Assume a script master.sh, which is called as
./foo/bar/master.sh
and contains the lines
#!/bin/zsh
. ./x/y/slave.sh
Is it possible to find out from within slave.sh, that the script which is doing the sourcing, is ./foo/bar/master.sh ?
I can not use $0 here, because this would return ./x/y/slave.sh.
I'm using zsh 5.0.6
one way you can achieve this is that for the child script to take as optional argument the name of the caller. Thus this would be accessible with `$1``
ex:
#!/bin/zsh
# master/leader
. ./x/y/slave.sh $0 # or hardcoded path
#!/bin/zsh
# slave/worker
echo "Here is my master $1"
(you can also do another custom protocol using a environment variable set by the master)
(this solution would also works on bash, and other shell)
The information can already be obtained in zsh right now (thanks to Bart Schaefer, who pointed out to me the existence of the variable functrace in the zsh/parameter module):
#!/bin/zsh
# slave/worker
zmodload zsh/parameter
echo "Here is my master ${functrace[$#functrace]%:*}"
The '%:*' is necessary, because the entries in the functrace array also contain the line number of the call.
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 wrote hook for command line:
# Transforms command 'ls?' to 'man ls'
function question_to_man() {
if [[ $2 =~ '^\w+\?$' ]]; then
man ${2[0,-2]}
fi
}
autoload -Uz add-zsh-hook
add-zsh-hook preexec question_to_man
But when I do:
> ls?
After exiting from man I get:
> zsh: no matches found: ls?
How can I get rid of from message about wrong command?
? is special to zsh and is the wildcard for a single character. That means that if you type ls? zsh tries find matching file names in the current directory (any three letter name starting with "ls").
There are two ways to work around that:
You can make "?" "unspecial" by quoting it: ls\?, 'ls?' or "ls?".
You make zsh handle the cases where it does not match better:
The default behaviour if no match can be found is to print an error. This can be changed by disabling the NOMATCH option (also NULL_GLOB must not be set):
setopt NO_NOMATCH
setopt NO_NULL_GLOB
This will leave the word untouched, if there is no matching file.
Caution: In the (maybe unlikely) case that there is a file with a matching name, zsh will try to execute a command with the name of the first matching file. That is if there is a file named "lsx", then ls? will be replaced by lsx and zsh will try to run it. This may or may not fail, but will most likely not be the desired effect.
Both methods have their pro and cons. 1. is probably not exactly what you are looking for and 2. does not work every time as well as changes your shells behaviour.
Also (as #chepner noted in his comment) preexec runs additionally to not instead of a command. That means you may get the help for ls but zsh will still try to run ls? or even lsx (or another matching name).
To avoid that, I would suggest defining a command_not_found_handler function instead of preexec. From the zsh manual:
If no external command is found but a function command_not_found_handler exists the shell executes this function with all command line arguments. The function should return status zero if it successfully handled the command, or non-zero status if it failed. In the latter case the standard handling is applied: ‘command not found’ is printed to standard error and the shell exits with status 127. Note that the handler is executed in a subshell forked to execute an external command, hence changes to directories, shell parameters, etc. have no effect on the main shell.
So this should do the trick:
command_not_found_handler () {
if [[ $1 =~ '\?$' ]]; then
man ${1%\?}
return 0
else
return 1
fi
}
If you have a lot of matching file names but seldomly misstype commands (the usual reason for "Command not found" errors) you might want to consider using this instead:
command_not_found_handler () {
man ${1%?}
}
This does not check for "?" at the end, but just cuts away any last character (note the missing "\" in ${1%?}) and tries to run man on the rest. So even if a file name matches, man will be run unless there is indeed a command with the same name as the matched file.
Note: This will interfere with other tools using command_not_found_handler for example the command-not-found tool from Ubuntu (if enabled for zsh).
That all being said, zsh has a widget called run-help which can be bound to a key (in Emacs mode it is by default bound to Alt+H) and than runs man for the current command.
The main advantages of using run-help over the above are:
You can call it any time while typing a longer command, as long as the command name is complete.
After you leave the manpage, the command is still there unchanged, so you can continue writing on it.
You can even bind it to Alt+? to make it more similar: bindkey '^[?' run-help
In /etc/init.d/functions#_pids_var_run (which will be there in most Linux distros), i see a local variable initialization like this,
local base=${1##*/}
what does,
${1##*/}
get resolved to?
