I need some help:
(On macos, bash shell)
If I run a .sh file which calls e.g. exit 1 (any exit code) my terminal session ends (and the iterm2 tab/window closes).
I'm calling the script like this $ . myscript.sh
I'm pretty sure it should not be like that or was not like this a while before.
Using:
. myscript.sh
You are actually running the script in the existing shell or "sourcing" the script. With exit at the end of the script, this means that the terminal session will also exit
Alternatively:
./myscript.sh
or
bash myscript.sh
Will run the script in a separate bash shell and stop the terminal session from exiting.
Instead of . myscript.sh you can run ./myscript.sh which will run it in a separate bash shell and will not exit the current session.
If you control the content of this .sh file, and you do want to source the script - simply return 1 instead of exit 1, and use proper error handling.
bash scripting noob here. I've found this article: https://www.shellhacks.com/print-usage-exit-if-arguments-not-provided/ that suggests putting
[ $# -eq 0 ] && { echo "Usage: $0 argument"; exit 1; }
at the top of a script to ensure arguments are passed. Seems sensible.
However, when I do that and test that that line does indeed work (by running the script without supplying any arguments: . myscript.sh) then the script does indeed exit but so does the bash session that I was calling the script from. This is very irritating.
Clearly I'm doing something wrong but I don't know what. Can anyone put me straight?
. myscript.sh is a synonym for source myscript.sh, which runs the script in the current shell (rather than as a separate process). So exit terminates your current shell. (return, on the other hand, wouldn't; it has special behaviour for sourced scripts.)
Use ./myscript.sh to run it "the normal way" instead. If that gives you a permission error, make it executable first, using chmod a+x myscript.sh. To inform the kernel that your script should be run with bash (rather than /bin/sh), add the following as the very first line in the script:
#!/usr/bin/env bash
You can also use bash myscript.sh if you can't make it executable, but this is slightly more error-prone (somebody might do sh myscript.sh instead).
Question seems not clear if you're sourcing script source script_name or . script_name it's interpreted in current bash process, if you're running a function it's the same it's running in same process, otherwise, calling a script, caller bash forks a new bash process and waits until it terminates (so running exit doesn't exit caller process), but when running exit builtin in in current bash it exits current process.
I have two shell scripts, a.sh and b.sh.
How can I call b.sh from within the shell script a.sh?
There are a couple of different ways you can do this:
Make the other script executable with chmod a+x /path/to/file(Nathan Lilienthal's comment), add the #!/bin/bash line (called shebang) at the top, and the path where the file is to the $PATH environment variable. Then you can call it as a normal command;
Or call it with the source command (which is an alias for .), like this:
source /path/to/script
Or use the bash command to execute it, like:
/bin/bash /path/to/script
The first and third approaches execute the script as another process, so variables and functions in the other script will not be accessible.
The second approach executes the script in the first script's process, and pulls in variables and functions from the other script (so they are usable from the calling script).
In the second method, if you are using exit in second script, it will exit the first script as well. Which will not happen in first and third methods.
Check this out.
#!/bin/bash
echo "This script is about to run another script."
sh ./script.sh
echo "This script has just run another script."
There are a couple of ways you can do this. Terminal to execute the script:
#!/bin/bash
SCRIPT_PATH="/path/to/script.sh"
# Here you execute your script
"$SCRIPT_PATH"
# or
. "$SCRIPT_PATH"
# or
source "$SCRIPT_PATH"
# or
bash "$SCRIPT_PATH"
# or
eval '"$SCRIPT_PATH"'
# or
OUTPUT=$("$SCRIPT_PATH")
echo $OUTPUT
# or
OUTPUT=`"$SCRIPT_PATH"`
echo $OUTPUT
# or
("$SCRIPT_PATH")
# or
(exec "$SCRIPT_PATH")
All this is correct for the path with spaces!!!
The answer which I was looking for:
( exec "path/to/script" )
As mentioned, exec replaces the shell without creating a new process. However, we can put it in a subshell, which is done using the parantheses.
EDIT:
Actually ( "path/to/script" ) is enough.
