processing command args in bash

slzckboy · 09-27-2005, 01:57 PM

#!/bin/bash noob here.

i'm comfortable with C but i think some simpler tasks on my system will be more easily achived with bash.
I'd also like to getter a better grasp of all my config files and initialisation scripts so here i go..

Code:

if [ "$#" -eq 0 ];then
	echo "pls state the name of the file "
	else
	echo "on we go" 
	torrent=$#
	echo $torrent # i wanted this to print out the name of the parsed cmd line arg 0:(
	echo $1          #this obviously printed the name of the file i parsed at cmd line
	fi
	if [ -f $torrent ];then 
	echo "the file is here"
	else
	echo "no such file"
	fi

pls see the comments in the middle of the code.
the program will process args of an indeterminate number but it will stipulate what
the last arg will be and thus i need acces to it,i.e test that it is there etc

i cant code [ -f $n .... where n is some integer value
because i will not know what the integer value will be ahead of time.
how do i achieve this pls?

Hko · 09-27-2005, 04:21 PM

Quote:

i cant code [ -f $n .... where n is some integer value
because i will not know what the integer value will be ahead of time.
how do i achieve this pls?

Code:

#!/bin/bash

# Processing of command line arguments in bash
# "shift" shifts all arguments to the left ($2=$1, $3=$2 etc.)
# "shift" returns non-zero exit-code (indicating "false")
# when no more arguments available.
# (note that $0 will not get "shifted").


ARGV0=$0 # First argument is shell command (as in C)
echo "Command: $ARGV0"

ARGC=$#  # Number of args, not counting $0
echo "Number of args: $ARGC"

i=1  # Used as argument index
while true; do
	if [ $1 ]; then
		echo "Argv[$i] = $1" # Always print $1, shifting $2 into $1 later
		shift
	else
		break # Stop loop when no more args.
	fi
	i=$((i+1))
done
echo "Done."

Hko · 09-27-2005, 04:30 PM

Or, without "shift":

("${!i}" is actually the answer to your question "i cant code [ -f $n .... where n is some integer value")

Code:

#!/bin/bash

# Or, without "shift": use loop counter until $# ( = number of args)

ARGV0=$0 # First argument is shell command (as in C)
echo "Command: $ARGV0"

ARGC=$#  # Number of args, not counting $0
echo "Number of args: $ARGC"

i=1  # Used as argument index
while [ $i -le $ARGC ]; do # "-le" means "less or equal", see "man test".
	# "${!i} "expands" (resolves) to $1, $2,.. by first expanding i, and
	# using the result (1,2,3..) as a variable which is then expanded.
	echo "Argv[$i] = ${!i}"
	i=$((i+1))
done
echo "Done."

koodoo · 09-27-2005, 05:41 PM

Hey Slzckboy,

Didn't quite understand what you were trying to do. The first comment seemed to be confusing. But I guess your actual problem was parsing the last argument. Yes, the solution to that is using 'shift' or "${!i}" as Hko suggested. I guess you wanted this :
a) Invoke the script with n no. of arguments,
b) If n = 0 ask the user to input the name of the file, else
c) if n != 0 assume that the last argument is the name of a file, and check to see if that file exists or not.

Here's how the above script would look with shift :

Code:

#!/bin/bash
        if [ "$#" -eq 0 ];then
	echo "pls state the name of the file "
	
	
	
	else
		echo "on we go" 
				
			while [ "$2" != "" ]; do
			shift
			done
				
		if [ -f $1 ];then 
		echo "Searching for file $1 : the file is here"
		else
		echo "Searching for file $1 : no such file"
		fi

	fi

exit 0

If there's something more you wanted please state that.
Hope this helps !

slzckboy · 09-28-2005, 03:20 AM

no..thats fine . That is exactly what i meant.

thnks

slzckboy · 09-28-2005, 04:36 AM

i think i will go the non "shift" way or else I will loose the data passed in on the previous args.!!?!

