How to programmatically get the name of current gnuplot script? I know that I can call gnuplot script from bash and get it file name but I am wondering if it is possible from inside gnuplot. My goal is to make something like:
date=system("date +%F_%T | sed 's/:/-/g'")
my_name=$0 # THIS IS HOW TO DO IT IN BASH
set term png
set output my_name.date.".png"
I've tried:
my_name=system("cat /proc/$$/cmdline")
but it returned sh instead of script name
Not quite an answer to your question, but this might help with what you want to do:
You can leave my_name unset in the script, and set it either inside gnuplot, just before you load the script (where you need to know the script name anyway):
my_name=...
load(my_name)
or set it when you invoke gnuplot from the shell:
$ gnuplot -e "my_name=${FILE}" ${FILE}
A few more things:
date=system("date +%F_%T | sed 's/:/-/g'")
can be replaced with
date=system("date +%F_%H-%M-%S")
(which is shorter and doesn't need to be parsed through sed) or without any forking at all:
date=strftime("%F_%H-%M-%S",time(0.0))
Using gnuplot version 5 you have access to the file called with load via the variable ARG0
Consider the script test.gp which contains only
print ARG0
Now, calling this with
gnuplot -e "load 'test.gp'"
prints you test.gp on the screen. With earlier versions you don't have access to a similar variable (also not when using call). For earlier versions you must stick to one of the solutions given by #chw21
Related
I am working on Bash 5.0 from GNU repository. I wanted to find the place where Bash reads a string with ASCII colors and convert it to colors, like in the following case where it convert "Hello" to red:
root#ubuntu:~/Desktop/bash-5.0# ./bash
root#ubuntu:~/Desktop/bash-5.0# echo $BASH_VERSION
5.0.0(8)-release
root#ubuntu:~/Desktop/bash-5.0# ./bash -c 'echo -e "\033[31mHello\e[0m World"'
Hello World
I searched inside the source code and found two files that seems to be related:
bash-5.0/lib/readline/colors.c - link
bash-5.0/lib/readline/parse-colors.c - link
But they are not, they work only on the first time I load Bash and you need to write the following rows in the file ~/.inputrc for it to work:
set colored-completion-prefix on
set colored-stats on
Any idea where in the code Bash takes string like that "\033[31mHello" and convert it to red?
It's not the shell that's converting anything to colors, it is your terminal. The shell only outputs ANSI escape codes which are then picked up by the terminal.
Depending on your point of view and philosophical interpretations, \033[31mHello already is a colored string (for the shell, at least, it is)
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 $*?
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.
What I'm trying to accomplish is having a central configuration file, in bash, that defines some variables that are re-used in different bash files. The example below attempts to generate a file name with the current date included in the file name as well as a variable defined in another shell script. However whenever I try to concatenate this external variable it doesn't work. I can concatenate the variable in any other situation.
Example Code:
../config/vars.sh
#!/bin/bash
mysqlUser="backupuser"
mysqlPwd="fakePwd"
mysqlSocket="/var/run/mysqld/mysqld.sock"
mysqlPort="3306"
serverName="s01.catchyservername.com"
./dbBackup.sh
#!/bin/bash
source ../config/vars.sh
tempName=$(date +"%Y%m%d.sql.gz")
fileName="mysqld_${mysqlPort}_${tempName}"
echo "mysqld_${mysqlPort}"
echo ${tempName}
echo ${fileName}
output of dbBackup.sh
mysqld_3306
20140926.sql.gz
_20140926.sql.gz
As you can see when echoing "mysqld_${mysqlPort}" I get the expected output, but when echoing ${fileName} the entire first half of the string is ignored. What am I misunderstanding?
Your vars.sh file was probably created with a DOS/windows text editor:
$ ./dbBackup.sh
mysqld_3306
20140926.sql.gz
_20140926.sql.gz
$ dos2unix vars.sh
dos2unix: converting file vars.sh to Unix format ...
$
$ ./dbBackup.sh
mysqld_3306
20140926.sql.gz
mysqld_3306_20140926.sql.gz
$
As you can see above, I use the dos2unix utility to convert the line separators to Unix style.
I'm using Gnuplot as my back-end plotter and I often use the following setup :
#Filename : my_plot.gnuplot
set terminal pdfcairo [my_options]
set output 'my_plot.pdf'
....
coupled with a Makefile :
%.pdf : %.gnuplot
gnuplot $<
My question is simple : is there a command / way to refer to the name of the script inside the script (the equivalent of bash's $0) and set the output with a clever sprintf or equivalent ?
You cannot access the script name from gnuplot, but you can give a parameter when calling the script, which should work fine, especially when using Makefiles.
%.pdf: %.gnuplot
gnuplot -e "scriptname='$<'" $<