I see a weird behavior in shell scripts when I pass a variable with parameters to external ruby script
For example:
params="--val1=test --val2='test'"
ruby ./script.rb
causes ruby to output 'test' for var2 instead of test.
If I just pass params directly without using a variable everything works just fine.
Could you please clarify your question a bit?
From what I understand you have a shell script, something like:
#!/bin/bash
PARAMS="--val1=test --val2='test'"
ruby ./script.rb $PARAMS
And in script.rb you print out the value for the command line parameter val2. In this case it's expected that it prints out test instead of 'test', because the following steps are happening:
bash replaces $PARAMS with its value
bash tries to execute the line ruby ./script.rb --val1=test --val2='test'
now bash sees the quoted value 'test' and replaces it with test, so ruby / your script sees test
Related
I have a BASH script that has a long set of arguments and two ways of calling it:
my_script --option1 value --option2 value ... etc
or
my_script val1 val2 val3 ..... valn
This script in turn compiles and runs a large FORTRAN code suite that eventually produces a netcdf file as output. I already have all the metadata in the netcdf output global attributes, but it would be really nice to also include the full run command one used to create that experiment. Thus another user who receives the netcdf file could simply reenter the run command to rerun the experiment, without having to piece together all the options.
So that is a long way of saying, in my BASH script, how do I get the last command entered from the parent shell and put it in a variable? i.e. the script is asking "how was I called?"
I could try to piece it together from the option list, but the very long option list and two interface methods would make this long and arduous, and I am sure there is a simple way.
I found this helpful page:
BASH: echoing the last command run
but this only seems to work to get the last command executed within the script itself. The asker also refers to use of history, but the answers seem to imply that the history will only contain the command after the programme has completed.
Many thanks if any of you have any idea.
You can try the following:
myInvocation="$(printf %q "$BASH_SOURCE")$((($#)) && printf ' %q' "$#")"
$BASH_SOURCE refers to the running script (as invoked), and $# is the array of arguments; (($#)) && ensures that the following printf command is only executed if at least 1 argument was passed; printf %q is explained below.
While this won't always be a verbatim copy of your command line, it'll be equivalent - the string you get is reusable as a shell command.
chepner points out in a comment that this approach will only capture what the original arguments were ultimately expanded to:
For instance, if the original command was my_script $USER "$(date +%s)", $myInvocation will not reflect these arguments as-is, but will rather contain what the shell expanded them to; e.g., my_script jdoe 1460644812
chepner also points that out that getting the actual raw command line as received by the parent process will be (next to) impossible. Do tell me if you know of a way.
However, if you're prepared to ask users to do extra work when invoking your script or you can get them to invoke your script through an alias you define - which is obviously tricky - there is a solution; see bottom.
Note that use of printf %q is crucial to preserving the boundaries between arguments - if your original arguments had embedded spaces, something like $0 $* would result in a different command.
printf %q also protects against other shell metacharacters (e.g., |) embedded in arguments.
printf %q quotes the given argument for reuse as a single argument in a shell command, applying the necessary quoting; e.g.:
$ printf %q 'a |b'
a\ \|b
a\ \|b is equivalent to single-quoted string 'a |b' from the shell's perspective, but this example shows how the resulting representation is not necessarily the same as the input representation.
Incidentally, ksh and zsh also support printf %q, and ksh actually outputs 'a |b' in this case.
If you're prepared to modify how your script is invoked, you can pass $BASH_COMMANDas an extra argument: $BASH_COMMAND contains the raw[1]
command line of the currently executing command.
For simplicity of processing inside the script, pass it as the first argument (note that the double quotes are required to preserve the value as a single argument):
my_script "$BASH_COMMAND" --option1 value --option2
Inside your script:
# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
myInvocation=$1 # Save the command line in a variable...
shift # ... and remove it from "$#".
# Now process "$#", as you normally would.
Unfortunately, there are only two options when it comes to ensuring that your script is invoked this way, and they're both suboptimal:
The end user has to invoke the script this way - which is obviously tricky and fragile (you could however, check in your script whether the first argument contains the script name and error out, if not).
Alternatively, provide an alias that wraps the passing of $BASH_COMMAND as follows:
alias my_script='/path/to/my_script "$BASH_COMMAND"'
The tricky part is that this alias must be defined in all end users' shell initialization files to ensure that it's available.
