I have a small shell script as follows that I am using to login to multiple servers to capture whether the target server is using Redhat or Ubuntu as the OS version.
#!/bin/ksh
if [ -f $HOME/osver.report.txt ];then
rm -rf $HOME/osver.report.txt
fi
for x in `cat hostlist`
do
OSVER=$(ssh $USER#${x} "cat /etc/redhat-release 2>/dev/null || grep -i DISTRIB_DESCRIPTION /etc/lsb-release 2>/dev/null")
echo -e "$x \t\t $OSVER" >> osver.report.txt
done
The above script works, however, if I attempt to add in some awk as shown below and the server is a redhat server...my results in the osver.report.txt will only show the hostname and no OS version. I have played around with the quoting, but nothing seems to work.
OSVER=$(ssh $USER#${x} "cat /etc/redhat-release | awk {'print $1,$2,$6,$7'} 2>/dev/null || grep -i DISTRIB_DESCRIPTION /etc/lsb-release 2>/dev/null")
If I change the script as suggested to the following:
#!/bin/bash
if [ -f $HOME/osver.report.txt ];then
rm -rf $HOME/osver.report.txt
fi
for x in cat hostlist
do
OSVER=$(
ssh $USER#${x} bash << 'EOF'
awk '{print "$1,$2,$6,$7"}' /etc/redhat-release 2>/dev/null || grep -i DISTRIB_DESCRIPTION /etc/lsb-release 2>/dev/null
EOF
)
echo -e "$x \t\t $OSVER" >> osver.report.txt
done
Then I get the following errors:
./test.bash: line 9: unexpected EOF while looking for matching `)'
./test.bash: line 16: syntax error: unexpected end of file
You're suffering from a quoting problem. When you pass a quoted command to ssh, you effectively lose one level of quoting (as if you passed the same arguments to sh -c "..."). So the command that you're running on the remote host is actually:
cat /etc/redhat-release | awk '{print ,,,}' | grep -i DISTRIB_DESCRIPTION /etc/lsb-release
One way of resolving this is to pipe your script into a shell, rather than passing it as arguments:
OSVER=$(
ssh $USER#${x} bash <<'EOF'
awk '{print "$1,$2,$6,$7"}' /etc/redhat-release 2>/dev/null ||
grep -i DISTRIB_DESCRIPTION /etc/lsb-release 2>/dev/null
EOF
)
The use of <<'EOF' here inhibits any variable expansion in the here document...without that, expressions like $1 would be expanded locally.
A better solution would be to look into something like ansible which has built-in facilities for sshing to groups of hosts and collecting facts about them, including distribution version information.
I want to set up a teamspeak bot, and I have this script to start this.
#!/bin/bash
if [ $1 = 'stop' ]
then
echo stop >> /root/ts3bot/tmp/log.txt
date >>/root/ts3bot/tmp/log.txt
echo ======================
screen -S bot -X quit
fi
if [ $1 = 'start' ]
then
echo start >> /root/ts3bot/tmp/log.txt
date >> /root/ts3bot/tmp/log.txt
echo ======================
screen -dmS bot php core.php
ps ax | grep -v grep | grep -v -i SCREEN | grep links >> /root/ts3bot/tmp/log.txt
fi
<here is an extra blank line>
but when I type bash bot.sh it says syntax error: unexpected end of file
I don't know what I did wrong :/ the chmod is set on 755
Thanks!
I suspect you may have copied this shell script from a Microsoft Windows box over to a Linux or Unix server. If so, the problem might be that you have DOS/Windows line endings, which can cause unpredictable results in scripts.
To check the script for bad line endings on a Linux or Unix server, you can dump the file (sort of like a hex dump) by typing the following at the shell prompt:
$ od -c bot.sh | less
And look for \n or \r or \r\n. If lines appear to have a \r at the end, then you've found the problem.
To FIX this line-ending problem, you can use a tool like dos2unix if it's installed on your system. If you don't have dos2unix but you're on a Linux server, you may be able to do this instead:
$ sed -i 's/\r//' bot.sh
to convert the file.
Lastly ... see the first line of the script, #!/bin/bash? Because of that, you don't need to run this with bash bot.sh, you can just execute it directly with ./bot.sh.
starting bash with -v option produces a long output to the console
$ bash -v
source ~/Dropbox/bin/tim_functions.sh
\#!/bin/bash
...several hundred more lines
I would like to capture the output to a file to make it easier to browse through, but I have tried bash -v 2>&1 > out_bash.txt and bash -v | tee out_bash.txt and cannot capture the information on the terminal screen within a file. It is as if the verbose output is neither stderr or stdout. How can this be?
