Displaying all the variables declared in a sh file - linux

I have a test.sh file which contains something like the following
source lib.sh
input=$0
A=aa/bb/cc
B=$A/dd
C=$A/ee
doSomething $B $C
Is it possible to display the values of all the variables so that when I read the code, I don't have to trace their final value row by row.
In my case, I mean to display
input=$0
A=aa/bb/cc
B=aa/bb/cc/dd
C=aa/bb/cc/ee
I know
bash -x test.sh
can achieve this goal; but I don't want test.sh to be executed. ( Executing the functions in test.sh would delete some important files on my disk.) I only want to know the values of all the variables.

The concept of 'dry run' in bash actually is not possible. What would be the ouptput e.g in cases like the following:
if some_command; then
variable=output_1
else
variable=output_2
fi
In order to determine the script flow you have to execute the some_command which may require running some command and getting its output. This will modify your system (it will depend on the command). Thus, there's no 'safe' way to do what you need without executing the bash script.
In order to test your script will have to simulate the commands which could modify your system (by adding an echo at the beginning e.g). This way you can run the script and see the values for your variables.

Related

$_ variable behaves differently when script is called from konsole and when called from another script

I am not much experienced with tch. To simplify the scenario, I created a script and just put "echo $_" there.
if I source the script like this:
source set_env_vars.csh
I get "source set_env_vars.csh"
but if I run the script like:
./set_env_vars.csh
I get a different output 'some paths'
I want to understand what exactly $_ holds and why the outputs are different in two cases.
Thanks
Environmental variables propagate from the shell to the sub-shell only.
source set_env_vars.csh adds the environmental variables to the shell.
./set_env_vars.csh adds the environmental variables to a sub-shell that exists only for the duration of this command. These changes are lost when the sub-shell exits and are not seen by the shell. This is why the outputs are different.

Bash: Only allow script to run by being called from another script

We have two bash scripts to start up an application. The first (Start-App.sh) one sets up the environment and the second (startup.sh) is from a 3rd party that we are trying not to heavily edit. If someone runs the second script before the first the application does not come up correctly.
Is there a way to ensure that the startup.sh can only be called from the Start-App.sh script?
They are both in the same directory and run via bash on Red Hat Linux.
Is there a way to ensure that the startup.sh can only be called from the Start-App.sh script?
Ensure? No. And even less so without editing startup.sh at all. But you can get fairly close.
Below are three suggestions − you can either use one of them, or any combination of them.
The simplest, and probably the best, way is to add a single line at the top of startup.sh:
[ -z $CALLED_FROM_START_APP ] && { echo "Not called from Start-App.sh"; exit 42; }
And then call it from Start-App.sh like so:
export CALLED_FROM_START_APP=yes
sh startup.sh
of course, you can set this environment variable from the shell yourself, so it won't actually ensure anything, but I hope your engineering staff is mature enough not to do this.
You can also remove the execute permissions from startup.sh:
$ chmod a-x startup.sh
This will not prevent people from using sh startup.sh, so there is a very small guarantee here; but it might prevent auto-completion oopsies, and it will mark the file as "not intended to be executed" − if I see a directory with only one executable .sh file, I'll try and run that one, and not one of the others.
Lastly, you could perhaps rename the startup.sh script; for example, you could rename it to do_not_run, or "hide" it by renaming it to .startup. This probably won't interfere with the operation of this script (although I can't check this).
TL;DR:
[ $(basename "$0") = "Start-App.sh" ] || exit
Explanation
As with all other solutions presented it's not 100% bulletproof but this covers most common instances I've come across for preventing accidentally running a script directly as opposed to calling it from another script.
Unlike other approaches presented, this approach:
doesn't rely on manually set file names for each included/sourced script (i.e. is resilient to file name changes)
behaves consistently across all major *nix distros that ship with bash
introduces no unnecessary environment variables
isn't tied to a single parent script
prevents running the script through calling bash explicitly (e.g. bash myscript.sh)
The basic idea is having something like this at the top of your script:
[ $(basename "$0") = $(basename "$BASH_SOURCE") ] && exit
$0 returns the name of the script at the beginning of the execution chain
$BASH_SOURCE will always point to the file the currently executing code resides in (or empty if no file e.g. piping text directly to bash)
basename returns only the main file name without any directory information (e.g. basename "/user/foo/example.sh" will return example.sh). This is important so you don't get false negatives from comparing example.sh and ./example.sh for example.
To adapt this to only allow running when sourced from one specific file as in your question and provide a helpful error message to the end user, you could use:
[ $(basename "$0") = "Start-App.sh" ] || echo "[ERROR] To start MyApplication please run ./Start-App.sh" && exit
As mentioned from the start of the answer, this is not intended as a serious security measure of any kind, but I'm guessing that's not what you're looking for anyway.
You can make startup.sh non-executable by typing chmod -x startup.sh. That way the user would not be able to run it simply by typing ./startup.sh.
Then from Start-App.sh, call your script by explicitly invoking the shell:
sh ./startup.sh arg1 arg2 ...
or
bash ./startup.sh arg1 arg2 ...
You can check which shell it's supposed to run in by inspecting the first line of startup.sh, it should look like:
#!/bin/bash
You can set environment variable in your first script and before running second script check if that environment variable is set properly.
Another alternative is checking the parent process and finding the calling script. This also needs adding some code to the second script.
For example, in the called script, you can check the exit status of this and terminate.
ps $PPID | tail -1 | awk '$NF!~/parent/{exit 1}'
As others have pointed out, the short answer is "no", although you can play with permissions all day but this is still not bulletproof. Since you said you don't mind editing (just not heavily editing) the second script, the best way to accomplish this would be something along the lines of:
1) in the parent/first script, export an environment variable with its PID. This becomes the parent PID. For example,
# bash store parent pid
export FIRST_SCRIPT_PID = $$
2) then very briefly, in the second script, check to see if the calling PID matches the known acceptable parent PID. For example,
# confirm calling pid
if [ $PPID != $FIRST_SCRIPT_PID ] ; then
exit 0
fi
Check out these links here and here for reference.
To recap: the most direct way to do this is adding at least a minimal line or two to the second script, which hopefully doesn't count as "heavily editing".
You can create a script, let's call it check-if-my-env-set containing
#! /bin/bash
source Start-App.sh
exec /bin/bash $#
and replace the shebang (see this) on startup.sh by that script
#! /abs/path/to/check-if-my-env-set
#! /bin/bash
...
then, every time you run startup.sh it will ensure the environment is set correctly.
To the best of my knowledge, there is no way to do this in a way that it would be impossible to get around it.
However, you could stop most attempts by using permissions.
Change the owner of the startup.sh file:
sudo chown app_specific_user startup.sh
Make startup.sh only executable by the owner:
chmod u+x startup.sh
Run startup.sh as the app_specific_user from Start-App.sh:
sudo -u app_specific_user ./startup.sh