In bash scripts, ${varname##pattern} removes the longest prefix that matches pattern from varname. For example,
foo=bar/baz/qux
echo "${foo##*/}"
will print
qux
In the case of
${1##*/}
it will remove the longest prefix that matches */ from $1, which is the first parameter passed to the script/function.
In essence, it is a poor man's $(basename "$1").
If I am writing a bash script, and I choose to use a config file for parameters. Can I still pass in parameters for it via the command line? I guess I'm asking can I do both on the same command?
The watered down code:
#!/bin/bash
source builder.conf
function xmitBuildFile {
for IP in "{SERVER_LIST[#]}"
do
echo $1#$IP
done
}
xmitBuildFile
builder.conf:
SERVER_LIST=( 192.168.2.119 10.20.205.67 )
$bash> ./builder.sh myname
My expected output should be myname#192.168.2.119 and myname#10.20.205.67, but when I do an $ echo $#, I am getting 0, even when I passed in 'myname' on the command line.
Assuming the "config file" is just a piece of shell sourced into the main script (usually containing definitions of some variables), like this:
. /etc/script.conf
of course you can use the positional parameters anywhere (before or after ". /etc/..."):
echo "$#"
test -n "$1" && ...
you can even define them in the script or in the very same config file:
test $# = 0 && set -- a b c
Yes, you can. Furthemore, it depends on your architecture of script. You can overwrite parametrs with values from config and vice versa.
By the way shflags may be pretty useful in writing such script.
I'm writing a cPanel postwwwact script, if you're not familiar with the script its run after a new account is created. it relies on the user account variable being passed to the script which i then use for various things (creating databases etc). However, I can't seem to find the right way to access the variable i want. I'm not that good with shell scripts so i'd appreciate some advice. I had read somewhere that the value i wanted would be included in $ARGV{'user'} but this simply gives "root" as opposed to the value i need. I've tried looping through all the arguments (list of arguments here) like this:
#!/bin/sh
for var
do
touch /root/testvars/$var
done
and the value i want is in there, i'm just not sure how to accurately target it. There's info here on doing this with PHP or Perl but i have to do this as a shell script.
EDIT Ideally i would like to be able to call the variable by something other than $1 or $2 etc as this would create issues if an argument is added or removed
..for example in the PHP code here:
function argv2array ($argv) {
$opts = array();
$argv0 = array_shift($argv);
while(count($argv)) {
$key = array_shift($argv);
$value = array_shift($argv);
$opts[$key] = $value;
}
return $opts;
}
// allows you to do the following:
$opts = argv2array($argv);
echo $opts[‘user’];
Any ideas?
The parameters are passed to your script as a hash:
/scripts/$hookname user $user password $password
You can use associative arrays in Bash 4, or in earlier versions of Bash you can use built up variable names.
#!/bin/bash
# Bash >= 4
declare -A argv
for ((i=1;i<=${##};i+=2))
do
argv[${#:i:1}]="${#:$((i+1)):1}"
done
echo ${argv['user']}
Or
#!/bin/bash
# Bash < 4
for ((i=1;i<=${##};i+=2))
do
declare ARGV${#:i:1}="${#:$((i+1)):1}"
done
echo ${!ARGV*} # outputs all variable names that begin with ARGV
echo $ARGVuser
Running either:
$ ./argvtest user dennis password secret
dennis
Note: you can also use shift to step through the arguments, but it's destructive and the methods above leave $# ($1, $2, etc.) in place.
#!/bin/bash
# Bash < 4
# using shift (can use in Bash 4, also)
for ((i=1;i<=${##}+2;i++))
do
declare ARGV$1="$2"
# Bash 4: argv[$1}]="$2"
shift 2
done
echo ${!ARGV*}
echo $ARGVuser
If it's passed as a command-line parameter to the script, it's available as $1 if it's first parameter, $2 for the second, and so on.
Why not start off your script with something like
ARG_USER=$1
ARG_FOO=$2
ARG_BAR=$3
And then later in your script refer to $ARG_USER, $ARG_FOO and $ARG_BAR instead of $1, $2, and $3. That way, if you decide to change the order of arguments, or insert a new argument somewhere other than at the end, there is only one place in your code that you need to update the association between argument order and argument meaning.
You could even do more complex processing of $* to set your $ARG_WHATEVER variables, if it's not always going to be that all of the are specified in the same order every time.
You can do the following:
#!/bin/bash
for var in $argv; do
<do whatver you want with $var>
done
And then, invoke the script as:
$ /path/to/script param1 arg2 item3 item4 etc