Using same getopts arguments for different options - linux

I created a bash script and i need help with getopts
Each options have arguments :
while getopts ":a:b:c:" opt
do
case $opt in
a) echo "$OPTARG";;
b) echo "$OPTARG";;
c) echo "$OPTARG";;
/?) echo "wrong option $OPTARG";;
esac
done
If i want to use the same argument, i need to type -option arg for each options
./script.sh -a hello -b hello -c hello
hello
hello
hello
I would like to be able to use the script like a command, typing options all together when i have the same argument :
./script.sh -abc hello
Unfortunately bc is now the argument :
bc
Does anyone could help me please ?
Regards,

Related

Combining multiple options into one single option (Getopts)

Due to my lack of thorough understanding using getopts, the title is definitely vague :0. I am currently writing a bash script and I would like to add an option that outputs the other options within the case statement in getopts. For the sake of scaling, I have shortened the program.
#!/bin/bash
while getopts :abc opt
do
case $opt in
a)
echo "Hello"
;;
b)
echo "Goodbye"
c)
:ab #****I WANT -c TO OUTPUT THE RESULTS OF a and b************
;;
esac
done
As you can see in option c, I would like this particular option (-c) to put out both the results of -a and -b. Is there a way to go about this by simply making c call on option a and b?
you can introduce functions to reduce duplications, something like this:
#!/bin/bash
do_a() {
echo "Hello"
}
do_b() {
echo "Goodbye"
}
while getopts :abc opt
do
case $opt in
a)
do_a
;;
b)
do_b
;;
c)
do_a
do_b
;;
esac
done
If you are using a recent version of Bash, instead of terminating case clauses with ;; you could use bash specific ;;& with multiple patterns:
#!/bin/bash
while getopts :abc opt
do
case $opt in
a|c)
echo "Hello"
;;&
b|c)
echo "Goodbye"
;;&
esac
done
And:
$ bash script.bash -a
Hello
$ bash script.bash -c
Hello
Goodbye
Using ‘;;&’ in place of ‘;;’ causes the shell to test the patterns in the next clause, if any, and execute any associated command-list on a successful match.

getopts: How to accept arguments that aren't tied to an option in my script? [duplicate]

This question already has answers here:
How do I parse command line arguments in Bash?
(40 answers)
Closed 6 years ago.
I'm writing a Bash script that has the following usage:
ci-badge -t|-a [-b branch] [-m] [-d description] [-l [link] ] repo
Examples:
$ ci-badge -t foo/bar
$ ci-badge -ab dev foo/bar -m
$ ci-badge quux/bar -md 'Hello, world.'
More samples can be found here on GitHub. Anyway, I'm wondering exactly how to implement argument parsing for this script using getopts. After a few hours looking at this basic getopts guide and scouring SO, this is what my code looks so far:
#!/usr/bin/env bash
generate_url=false # set to true if -l is passed with no arg
markdown=false # true -> surround the URL with markdown
# Parse options
while getopts ':tab:md:l:' opt
do
case "$opt" in
a) service=appveyor ;;
b) branch=$OPTARG ;;
d) description=$OPTARG ;;
l) url=$OPTARG ;;
m) markdown=true ;;
t) service=travis ;;
# Colon: runs if no args are passed to
# an option that normally requires parameters.
:) test "$OPTARG" = l && generate_url=true ;;
# ?: Runs if an invalid option is passed.
'?') die ;;
esac
done
As you can see, most of the functionality is there, but I'm wondering how to accept repo as an argument to the script. Since getopts stops parsing after it encounters the first non-option argument, I'm wondering, how would I implement this (preferably with minimal complexity)? The guide I linked earlier doesn't seem to mention dealing with arguments that aren't associated with an option, so I'm a bit lost.
Thanks for helping!
Use $OPTIND value. After getopts cycle:
shift $((OPTIND-1))
echo $#
echo $1 $2 ...

How get next to next argument in linux shell script?

I have wrote simple shell script test.sh as follows:
while getopts ":A:" OPTION
do
case $OPTION in
A)
echo $OPTARG
?)
echo "no option"
esac
done
And executed the scripts as follows
$ ./test.sh -A 1 2
Now if got argument 1 by $OPTARG but how can i access the second argument ( 2 in this case)?
Regards
Jayesh
There are several options.
(1) You can use shift and take $1
while -n "$1"
do
# do something with $1
shift
done
(2) You can iterate through the args:
for i
do
# do something with $i
done
There are other alternatives also.

switches for shell script

what is the best way to implement switches [ex: -m] for shell scripts?
I can do it via the switch case statement. But i am curious to know is there any other standard way to get all the arguments into a variable via a switch.
Ex:
-m A1 A2 -c c1 c2
So that,
M[] can take -m
and C[] can all take -c
The best known way is to use getopts, see http://wiki.bash-hackers.org/howto/getopts_tutorial
An example :
#!/bin/bash
while getopts ":a" opt; do
case $opt in
a)
echo "-a was triggered!" >&2
;;
\?)
echo "Invalid option: -$OPTARG" >&2
;;
esac
done

need help parsing shell script command line arguments

I am new to Unix shell scripting and would like some help with writing small script.
I have defined a following synopsis for my script:
install.sh [-h|-a path|[-k path][-f path][-d path][-e path]]
ie user can request some help (-h), install all to a specified place (-a path), or install one or more of a components (-k, -f, -d -e) to a appropriate paths. If there is no arguments, the help should be shown.
Thanks in advance.
You can use getopts to parse a command line with bash. Here is an example taken from Bash/Parsing command line arguments using getopts (obviously you'd have to adjust the options to your needs).
#!/bin/bash
#Set a default value for the $cell variable
cell="test"
#Check to see if at least one argument was specified
if [ $# -lt 1 ] ; then
echo "You must specify at least 1 argument."
exit 1
fi
#Process the arguments
while getopts c:hin: opt
do
case "$opt" in
c) cell=$OPTARG;;
h) usage;;
i) info="yes"
n) name=$OPTARG;;
\?) usage;;
esac
done
Related SO question How do I parse command line arguments in bash?
For more information search for getopts on this man page for bash.

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