If you have another file in same directory, you can either do:
bash another_script.sh
or
source another_script.sh
or
. another_script.sh
When you use bash instead of source, the script cannot alter environment of the parent script. The . command is POSIX standard while source command is a more readable bash synonym for . (I prefer source over .). If your script resides elsewhere just provide path to that script. Both relative as well as full path should work.
Depends on.
Briefly...
If you want load variables on current console and execute you may use source myshellfile.sh on your code. Example:
#!/bin/bash
set -x
echo "This is an example of run another INTO this session."
source my_lib_of_variables_and_functions.sh
echo "The function internal_function() is defined into my lib."
returned_value=internal_function()
echo $this_is_an_internal_variable
set +x
If you just want to execute a file and the only thing intersting for you is the result, you can do:
#!/bin/bash
set -x
./executing_only.sh
bash i_can_execute_this_way_too.sh
bash or_this_way.sh
set +x
You can use /bin/sh to call or execute another script (via your actual script):
# cat showdate.sh
#!/bin/bash
echo "Date is: `date`"
# cat mainscript.sh
#!/bin/bash
echo "You are login as: `whoami`"
echo "`/bin/sh ./showdate.sh`" # exact path for the script file
The output would be:
# ./mainscript.sh
You are login as: root
Date is: Thu Oct 17 02:56:36 EDT 2013
First you have to include the file you call:
#!/bin/bash
. includes/included_file.sh
then you call your function like this:
#!/bin/bash
my_called_function
Simple source will help you.
For Ex.
#!/bin/bash
echo "My shell_1"
source my_script1.sh
echo "Back in shell_1"
Just add in a line whatever you would have typed in a terminal to execute the script!
e.g.:
#!bin/bash
./myscript.sh &
if the script to be executed is not in same directory, just use the complete path of the script.
e.g.:`/home/user/script-directory/./myscript.sh &
This was what worked for me, this is the content of the main sh script that executes the other one.
#!/bin/bash
source /path/to/other.sh
The top answer suggests adding #!/bin/bash line to the first line of the sub-script being called. But even if you add the shebang, it is much faster* to run a script in a sub-shell and capture the output:
$(source SCRIPT_NAME)
This works when you want to keep running the same interpreter (e.g. from bash to another bash script) and ensures that the shebang line of the sub-script is not executed.
For example:
#!/bin/bash
SUB_SCRIPT=$(mktemp)
echo "#!/bin/bash" > $SUB_SCRIPT
echo 'echo $1' >> $SUB_SCRIPT
chmod +x $SUB_SCRIPT
if [[ $1 == "--source" ]]; then
for X in $(seq 100); do
MODE=$(source $SUB_SCRIPT "source on")
done
else
for X in $(seq 100); do
MODE=$($SUB_SCRIPT "source off")
done
fi
echo $MODE
rm $SUB_SCRIPT
Output:
~ ❯❯❯ time ./test.sh
source off
./test.sh 0.15s user 0.16s system 87% cpu 0.360 total
~ ❯❯❯ time ./test.sh --source
source on
./test.sh --source 0.05s user 0.06s system 95% cpu 0.114 total
* For example when virus or security tools are running on a device it might take an extra 100ms to exec a new process.
pathToShell="/home/praveen/"
chmod a+x $pathToShell"myShell.sh"
sh $pathToShell"myShell.sh"
#!/bin/bash
# Here you define the absolute path of your script
scriptPath="/home/user/pathScript/"
# Name of your script
scriptName="myscript.sh"
# Here you execute your script
$scriptPath/$scriptName
# Result of script execution
result=$?
chmod a+x /path/to/file-to-be-executed
That was the only thing I needed. Once the script to be executed is made executable like this, you (at least in my case) don't need any other extra operation like sh or ./ while you are calling the script.
Thanks to the comment of #Nathan Lilienthal
Assume the new file is "/home/satya/app/app_specific_env" and the file contents are as follows
#!bin/bash
export FAV_NUMBER="2211"
Append this file reference to ~/.bashrc file
source /home/satya/app/app_specific_env
When ever you restart the machine or relogin, try echo $FAV_NUMBER in the terminal. It will output the value.
Just in case if you want to see the effect right away, source ~/.bashrc in the command line.
There are some problems to import functions from other file.