Can anyone suggest a way to capture the output of bash -v ?
bash -v 2>&1 > out_bash.txt
is not what you want, it should be
bash -v >out_bash.txt 2>&1
I poked around and found this http://www.commandlinefu.com/commands/view/3310/run-a-bash-script-in-debug-mode-show-output-and-save-it-on-a-file
On the website they use
bash -x test.sh 2>&1 | tee out.test, but I tested it with
bash -v test.sh 2>&1 | tee out.test and it worked fine.
you can also use the exec command in the script to redirect all output:
#!/bin/bash
exec >> out.txt 2>> out.txt
set -x
set -v
echo "testing debug of shell scripts"
ls
After reading other helpful answers, I believe this issue has to do with how bash is sending the verbose information to tty--which is somehow different than stderr or stdout. It can be caught with the following work around:
$ screen -L
$ bash -v
$ exit #from the bash session
$ exit #from the screen session
This results in a screenlog.0 file being generated containing the output.
The bash -v output of interest was on a mac running 10.7.3 (Lion) with
$ bash --version
GNU bash, version 3.2.48(1)-release (x86_64-apple-darwin11)
Copyright (C) 2007 Free Software Foundation, Inc.)
Another 10.6.8 mac I tried had a less (interesting/verbose) output, despite a similar .bashrc file.
You can use,
bash -v 2>&1 | tee file.txt
or
bash -v 2>&1 | grep search_string
Have you tried wrapping your child bash in a subshell?
( bash -v ) 2>&1 > out_bash.txt
How can I determine the current shell I am working on?
Would the output of the ps command alone be sufficient?
How can this be done in different flavors of Unix?
There are three approaches to finding the name of the current shell's executable:
Please note that all three approaches can be fooled if the executable of the shell is /bin/sh, but it's really a renamed bash, for example (which frequently happens).
Thus your second question of whether ps output will do is answered with "not always".
echo $0 - will print the program name... which in the case of the shell is the actual shell.
ps -ef | grep $$ | grep -v grep - this will look for the current process ID in the list of running processes. Since the current process is the shell, it will be included.
This is not 100% reliable, as you might have other processes whose ps listing includes the same number as shell's process ID, especially if that ID is a small number (for example, if the shell's PID is "5", you may find processes called "java5" or "perl5" in the same grep output!). This is the second problem with the "ps" approach, on top of not being able to rely on the shell name.
echo $SHELL - The path to the current shell is stored as the SHELL variable for any shell. The caveat for this one is that if you launch a shell explicitly as a subprocess (for example, it's not your login shell), you will get your login shell's value instead. If that's a possibility, use the ps or $0 approach.
If, however, the executable doesn't match your actual shell (e.g. /bin/sh is actually bash or ksh), you need heuristics. Here are some environmental variables specific to various shells:
$version is set on tcsh
$BASH is set on bash
$shell (lowercase) is set to actual shell name in csh or tcsh
$ZSH_NAME is set on zsh
ksh has $PS3 and $PS4 set, whereas the normal Bourne shell (sh) only has $PS1 and $PS2 set. This generally seems like the hardest to distinguish - the only difference in the entire set of environment variables between sh and ksh we have installed on Solaris boxen is $ERRNO, $FCEDIT, $LINENO, $PPID, $PS3, $PS4, $RANDOM, $SECONDS, and $TMOUT.
ps -p $$
should work anywhere that the solutions involving ps -ef and grep do (on any Unix variant which supports POSIX options for ps) and will not suffer from the false positives introduced by grepping for a sequence of digits which may appear elsewhere.
Try
ps -p $$ -oargs=
or
ps -p $$ -ocomm=
If you just want to ensure the user is invoking a script with Bash:
if [ -z "$BASH" ]; then echo "Please run this script $0 with bash"; exit; fi
or ref
if [ -z "$BASH" ]; then exec bash $0 ; exit; fi
You can try:
ps | grep `echo $$` | awk '{ print $4 }'
Or:
echo $SHELL
$SHELL need not always show the current shell. It only reflects the default shell to be invoked.
To test the above, say bash is the default shell, try echo $SHELL, and then in the same terminal, get into some other shell (KornShell (ksh) for example) and try $SHELL. You will see the result as bash in both cases.
To get the name of the current shell, Use cat /proc/$$/cmdline. And the path to the shell executable by readlink /proc/$$/exe.
There are many ways to find out the shell and its corresponding version. Here are few which worked for me.
Straightforward
$> echo $0 (Gives you the program name. In my case the output was -bash.)
$> $SHELL (This takes you into the shell and in the prompt you get the shell name and version. In my case bash3.2$.)