bigearsbilly · 09-28-2005, 08:00 AM

for parsing of command lines with flags,
man getopts

make life easy for yourself!

e.g parsing -d and -f flags, where -f requires an argument:

Code:

    while getopts df: flag
    do
        case $flag in

            d)
                  echo debugging on
                  ;;
            f)
                 file=$OPTARG
                 echo filename is $file
                 ;;
            ?)
                exit
                ;;
        esac
   done
shift $(( OPTIND - 1 ))  # shift past the last flag or argument

echo parameters are $*

Result

Code:

$ bash opt.sh -df  blah one two three
debugging on
filename is blah
parameters are one two three

$ bash opt.sh -f 
opt.sh: option requires an argument -- f

$ bash opt.sh -r 
opt.sh: illegal option -- r

Hko · 09-28-2005, 10:51 AM

Quote:

Originally posted by slzckboy
i think i will go the non "shift" way or else I will loose the data passed in on the previous args.!!?!

...unless you store them in shell-variables of course.

slzckboy · 09-28-2005, 11:04 AM

hmm...

ok..

well i'd value an opinion on what I have done thus far

i'm getting
"parameter passed in at end with no value" spat back at me.

Code:

#!/bin/sh

	argv[0]=$0
	bittorrent="btdownloadheadless.py"
	#   set default values
	up_speed=2
	min_port=49152
	max_port=65535
	
	if [ "$#" -eq 0 ];then
	echo "pls state the name of the file"
	exit 0
	else
	echo "on we go"
	torrent=$#
	fi

	i=1	
	if  [ -f "${!torrent}" ]; then
	echo "${!torrent} file found"
	while [ $i -lt $#  ];do
	case "${!i}" in
	"--max_upload_rate")
	i=$((i+1))
	up_speed=${!i};;
	esac
	i=$((i+1))
	done

	exec /usr/local/bin/$bittorrent --max_upload_rate=$up_speed		
	
	else
	echo "${!torrent} not found"
	fi

basically wots wrong with doing.....
exec /usr/local/bin/$bittorrent --max_upload_rate=$up_speed ????

i tried "man getopts" but strangely the man page is not on my system.
will have to google 4 it.

koodoo · 09-28-2005, 02:06 PM

Quote:

Originally posted by slzckboy

i tried "man getopts" but strangely the man page is not on my system.
will have to google 4 it.

Try "man getopt"

slzckboy · 09-28-2005, 04:49 PM

hey ..

thnks 4 everyones help.

i have a working base now that I can expand on.
I have worked out the bugs in the above posted code
No need for getopts for now but I will keep it in mind for later use.

tony4260 · 09-27-2017, 09:10 PM

I'm having some trouble wrapping my head around the {!1} part. I love seeing different ways to do the same thing (comparing this to the earlier shift example) it really helps me gain a deeper understanding of said language (and programming in general).

ARGV[$i] is already the name so what different is the {$i} doing? Is it like command substitution?
I've done some research before asking and I've come across some info about hash keys, but I'm not sure that's really whats going on here. Thanks so much!

Quote:

Originally Posted by Hko

Or, without "shift":

("${!i}" is actually the answer to your question "i cant code [ -f $n .... where n is some integer value")

Code:

#!/bin/bash

# Or, without "shift": use loop counter until $# ( = number of args)

ARGV0=$0 # First argument is shell command (as in C)
echo "Command: $ARGV0"

ARGC=$#  # Number of args, not counting $0
echo "Number of args: $ARGC"

i=1  # Used as argument index
while [ $i -le $ARGC ]; do # "-le" means "less or equal", see "man test".
	# "${!i} "expands" (resolves) to $1, $2,.. by first expanding i, and
	# using the result (1,2,3..) as a variable which is then expanded.
	echo "Argv[$i] = ${!i}"
	i=$((i+1))
done
echo "Done."

