Setting environment variable to shell executed from perl script - shell

I am running a perl file a.pm which invokes b.sh via system command.
Here, b.sh is using find utility whose path is /usr/local/bin.
If I run env on shell directly on machine, I get output as below for PATH variable.
PATH=/sbin:/usr/sbin:/usr/local/bin:/usr/local/sbin:/home/bin:/home/bin/samba::/home/venv/bin/:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin`
Thats why if I run the b.sh directly from shell, it is able to execute find utility.
Now, If I run b.sh via a.pm as mentioned earlier using system(), and when I print PATH env variable in b.sh, its coming as
/bin:/usr/bin:/usr/X11R6/bin:/home/bin:/home/perl5/bin
which does not have /usr/local/bin, and thats why find command is failing.
If I tried to print all ENV variables in perl before invoking system(b.sh), PATH variable is not printed.
Now, I tried adding path variable in a.pm file as follows just before invoking system(b.sh).
$ENV{'PATH'} = '/usr/local/bin:/sbin:/usr/sbin:/usr/local/sbin:/home/bin:/usr/local/sbin:/sbin:/bin:/usr/sbin:/usr/bin:/usr/local/';
Now, if I try to print all ENV variables in perl before invoking system(b.sh), PATH variable is printed with above value.
Still executing the a.pm file, the PATH variable printed in b.sh is same:
/bin:/usr/bin:/usr/X11R6/bin:/home/bin:/home/perl5/bin
How can I add corresponding path /usr/local/bin to shell of b.sh invoked using a.pm?

I suspect that the Perl program is either modifying the path it gets from the shell that invokes it, or that you have left out a step somewhere. For example, if you invoke the Perl program from a different environment, it will likely have a different PATH.
You seemed to have found your answer though. Add the necessary directory to the PATH in the Perl program. But, you say this doesn't work. Again, I think there's some step that you haven't included. I suspect that the way in which you run system overwrites the PATH inherited from the parent.
For example, here's a small Perl program the merely runs a shell script:
#!perl
use v5.10;
$ENV{PATH} = '/bin:/usr/bin';
say "PATH in Perl is $ENV{PATH}";
system( "sh ./pather.sh" );
The shell script echos the PATH:
#!/bin/sh
echo "PATH in shell:" $PATH
When I run this, both PATHs match:
PATH in Perl is /bin:/usr/bin
PATH in shell: /bin:/usr/bin
But, maybe the command in system is something else. The -l switch treats the shell as a login shell, so it will load the various profiles and whatnot:
system( "sh -l ./pather.sh" );
Now the PATH is different in the shell script because my particular profiles overwrote PATH:
PATH in Perl is /bin:/usr/bin
Path in shell: /usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/Library/TeX/texbin:/usr/local/go/bin:...
Our answers can be more targeted you can produce a minimal working example where we see actual code that demonstrates the problem. Since we don't see what you actually run in system, we can only guess.

Related

Running a bash script stored in a variable

I want to send the path to a bash script which sources some environmental variables as an argument to another bash script to run it and use the environmental variables. It works well with no arguments if I hard coded the path to the bash script to run it works and I can retrieve the environmental variables in the main script. the problem happens when I send the path as an argument it does not want to run it.
for example if the path is /path/script.bash and I send the path as an argument I get the error that /path/env_set: No such a file or directory
I run the script by this line
. $1 (this doesn't work)
. /path/script.bash (this works)
if I use
bash -c $1
the bash file runs but it does not set the environmental variables to use it in the main script
I don't know why env_set replaces the script name when I use arguments. Is there any approach to achieve this or any work around to achieve my goal?
It sounds like the problem could be either with your quoting, or with relative paths.
Quoting isn't just about spaces, it's also about pathname expansion (ie. []?* characters).
Do
. "$1"
(not . $1)
And remember, if you're giving a relative path for the environment script (or that script uses some relative paths), you will have a problem. Those paths are relative to the pwd - which is wherever you happen to be when you execute the main script (not where any of the script files themselves happen to be located, for example).
Finally, you can debug this problem by throwing echo at the start, and running the script (if it's safe to do that):
echo . "$1"
exit # Add exit here if you don't want to run w/o the vars.
Now you can see what you're actually trying to source.
In script 1, in your main code, you can call and run script 2,
. ./script 2
The first . stands for current shell, and the second . for current directory.
which will create the environment variables for you, and configure any other settings as well in the same terminal.
Afterwards when script 2 has finished running, script 1 would continue to run, and your environment variables which was created in script 2, will be accessible for script 1 to use in the same session.