$> echo $SHELL (This will give you executable path. In my case /bin/bash.)
$> $SHELL --version (This will give complete info about the shell software with license type)
Hackish approach
$> ******* (Type a set of random characters and in the output you will get the shell name. In my case -bash: chapter2-a-sample-isomorphic-app: command not found)
ps is the most reliable method. The SHELL environment variable is not guaranteed to be set and even if it is, it can be easily spoofed.
I have a simple trick to find the current shell. Just type a random string (which is not a command). It will fail and return a "not found" error, but at start of the line it will say which shell it is:
ksh: aaaaa: not found [No such file or directory]
bash: aaaaa: command not found
I have tried many different approaches and the best one for me is:
ps -p $$
It also works under Cygwin and cannot produce false positives as PID grepping. With some cleaning, it outputs just an executable name (under Cygwin with path):
ps -p $$ | tail -1 | awk '{print $NF}'
You can create a function so you don't have to memorize it:
# Print currently active shell
shell () {
ps -p $$ | tail -1 | awk '{print $NF}'
}
...and then just execute shell.
It was tested under Debian and Cygwin.
The following will always give the actual shell used - it gets the name of the actual executable and not the shell name (i.e. ksh93 instead of ksh, etc.). For /bin/sh, it will show the actual shell used, i.e. dash.
ls -l /proc/$$/exe | sed 's%.*/%%'
I know that there are many who say the ls output should never be processed, but what is the probability you'll have a shell you are using that is named with special characters or placed in a directory named with special characters? If this is still the case, there are plenty of other examples of doing it differently.
As pointed out by Toby Speight, this would be a more proper and cleaner way of achieving the same:
basename $(readlink /proc/$$/exe)
My variant on printing the parent process:
ps -p $$ | awk '$1 == PP {print $4}' PP=$$
Don't run unnecessary applications when AWK can do it for you.
Provided that your /bin/sh supports the POSIX standard and your system has the lsof command installed - a possible alternative to lsof could in this case be pid2path - you can also use (or adapt) the following script that prints full paths:
#!/bin/sh
# cat /usr/local/bin/cursh
set -eu
pid="$$"
set -- sh bash zsh ksh ash dash csh tcsh pdksh mksh fish psh rc scsh bournesh wish Wish login
unset echo env sed ps lsof awk getconf
# getconf _POSIX_VERSION # reliable test for availability of POSIX system?
PATH="`PATH=/usr/bin:/bin:/usr/sbin:/sbin getconf PATH`"
[ $? -ne 0 ] && { echo "'getconf PATH' failed"; exit 1; }
export PATH
cmd="lsof"
env -i PATH="${PATH}" type "$cmd" 1>/dev/null 2>&1 || { echo "$cmd not found"; exit 1; }
awkstr="`echo "$#" | sed 's/\([^ ]\{1,\}\)/|\/\1/g; s/ /$/g' | sed 's/^|//; s/$/$/'`"
ppid="`env -i PATH="${PATH}" ps -p $pid -o ppid=`"
[ "${ppid}"X = ""X ] && { echo "no ppid found"; exit 1; }
lsofstr="`lsof -p $ppid`" ||
{ printf "%s\n" "lsof failed" "try: sudo lsof -p \`ps -p \$\$ -o ppid=\`"; exit 1; }
printf "%s\n" "${lsofstr}" |
LC_ALL=C awk -v var="${awkstr}" '$NF ~ var {print $NF}'
My solution:
ps -o command | grep -v -e "\<ps\>" -e grep -e tail | tail -1
This should be portable across different platforms and shells. It uses ps like other solutions, but it doesn't rely on sed or awk and filters out junk from piping and ps itself so that the shell should always be the last entry. This way we don't need to rely on non-portable PID variables or picking out the right lines and columns.
I've tested on Debian and macOS with Bash, Z shell (zsh), and fish (which doesn't work with most of these solutions without changing the expression specifically for fish, because it uses a different PID variable).
If you just want to check that you are running (a particular version of) Bash, the best way to do so is to use the $BASH_VERSINFO array variable. As a (read-only) array variable it cannot be set in the environment,
so you can be sure it is coming (if at all) from the current shell.
However, since Bash has a different behavior when invoked as sh, you do also need to check the $BASH environment variable ends with /bash.