Also, inside your script, you'd have to do extra work to re-transform the alias-expanded version of the command line into its aliased form:
# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
# Here we also re-transform the alias-expanded command line to
# its original aliased form, by replacing everything up to and including
# "$BASH_COMMMAND" with the alias name.
myInvocation=$(sed 's/^.* "\$BASH_COMMAND"/my_script/' <<<"$1")
shift # Remove the first argument from "$#".
# Now process "$#", as you normally would.
Sadly, wrapping the invocation via a script or function is not an option, because the $BASH_COMMAND truly only ever reports the current command's command line, which in the case of a script or function wrapper would be the line inside that wrapper.
[1] The only thing that gets expanded are aliases, so if you invoked your script via an alias, you'll still see the underlying script in $BASH_COMMAND, but that's generally desirable, given that aliases are user-specific.
All other arguments and even input/output redirections, including process substitutiions <(...) are reflected as-is.
"$0" contains the script's name, "$#" contains the parameters.
Do you mean something like echo $0 $*?
I've written a small bash script to invoke a make command.
The makefile expects args='<arguments to be passed to program>' argument. (i.e. make a args='--aaa 5 --bbb 6')
The problem is when I use the script to invoke the make command, the argument is truncated on its first white space.
For example, when executing make a args='--aaa 5' through the script, the args variable is '--aaa and not '--aaa 5'
I've also tried adding quotes and single quotes but the result is the same.
When I invoke the make command manually through terminal the, the args variable gets all the arguments and their values as expected.
Here is the script call :
args="args='--aaa 5 --bbb 6'"
make a ${args}
You have to double-quote $args to preserve the whitespace.
make a "$args"
I'm assuming the Makefile correctly handles the value once it receives it.
I am new to shell scripting just started off.
I have written this script
#!/bin/sh
profile_type= cat /www/data/profile.conf
echo $profile_type
the o/p of this script is
. /tmp/S_panicA1.txt
. /tmp/S_panicA0.txt
away_Def="panicA1 panicA0"
away_Byp=0
away_Sts=$((panicA1+panicA0-away_Byp))
In this i want to get panicA1 panicA0 and 0 and store it in other variable how to do this?
When you want to assign the output of a command to a variable, you use the dollar parenthesis syntax.
foo=$(cat /my/file)
You can also use the backticks syntax.
foo=`cat /my/file`
In your script, you simply run the command cat and assign its result, nothing, to your variable. Hence the output consisting of the content of your file, result of cat, followed by an empty line, result of echo with an empty variable.
I'm trying to run the Apache startup script, /etc/init.d/httpd. Environment variable definitions like this one give an error:
CONF_FILE=$(APACHE_HOME)/conf/httpd.conf
It says "/etc/init.d/httpd: line 15: APACHE_HOME: command not found"
So, I replaced the parentheses with curly brackets, and the script worked swimmingly. What gives? I'm really just asking this question because I want to understand why it's wrong, not how to fix it. The shebang is there, and it's unmodified from the original shell script, so why's it misinterpreting things?
In unix systems:
$SOMETHING /* variable */
$(SOMETHINGELSE) /* command */
${FOO} */ variable substitution */
$(...) executes its contents in a subshell, it doesn't get the value of a variable. You can use just plain $APACHE_HOME or ${APACHE_HOME}, which it sounds like you switched to.
$(something) tells the shell to execute command something and substitute the command's output.1
You want to substitute a variable's output, so you just need a $ in front of the variable, like so: CONF_FILE=$APACHE_HOME/conf/httpd.conf
Alternatively, you could use CONF_FILE=${APACHE_HOME}/conf/httpd.conf (note the curly braces instead of parenthesis), but it's not really necessary for your situation.
1This is useful when you want to assign a command's output to a variable. For example:
MY_VAR="$(egrep 'someline' somefile.txt)"
I have a groovy file named test.groovy and have a single line of coding in it :
println args[0];
when I run this program like this groovy test ants, output is ants.
But when I run the program with the argument ants( then I'm getting error like this :
bash: syntax error near unexpected token (
1)If I escape the character ( then I'm getting the output as ants(. But why ( is needed to be escaped?
And when I run the program with the argument ant's, then clicking enter would make my terminal look like this :
>
>
>
2)And I terminate the program only using ctrl+c. What actually happens in this case? Why my terminal look like this?
3)After seeing these, what are the rules and condition that are to be followed in Groovy with accordance with Command-line arguments and the same holds for Java?
Thanks in advance.
You need to escape it as ( has a meaning in the bash shell which you are using.
The same goes for '
Try other commands:
ls (
Or
ls '
You'll get the same effect
Another option (other than escaping) is to put your arguments inside quote chars like so:
groovy test 'ants('