How to use the source command with the system() function?

I need to source a few environment variables in another file. If I use the source command with system() function, it's complaining about "No such file or directory". Am I missing something?
My code looks like below. In my code, I have only the system() function running the source command. The source file has just only one command: pwd (Present working directory).
perl_system.pl
#!/usr/bin/perl
system "source env.mk"
env.mk (contents of env.mk which I want to source has just pwd for now"
pwd
When I run this command, I see the below error
$ perl -w perl_system.pl
Can't exec "source": No such file or directory at perl_system.pl line 2.
source is a shell built-in that executes a shell script using the current shell interpreter. So it doesn't work as an external command and won't change the environment of your perl process even if you change your system call to invoke a shell instead of it trying to run an external program directly.
You could run your env.mk and then output the resulting environment and update perl's environment accordingly, though:
for my $env (`bash -c 'source env.mk;env'`) {
chomp $env;
my ($var,$val) = split /=/, $env, 2;
$ENV{$var} = $val;
}
(with obvious problems if environment variables contain newlines).
Update: just read all of your question, not just the beginning. If all you want to do is execute a shell script, just do:
system "sh env.mk";
source is completely unnecessary for this.

How can I store and execute the command "export PATH=$PREFIX/bin" from a script?

I would like to write a script that has several commands of the kind
> export PATH=$PREFIX/bin
Where
> $PREFIX = /home/usr
or something else. Instead of typing it into the the Shell (/bin/bash) I would run the script to execute the commands.
Tried it with sh and then with a .py script having the line,
> commands.getstatusoutput('export PATH=$PREFIX/bin')
but these result into the error "bad variable name".
Would be thankful for some ideas!
If you need to adjust PATH (or any other environment variable) via a script after your .profile and equivalents have been run, you need to 'dot' or 'source' the file containing the script:
. file_setting_path
source file_setting_path
The . notation applies to all Bourne shell derivatives, and is standardized by POSIX. The source notation is used in C shell and has infected Bash completely unnecessarily.
Note that the file (file_setting_path) can be specified as a pathname, or if it lives in a directory listed on $PATH, it will be found. It only needs to be readable; it does not have to be executable.
The way the dot command works is that it reads the named file as part of the current shell environment, rather than executing it in a sub-shell like a normal script would be executed. Normally, the sub-shell sets its environment happily, but that doesn't affect the calling script.
The bad variable name is probably just a complaint that $PREFIX is undefined.
Usually a setting of PATH would look something like
export PATH=$PATH:/new/path/to/programs
so that you retain the old PATH but add something onto the end.
You are best off putting such things in your .bashrc so that they get run every time you log in.