In a script I wrote that uses function names with - (not underscore), and depends on associative arrays (added in Bash 4), I have the following sanity check (with helpful user error message):
case `eval 'echo $BASH#${BASH_VERSINFO[0]}' 2>/dev/null` in
*/bash#[456789])
# Claims bash version 4+, check for func-names and associative arrays
if ! eval "declare -A _ARRAY && func-name() { :; }" 2>/dev/null; then
echo >&2 "bash $BASH_VERSION is not supported (not really bash?)"
exit 1
fi
;;
*/bash#[123])
echo >&2 "bash $BASH_VERSION is not supported (version 4+ required)"
exit 1
;;
*)
echo >&2 "This script requires BASH (version 4+) - not regular sh"
echo >&2 "Re-run as \"bash $CMD\" for proper operation"
exit 1
;;
esac
You could omit the somewhat paranoid functional check for features in the first case, and just assume that future Bash versions would be compatible.
None of the answers worked with fish shell (it doesn't have the variables $$ or $0).
This works for me (tested on sh, bash, fish, ksh, csh, true, tcsh, and zsh; openSUSE 13.2):
ps | tail -n 4 | sed -E '2,$d;s/.* (.*)/\1/'
This command outputs a string like bash. Here I'm only using ps, tail, and sed (without GNU extesions; try to add --posix to check it). They are all standard POSIX commands. I'm sure tail can be removed, but my sed fu is not strong enough to do this.
It seems to me, that this solution is not very portable as it doesn't work on OS X. :(
echo $$ # Gives the Parent Process ID
ps -ef | grep $$ | awk '{print $8}' # Use the PID to see what the process is.
From How do you know what your current shell is?.
This is not a very clean solution, but it does what you want.
# MUST BE SOURCED..
getshell() {
local shell="`ps -p $$ | tail -1 | awk '{print $4}'`"
shells_array=(
# It is important that the shells are listed in descending order of their name length.
pdksh
bash dash mksh
zsh ksh
sh
)
local suited=false
for i in ${shells_array[*]}; do
if ! [ -z `printf $shell | grep $i` ] && ! $suited; then
shell=$i
suited=true
fi
done
echo $shell
}
getshell
Now you can use $(getshell) --version.
This works, though, only on KornShell-like shells (ksh).
Do the following to know whether your shell is using Dash/Bash.
ls –la /bin/sh:
if the result is /bin/sh -> /bin/bash ==> Then your shell is using Bash.
if the result is /bin/sh ->/bin/dash ==> Then your shell is using Dash.
If you want to change from Bash to Dash or vice-versa, use the below code:
ln -s /bin/bash /bin/sh (change shell to Bash)
Note: If the above command results in a error saying, /bin/sh already exists, remove the /bin/sh and try again.
I like Nahuel Fouilleul's solution particularly, but I had to run the following variant of it on Ubuntu 18.04 (Bionic Beaver) with the built-in Bash shell:
bash -c 'shellPID=$$; ps -ocomm= -q $shellPID'
Without the temporary variable shellPID, e.g. the following:
bash -c 'ps -ocomm= -q $$'
Would just output ps for me. Maybe you aren't all using non-interactive mode, and that makes a difference.
Get it with the $SHELL environment variable. A simple sed could remove the path:
echo $SHELL | sed -E 's/^.*\/([aA-zZ]+$)/\1/g'
Output:
bash
It was tested on macOS, Ubuntu, and CentOS.
On Mac OS X (and FreeBSD):
ps -p $$ -axco command | sed -n '$p'
Grepping PID from the output of "ps" is not needed, because you can read the respective command line for any PID from the /proc directory structure:
echo $(cat /proc/$$/cmdline)
However, that might not be any better than just simply:
echo $0
About running an actually different shell than the name indicates, one idea is to request the version from the shell using the name you got previously:
<some_shell> --version
sh seems to fail with exit code 2 while others give something useful (but I am not able to verify all since I don't have them):
$ sh --version
sh: 0: Illegal option --
echo $?
2
One way is:
ps -p $$ -o exe=
which is IMO better than using -o args or -o comm as suggested in another answer (these may use, e.g., some symbolic link like when /bin/sh points to some specific shell as Dash or Bash).
The above returns the path of the executable, but beware that due to /usr-merge, one might need to check for multiple paths (e.g., /bin/bash and /usr/bin/bash).
Also note that the above is not fully POSIX-compatible (POSIX ps doesn't have exe).
Kindly use the below command:
ps -p $$ | tail -1 | awk '{print $4}'
This one works well on Red Hat Linux (RHEL), macOS, BSD and some AIXes:
ps -T $$ | awk 'NR==2{print $NF}'
alternatively, the following one should also work if pstree is available,
pstree | egrep $$ | awk 'NR==2{print $NF}'
You can use echo $SHELL|sed "s/\/bin\///g"
And I came up with this:
sed 's/.*SHELL=//; s/[[:upper:]].*//' /proc/$$/environ