First: You needn't to do this file executable. Better not to do so!
just add
. file
to import all functions. And all of them will be as if they are defined in your file.
Second: You may be define the function with the same name. It will be overwritten. It's bad. You may declare like that
declare -f new_function_name=old_function_name
and only after that do import.
So you may call old function by new name.
Third: You may import only full list of functions defined in file.
If some not needed you may unset them. But if you rewrite your functions after unset they will be lost. But if you set reference to it as described above you may restore after unset with the same name.
Finally In common procedure of import is dangerous and not so simple. Be careful! You may write script to do this more easier and safe.
If you use only part of functions(not all) better split them in different files. Unfortunately this technique not made well in bash. In python for example and some other script languages it's easy and safe. Possible to make partial import only needed functions with its own names. We all want that in next bush versions will be done the same functionality. But now We must write many additional cod so as to do what you want.
Use backticks.
$ ./script-that-consumes-argument.sh `sh script-that-produces-argument.sh`
Then fetch the output of the producer script as an argument on the consumer script.
I'm using this bash script:
for a in `sort -u $HADOOP_HOME/conf/slaves`; do
rsync -e ssh -a "${HADOOP_HOME}/conf" ${a}:"${HADOOP_HOME}"
done
for a in `sort -u $HBASE_HOME/conf/regionservers`; do
rsync -e ssh -a "${HBASE_HOME}/conf" ${a}:"${HBASE_HOME}"
done
When I call this script directly from shell, there are no problems and it works fine. But when I call this script from another script, although the script does its job, I get this message at the end:
sort: open failed: /conf/slaves: No such file or directory
sort: open failed: /conf/regionservers: No such file or directory
I have set $HADOOP_HOME and $HBASE_HOME in /etc/profile and the script does the job right. But I don't understand why it gives this message in the end.
Are you sure it's doing it right? When you call this script from the shell it is acting as an interactive shell which reads and sources /etc/profile and ~/.bash_profile if it exists. When you call it from another script it is running as non-interactive and wont source those files. If you want a non-interactive shell to source a file you can do this by setting the BASH_ENV environment variable.
#!/bin/bash
export BASH_ENV=/etc/profile
./call/to/your/HADOOP/script.sh
Everything points to those variables not being defined when your script runs.
You should ensure that they are set for your script. Before the first loop, place the line:
echo "[${HADOOP_HOME}] [${HBASE_HOME}]"
and make sure that doesn't output "[] []" (or even one "[]").
Additionally, put a set +x line at the top of the script - this will output lines before executing them and you can see what's being done.
Keep in mind that some shells don't pass on environment variables to subshells unless you explicitly export them (setting them is not enough).
Is it possible to source a .bshrc file from .cshrc in a non-interactive session?
I'm asking because tcsh is our default shell at work and the .cshrc has to be used to set up the environment initially.
However, I am not really familiar with the tcsh and I have my own set-up in bash, so right now I have the following lines at the end of my .cshrc file:
if ( $?prompt && -x /bin/bash) then
exec /bin/bash
endif
This works fine, loading my environment from .bashrc and giving me a bash prompt for interactive sessions but now I also need the same set-up for non-interactive sessions, e.g. to run a command remotely via SSH with all the correct PATHs etc.
I can't use 'exec' in that case but I can't figure out how to switch to bash and load the bash config files "non-interactively".
All our machines share the same home directory, so any changes to my local *rc files will affect the remote machiens as well.
Any ideas welcome - thank you for your help!
After some more research I'm now quite sure that this won't work, but of course feel free to prove me wrong!
To load the environment in bash I have to switch to a bash shell. Even if that is possible "in the background", i.e. without getting a prompt, it would still break any tcsh commands which would then be attempted to execute under bash.
Hmmmm, back to the drawing board...
If $command is set there are arguments to csh, so it is a remote shell command. This works for me in .cshrc:
if ($?command) then
echo Executing non-interactive command in bash: $command $*
exec /bin/bash -c "${command} $*"
endif
echo Interactive bash shell
exec bash -l
Test:
$ ssh remotehost set | grep BASH
BASH=/bin/bash
...
proves that it ran in Bash.