LinuxQuestions.org - processing command args in bash

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- Programming (https://www.linuxquestions.org/questions/programming-9/)

- - processing command args in bash (https://www.linuxquestions.org/questions/programming-9/processing-command-args-in-bash-367563/)

slzckboy

09-27-2005 01:57 PM

processing command args in bash

#!/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 (Post 1873795)

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

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