start another bash script from main bash script

I have a bash script, from which I need to start another script with source. This is working fine, but I also need to pass the 2nd script parameters.
Example:
source /usr/local/scripts/parallel.sh test --gnu
So I need to start parallel.sh with a given data-source file called test, and I also need to assign a parameter --gnu at the end. But it is not recognizing the file and the parameter.
The source command is likely not what you're looking for.
When a script is run using source it runs within the existing shell, any variables created or modified by the script will remain available after the script completes. In contrast if the script is run just as filename, then a separate subshell (with a completely separate set of variables) would be spawned to run the script.
So, unless you need access to variables or functions inside parallel.sh, just call it directly:
/usr/local/scripts/parallel.sh test --gnu
As long as the script is executable (chmod +x /usr/local/scripts/parallel.sh) and set up to work with the options you're passing ($1 will contain "test" and $2 will contain "--gnu") it should work just fine.

incrementing an environmental variable

I need to increment an environmental variable by these steps:
envar=1
export envar
sh script_incrementation
echo $envar
where script_incrementation contains something like this:
#! /bin/sh
envar=$[envar+1] #I've tried also other methods of incrementation
export envar
Whatever I do, after exiting the script the variable remains with its initial value 1.
THanks for your time.
A shell script executes in its own shell, so you cannot affect the outer shell unless you source it. See this question for details of that discussion.
Consider the following script, which I will call Foo.sh.
#!/bin/bash
export HELLO=$(($HELLO+1))
Suppose in the outer shell, I define an environmental variable:
export HELLO=1
If I run the script like this, it run inside its own shell and will not affect the parent.
./Foo.sh
However, if I source it, it will just execute the commands in the current shell, and will achieve the desired affect.
. Foo.sh
echo $HELLO # prints 2
Your script can not change the environment of the calling process (shell), it merely inherits it.
So, if you export foo=bar, and then invoke sh (a new process) with your script, the script will see the value of $foo (which is "bar"), and it will be able to change its own copy of it – but that is not going to affect the environment of the parent process (where you exported the variable).
You can simply source your script in the original shell, i.e. run
source increment_script.sh
or
. increment_script.sh
and that will then change the value of the variable.
This is because sourceing a script avoids spawning a new shell (process).
Another trick is to have your script output the changed environment, and then eval that output, for example:
counter=$[counter+1]
echo "counter=$counter"
and then run that as
eval `increment_script.sh`