David the H. · 10-03-2017, 04:17 PM

The problem to solve is that we need to access a parameter by name (number), without knowing the name in advance. We need to perform an indirect substitution, in other words.

The "${!var}" pattern provides a way to do such indirect substitutions. It expands to the value of the variable given, then again to the value of the variable generated by it. So if "$2 = bar" and "i=2", then "${!i}" will expand to "bar". Read the bash man page under Parameter Expansion for more detail.

Note, by the way, that bash does not automatically have an ARGV array, that's why a loop like the one above is (was) needed to iterate through and print all the values. The output produced by it would look like this:

Code:

Argv[1] = foo
Argv[2] = bar
Argv[3] = baz
Argv[4] = bum

Assuming those four words were fed to the script as arguments.

Frankly, though, I would've just used a simple c-style for loop instead:

Code:

for ((i=1;i<=$#;i++)); do echo "ARGV[$i] = ${!i}"; done

Actually, most of what's written here is rather outdated now. Today's bash has better tools available. You can now call arguments directly with ranges, for example:

Code:

echo "${@: -1}"    # prints the last argument. Note that a space, or parentheses, is necessary for negative numbers.
echo "${@:(-2):2}" # prints the last two arguments.

There are also nameref variables that can provide more powerful indirection ability, although I'm not sure how useful they are in this specific case.

I myself would personally just go the simple route and copy the argument list into a new dedicated ARGV-style array. You can use both ranges and negative indexes that way.

Code:

ARGV=( "$0" "$@" )
echo "${ARGV[0]}"
echo "${ARGV[-1]}"
echo "${ARGV[@]: -1}"
echo "${ARGV[@]:(-2):2}"

With "${ARGV[0]}" being the script name, naturally, and "${ARGV[-1]}" being the last argument, just like "${ARGV[@]: -1}", but also making it possible to apply an additional parameter expansion to the output (e.g. "${ARGV[-1],,}" would lowercase the string at the same time).

tony4260 · 10-03-2017, 04:48 PM

That for loop makes it easier to see.

start at 1, until number of args, i++. Thanks

I'm taking it that these two lines are equivalent:

echo "${@: (-2):2}" # prints the last two arguments.
echo "${ARGV[@]: (-2):2}"

And in those lines the -2 is saying "start 2 back from the last" (the last is -1, the one before that is -2 so:

./argumentstuff.bash this that those many
many would be -1, so the -2 is starting at those.

Then the 2 thats after the -2 is saying do that to two items. so if it was ${@: (-2):1} would only delete "those"?

makyo · 10-09-2017, 09:49 AM

Hi.

Some additional sources that I have found useful:

Code:

Process command-line (CLI) options, arguments

        1) getopts, builtin, bash, ksh, zsh
           http://stackoverflow.com/questions/402377/using-getopts-in-bash-shell-script-to-get-long-and-short-command-line-options
           Also, search for option processing, including:
           http://mywiki.wooledge.org/ComplexOptionParsing

        2) perl: libgetopt-euclid-perl, creates man page, help automatically

        3) getopt, enhanced getopts, part of the util-linux, allows GNU "--"
           Examples: /usr/share/doc/util-linux/examples, 2016.03.27

        4) argp.sh, wrapper for getopt, creates man and help (text, XML), etc.
           Allows mixed options and arguments.
           Compiled argp.c -> argp for faster execution.
           https://sourceforge.net/projects/argpsh/, 2016.03.27

        5) shflags, wrapper for getopt, creates help, allow mixed options
           and arguments
           https://github.com/kward/shflags, 2016.08.01

        6) ksh getopts, enhanced, process GNU "--", creates man, help, etc.
           Examples: Learning the Korn Shell, O'Reilly, 2nd, p 380ff

        7) zsh zparseopts
           man zshmodules, part of zshutil

        8) Suggested option names:
           http://www.shelldorado.com/goodcoding/cmdargs.html#flagnames

Best wishes ... cheers, makyo