least in some environments. Assigning $0 in perl on linux does change what shows up in ps. I do not know how that is implemented, however. If I can find out, i'll update this.
edit:
Based on how perl does it, it is non-trivial. I doubt there is any bask built in way at runtime but don't know for sure. You can see how perl does sets the process name at runtime.
Copy the bash executable to a different name.
You can do this in the script itself...
cp /bin/bash ./new-name
PATH=$PATH:.
exec new-name $0
If you are trying to pretend you are not a shell script you can rename the script itself to something cool or even " " (a single space) so
exec new-name " "
Will execute bash your script and appears in the ps list as just new-name.
OK so calling a script " " is a very bad idea :)
Basically, to change the name
bash script
rename bash and rename the script.
If you are worried, as Mr McDoom. apparently is, about copying a binary to a new name (which is entirely safe) you could also create a symlink
ln -s /bin/bash ./MyFunkyName
./MyFunkyName
This way, the symlink is what appears in the ps list. (again use PATH=$PATH:. if you dont want the ./)

Resources