I am attempting to store the result of an echo command as a variable to be used in a shell script. Debian 4.19.0-6-amd64
The command works in terminal: echo $HOSTNAME returns debian-base, the correct hostname.
I attempt to run it in a shell script, such as:
#!/usr/bin/bash
CURRENT_HOSTNAME=`echo $HOSTNAME`
echo $CURRENT_HOSTNAME
I have tried expansion:
CURRENT_HOSTNAME=$(echo $HOSTNAME)
And just to cover some more bases, I tried things like:
CURRENT_HOSTNAME=$HOSTNAME
# or
CURRENT_HOSTNAME="$HOSTNAME"
# also, in case a problem with reserved names:
test=$HOSTNAME
test="$HOSTNAME"
Works great in the terminal! Output is as follows:
root#debian-base:/scripts# echo $HOSTNAME
debian-base
root#debian-base:/scripts# TEST_HOSTNAME=$HOSTNAME
root#debian-base:/scripts# echo $TEST_HOSTNAME
debian-base
root#debian-base:/scripts# TEST_TWO_HOSTNAME=$(echo $HOSTNAME)
root#debian-base:/scripts# echo $TEST_TWO_HOSTNAME
debian-base
As soon as I run the script (as above):
root#debian-base:/scripts# sh test.sh
root#debian-base:/scripts#
What am I doing wrong?
You are using bash as your terminal. Bash has the variable $HOSTNAME set. You run your script with sh. sh does not have a $HOSTNAME.
Options:
bash test.sh
Or run it as a program:
chmod +x test.sh
./test.sh
But I think you need to change your first line to:
#!/bin/bash
As I don't think bash is installed in /usr/bin in most cases. But you need to try. To figure out where bash is installed use which bash
Another option is to use the hostname binary:
CURRENT_HOSTNAME=$(hostname)
echo $CURRENT_HOSTNAME
Which works in both bash and sh.
You can start sh by just running sh. You will see it has a bash-like terminal. You can try to do echo $HOSTNAME. It will not show, because it's not there. You can use set to see all the variables that are there (as sh does not have tab completion it's harder to figure out).
Related
I have a simple bash script test.sh
#!/bin/sh
# This is a comment
echo "Hi"
It does not execute anything when I try to run ./test.sh
$ ./test.sh
$
It comes with empty output. The mac terminal is executing echo commands but not shell script. I am not sure what I am missing. Please suggest.
to execute command file , type sh test.sh
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'
#!/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.