How can I definitively determine if the currently running shell is bash or zsh?
(being able to disambiguate between additional shells is a bonus, but only bash & zsh are 100% necessary)
I've seen a few ways to supposedly do this, but they all have problems (see below).
The best I can think of is to run some syntax that will work on one and not the other, and to then check the errors / outputs to see which shell is running. If this is the best solution, what command would be best for this test?
The simplest solution would be if every shell included a read-only parameter of the same name that identified the shell. If this exists, however, I haven't heard of it.
Non-definitive ways to determine the currently running shell:
# default shell, not current shell
basename "${SHELL}"
# current script rather than current shell
basename "${0}"
# BASH_VERSINFO could be defined in any shell, including zsh
if [ -z "${BASH_VERSINFO+x}" ]; then
echo 'zsh'
else
echo 'bash'
fi
# executable could have been renamed; ps isn't a builtin
shell_name="$(ps -o comm= -p $$)"
echo "${shell_name##*[[:cntrl:][:punct:][:space:]]}"
# scripts can be sourced / run by any shell regardless of shebang
# shebang parsing
On $ prompt, run:
echo $0
but you can't use $0 within a script, as $0 will become the script's name itself.
To find the current shell (let's say BASH) if shebang / magic number executable was #!/bin/bash within a script:
#!/bin/bash
echo "Script is: $0 running using $$ PID"
echo "Current shell used within the script is: `readlink /proc/$$/exe`"
script_shell="$(readlink /proc/$$/exe | sed "s/.*\///")"
echo -e "\nSHELL is = ${script_shell}\n"
if [[ "${script_shell}" == "bash" ]]
then
echo -e "\nI'm BASH\n"
fi
Outputs:
Script is: /tmp/2.sh running using 9808 PID
Current shell used within the script is: /usr/bin/bash
SHELL is = bash
I'm BASH
This will work, if shebang was: #!/bin/zsh (as well).
Then, you'll get the output for SHELL:
SHELL is = zsh
While there is no 100% foolproof way to achieve it, it might help to do a
echo $BASH_VERSION
echo $ZSH_VERSION
Both are shell variables (not environment variables), which are set by the respective shell. In the respective other shell, they are empty.
Of course, if someone on purpose creates a variable of this name, or exports such a variable and then creates a subshell of the different kind, i.e.
# We are in bash here
export BASH_VERSION
zsh # the subshell will see BASH_VERSION even though it is zsh
this approach will fail; but I think if someone is really doing such a thing, he wants to sabotage your code on purpose.
This should work for most Linux systems:
cat /proc/$$/comm
Quick and easy.
Working from comments by #ruakh & #oguzismail, I think I have a solution.
\shopt -u lastpipe 2> /dev/null
shell_name='bash'; : | shell_name='zsh'
Related
I use zsh as the default shell, I run echo $0 and get -zsh on terminal, but the following code can't detect default interactive shell
#!/usr/bin/env bash
if [ -n "$ZSH_VERSION" ]; then
echo "zsh"
elif [ -n "$BASH_VERSION" ]; then
echo "bash"
else
echo "others"
fi
The detect result is always bash, why? Thank you.
Your code works for detecting the current shell. But scripts run in their own shell, independently from the interactive shell. Your script file always runs in bash due to its shebang. Without the shebang, the calling shell decides how to run the script (if at all).
Detecting the parent shell
To detect the shell that called your script, try
#!/usr/bin/env bash
ps -p $PPID -o comm=
When you run an interactive zsh and execute this file you should get zsh as output.
Detecting the default shell
Your question's title is about detecting the default interactive shell. To do so, you cannot check any processes, because even if your default shell is X you can always use Y. Instead, look at the file where the default is stored:
grep "^$USER:" /etc/passwd | cut -d: -f7
The standard way to do this is with the SHELL environment variable:
echo "Your default shell is $SHELL"
This is defined in the POSIX standard, section 8.3, "Other Environment Variables":
SHELL
This variable shall represent a pathname of the user's preferred command language interpreter. If this interpreter does not conform to the Shell Command Language in XCU Shell Command Language, utilities may behave differently from those described in POSIX.1-2017.
#!/bin/bash
if [ ! -f readexportfile ]; then
echo "readexportfile does not exist"
exit 0
fi
The above is part of my script. When the current shell is /bin/csh my script fails with the following error:
If: Expression Syntax
Then: Command not found
If I run bash and then run my script, it runs fine(as expected).
So the question is: If there is any way that myscript can change the current shell and then interpretate rest of the code.
PS: If i keep bash in my script, it changes the current shell and rest of the code in script doesn't get executed.
The other replies are correct, however, to answer your question, this should do the trick:
[[ $(basename $SHELL) = 'bash' ]] || exec /bin/bash
The exec builtin replaces the current shell with the given command (in this case, /bin/bash).
You can use SHEBANG(#!) to overcome your issue.
In your code you are already using she-bang but make sure it is first and foremost line.
$ cat test.sh
#!/bin/bash
if [ ! -f readexportfile ]; then
echo "readexportfile does not exist"
exit 0
else
echo "No File"
fi
$ ./test.sh
readexportfile does not exist
$ echo $SHELL
/bin/tcsh
In the above code even though I am using CSH that code executed as we mentioned shebang in the code. In case if there is no shebang then it will take the help of shell in which you are already logged in.
In you case you also check the location of bash interpreter using
$ which bash
or
$ cat /etc/shells |grep bash
When I run the following script, echo does not display anything and I don't know why. It works if I just type it into the terminal, but not from the shell script. Need some insight please. I might be tired but I'm very certain this should work:
#!/bin/sh
for n in `seq 1 10`
do
r=$RANDOM
t=$RANDOM
s=$RANDOM
f=$RANDOM
echo "$r $t $s $f"
done
echo "Done"
Your terminal probably runs a different shell than /bin/sh. For example, on Ubuntu, /bin/sh runs /bin/dash, but $RANDOM does not work there. You have to run /bin/bash or /bin/ksh to make it work.
When run from a terminal, you probably use bash, not sh.
Seems sh doesn't support $RANDOM and thus all variables you assign in your script will be assigned the empty string. Try changing the first line of your script to #!/bin/bash (or whereever bash is installed).
The default shell on the the system is csh but I want to write a script in bash. How do I write a script that will run bash and then convert back to csh at the end.
I tried this but it doesn't work:
bash
var=Hello
echo $var
csh
The command you are looking for is exit. When typing at the keyboard use exit instead of csh to get back to csh. When you enteredcsh, that just started a new csh session on top of the csh and bash sessions already running.
%bash
$ var=Hello
$ echo $var
Hello
$ exit
exit
%
As others have said, when using a script:
#! /bin/bash
var=Hello
echo $var
exit # You don't need exit; but it's okay here.
You don't need to change shells back again. When the script is run, it will be run by a sub-shell (which exits at the end of the script), and the parent shell is unchanged. So, as already suggested, the only thing you have to do is ensure the script is run by the correct shell, and the 'shebang' is the way to do that:
#!/bin/bash
var=Hello
echo $var
That's all it takes.
Define it using the sha bang
#!/bin/bash
at the starting of your file.
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.