I have a bash script and need to run some commands in cshell inside it.
#!/bin/bash
echo entering_to_cshell
csh
echo in_cshell
exit
echo exited_from_cshell
Why doesn't this script run as expected? It only prints entering_to_cshell and it doesn't exit from cshell.
By using
csh
you start a new subshell where your script isn't executed. That's why none of your following commands are executed. Your script waits for this subshell to end which, as you noted, never happens.
Try
csh -c "echo in_cshell"
This way you don't create a new subshell which isn't impacted by your script.
By simply calling csh in your script, you're starting an interactive csh subshell. You'll notice that once you quit from the csh session, your script will then continue to with the subsequent echo and quiting on exit.
To pass a series of commands to csh from you bash script, one approach would be to use the Here Document syntax to redirect commands to csh.
#!/bin/bash
echo entering_to_cshell
csh <<EOF
echo in_cshell
exit
EOF
echo exited_from_cshell
The lines between the EOF entries will be treated as a script that is interpreted by csh.
Related
how to run multiple bash scripts in 1 bash command ?
i use command
bash script1.sh
how to make it run multiple commands in 1 command ?
for example
bash script1.sh bash script2.sh bash script3.sh bash script4.sh
Please help
If you want to run all bash script in //:
for i in {1..4}; do bash "script${i}.sh" & done
If you put the control operator & at the end of a command, e.g. command &, the shell executes the command in the background in a subshell. The shell does not wait for the command to finish, and the return status is 0. Pid of the last backgrounded command is available via the special variable $!
If instead you want to run sequentially, use
printf '%s\n' script{1..4}.sh | xargs -n1 bash
or
for i in {1..4}; do bash "script${i}.sh"; done
I have a bash script that opens up a shell called salome shell and it should execute a command called as_run in that shell. The thing is that after entering the salome shell it doesn't execute the command until I exit the salome shell. This is the code that i got:
#!/bin/bash
cd /opt/salome/appli_V2018.0.1_public
./salome shell
eval "as_run /home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome"
What should I do in order to execute the command in the salome shell?
Might be this is what you want:
# call salome shell with commands in a specified script file
cd /opt/salome/appli_V2018.0.1_public
./salome shell <"/home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome"
Or might be this is what you want:
# pipe a command as_run... to salome shell
cd /opt/salome/appli_V2018.0.1_public
echo "as_run /home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome" | ./salome shell
Anyway, you have to read the salome guide about how salome shell call it's script.
Most shells implement a way to pass the commands as parameters, e.g.
dash -c 'x=1 ; echo $x'
You'll need to consult your shell's manual to see if it's possible.
You can also try sending the commands to the standard input of the shell:
echo 'set x = 1 ; echo $x' | tcsh
Using a HERE doc might be a bit more readable in case of complex commands:
tcsh << 'TCSH'
set x = 1
echo $x
TCSH
With my Bash session, I run this command:
(echo $$ $BASHPID $-)
and I get
22108 25602 himBH
So, my subshell runs as an interactive shell.
If I try to run the same command in background
(echo $$ $BASHPID $-) &
I get the same output. Why does the subshell run as an interactive shell?
A subshell inherits all the characteristics of the parent, and that includes settings. That's part of the definition of a subshell. It makes no difference if it is running in foreground or background - jobs can be swapped between foreground and background quite easily.
A non-subshell will not be interactive. For example, if I put
(echo $$ $BASHPID $-)
into a script called gash.sh and call it from a non-interactive shell:
$ bash gash.sh
73879 73880 hB
But if I source the file, then it is a subshell of an interactive shell:
$ . ./gash.sh
1130 himBH
date
ls
exit
On executing this script, the commands date and ls are executed but the terminal doesn't close. Why?
The script will exit only the shell it's running in. When you do ./abc then parent shell forks and execs a subshell in which ./abc is run.
Just to be bit more obvious about subshell exit, consider the following example:
#!/bin/bash
echo hi
(exit)
echo hello
prints both hi and hello. The exit happens in the subshell. So, the script won't exit and hence the last echo hello gets executed.
To see that the script itself exits, try
#!/bin/bash
echo hi
exit
echo hello
Running this script will output just "hi"; the exit command terminates the script (but not the shell running in your terminal) before it reaches echo hello.
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.