"Command not found" inside shell script

I have a shell script on a mac (OSX 10.9) named msii810161816_TMP_CMD with the following content.
matlab
When I execute it, I get
./msii810161816_TMP_CMD: line 1: matlab: command not found
However, when I type matlab into the shell directly it starts as normal. How can it be that the same command works inside the shell but not inside a shell script? I copy-pasted the command directly from the script into the shell and it worked ...
PS: When I replace the content of the script with
echo matlab
I get the desired result, so I can definitely execute the shell script (I use ./msii810161816_TMP_CMD)
Thanks guys!
By default, aliases are not expanded in non-interactive shells, which is what shell scripts are. Aliases are intended to be used by a person at the keyboard as a typing aid.
If your goal is to not have to type the full path to matlab, instead of creating an alias you should modify your $PATH. Add /Applications/MATLAB_R2014a.app/bin to your $PATH environment variable and then both you and your shell scripts will be able to simply say
matlab
This is because, as commenters have stated, the PATH variable inside of the shell executing the script does not include the directory containing the matlab executable.
When a command name is used, like "matlab", your shell looks at every directory in the PATH in order, searching for one containing an executable file with the name "matlab".
Without going into too much detail, the PATH is determined by the shell being invoked.
When you execute bash, it combines a global setting for basic directories that must be in the PATH with any settings in your ~/.bashrc which alter the PATH.
Most likely, you are not running your script in a shell where the PATH includes matlab's directory.
To verify this, you can take the following steps:
Run which matlab. This will show you the path to the matlab executable.
Run echo "$PATH". This will show you your current PATH settings. Note that the directory from which matlab is included in the colon-separated list.
Add a line to the beginning of your script that does echo "$PATH". Note that the directory from which matlab is not included.
To resolve this, ensure that your script is executed in a shell that has the needed directory in the PATH.
You can do this a few ways, but the two most highly recommended ones would be
Add a shebang line to the start of your script. Assuming that you want to run it with bash, do #!/bin/bash or whatever the path to your bash interpreter is.
The shebang line is not actually fully standardized by POSIX, so BSD-derived systems like OSX will happily handle multiple arguments to the shebanged executable, while Linux systems pass at most one argument.
In spite of this, the shebang is an easy and simple way to document what should be used to execute the script, so it's a good solution.
Explicitly invoke your script with a shell as its interpreter, as in bash myscript.sh or tcsh myscript.sh or even sh myscript.sh
This is not incompatible with using a shebang line, and using both is a common practice.
I believe that the default shell on OSX is always bash, so you should start by trying with that.
If these instructions don't help, then you'll have to dig deeper to find out why or how the PATH is being altered between the calling context and the script's internal context.
Ultimately, this is almost certainly the source of your issue.

How to run my own programm using command in Shell?

I just learned that I could use chmod make myscript.sh executable and the run it as $ ./myscript.sh But how can I attach a custom command to it, like $ connectme [options] ?
You need to do two things:
Give the name you want to use. Either just rename it, or establish a link (hard or symbolic). Make sure the correctly named object has the right permissions.
Make sure it is in you path. But putting "." in you PATH is a bad idea (tm), so copy it to $HOME/bin, and put that in you path.
A completely different approach. Most shells support aliases. You could define one to run your script.
Note: The environment variable PATH tells the shell where to look for programs to run (unless you specify a fully qualified path like /home/jdoe/scripts/myscript.sh or ./myscript.sh), it consists of a ":" seperated list of directories to examine. You can check yours with:
$ printenv PATH
resulting for me in
/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/usr/X11/bin:/usr/X11R6/bin
which are the usual directories for binaries. You can add a new path element with (in /bin/sh and derivatives):
$ export PATH=$PATH:$HOME/bin
in csh and derivatives use
$ setenv PATH $PATH:$HOME/bin
either of which which will result in the shell also searching ~/bin for things to run. Then move your script into that directory (giving ta new name if you want). Check that you execute permissions for the script, and just type its name like any other command.
Fianlly, the use of a ".sh" extension to denote a shell script is for human consumption only. Unix does not care about how you name your script: it is the so-called "shebang" ("#!") on the first line of the script that the OS uses to find the interpreter.
You need to learn about arguments in BASH PROGRAMMING. Here is a good tutorial on them. Check section #4 out.
Basically, you need to use special variables $1, $2, $3 to refer to first, second and third command line arguments respectively.
Example:
$ ./mycript.sh A-Rod
With myscript.sh being:
#!/bin/bash
echo "Hello $1"
Will print:
Hello A-Rod

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