How to run a csh script from a sh script

I was wondering if there is a way to source a csh script from a sh script. Below is an example of what is trying to be implemented:
script1.sh:
#!/bin/sh
source script2
script2:
#!/bin/csh -f
setenv TEST 1234
set path = /home/user/sandbox
When I run sh script1.sh, I get syntax errors generated from script2 (expected since we are using a different Shebang). Is there a way I can run script2 through script1?
Instead of source script2 run it as:
csh -f script2
Since your use case depends on retaining environment variables set by the csh script, try adding this to the beginning of script1:
#!/bin/sh
if [ "$csh_executed" -ne 1 ]; then
csh_executed=1 exec csh -c "source script2;
exec /bin/sh \"$0\" \"\$argv\"" "$#"
fi
# rest of script1
If the csh_executed variable is not set to 1 in the environment, run a csh script that sources script2 then executes an instance of sh, which will retain the changes to the environment made in script2. exec is used to avoid creating new processes for each shell instance, instead just "switching" from one shell to the next. Setting csh_executed in the environment of the csh command ensures that we don't get stuck in a loop when script1 is re-executed by the csh instance.
Unfortunately, there is one drawback that I don't think can be fixed, at least not with my limited knowledge of csh: the second invocation of script1 receives all the original arguments as a single string, rather than a sequence of distinct arguments.
You don't want source there; it runs the given script inside your existing shell, without spawning a subprocess. Obviously, your sh process can't run something like that which isn't a sh script.
Just call the script directly, assuming it is executable:
script2
The closest you can come to sourcing a script with a different executor than your original script is to use exec. exec will replace the running process space with the new process. Unlike source, however, when your exec-ed program ends, the entire process ends. So you can do this:
#!/bin/sh
exec /path/to/csh/script
but you can't do this:
#!/bin/sh
exec /path/to/csh/script
some-other-command
However, are you sure you really want to source the script? Maybe you just want to run it in a subprocess:
#!/bin/sh
csh -f /path/to/csh/script
some-other-command
You want the settings in your csh script to apply to the sh script that invokes it.
Basically, you can't do that, though there are some (rather ugly) ways you could make it work. If you execute your csh script, it will set those variables in the context of the process running the script; they'll vanish as soon as it returns to the caller.
Your best bet is simply to write a new version of your csh script as an sh script, and source or . it from the calling sh script.
You could translate your csh script:
#!/bin/csh -f
setenv TEST 1234
set path = /home/user/sandbox
to this:
export TEST=1234
export PATH=/home/user/sandbox
(csh treats the shell array variable $path specially, tying it to the environment variable $PATH. sh and its derivatives don't do that, they deal with $PATH itself directly.)
Note that a script intended to be sourced should not have a #! line at the top, since it doesn't make sense to execute it in its own process; you need to execute its contents in the context of the caller.
If maintaining two copies of the script, one to be sourced from csh or tcsh scripts and another to be sourced or .ed from sh/ksh/bash/zsh script, is not practical, there are other solutions. For example, your script can print a series of sh commands to be executed; you can then do something like
eval `./foo.csh`
(line endings will pose some issues here).
Or you can modify the csh script so it sets the required environment variables and then invokes some specified command, which could be a new interactive shell; this is inconvenient, since it doesn't set those variables in the interactive shell you're running.
If a software package requires some special environment variables to be set, it's common practice to provide scripts called, for example, setup.sh and setup.csh, so that sh/ksh/bash/zsh users can do:
. /path/to/package/setup.sh
and csh/tcsh users can do:
source /path/to/package/setup.csh
Incidentally, this command:
set path = /home/user/sandbox
in your sample script is probably not a good idea. It replaces your entire $PATH with just a single directory, which means you won't be able to execute simple commands like ls unless you specify their full paths. You'd usually want something like:
set path = ( $path /home/user/sandbox )
or, in sh:
PATH=$PATH:/home/user/sandbox

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