How to write a command line tool using bash - bash

I want to write a command line tool like git which will follow the POSIX standards. It will take the options like --help or -h , --version ..etc. But i am not getting how to do it. Can anybody tell me how to do this using bash scripting. Please help me. This is something very new to me.
Example : if the name of my tool is Check-code then i want to use the tool like ;
Check-code --help
or
Check-code --version

So far as I know, "long options", like --help and --version are not POSIX standard, but GNU standard. For command-line utilities the POSIX standard says:
The arguments that consist of hyphens and single letters or digits, such as 'a', are known as "options" (or, historically, "flags").
To support POSIX short options options it is worth getting to know getopts (there are tutorials on the web), but it does not support GNU long options.
For long options you have to roll your own:
filename=default
while (( $# > 0 ))
do
opt="$1"
shift
case $opt in
--help)
helpfunc
exit 0
;;
--version)
echo "$0 version $version"
exit 0
;;
--file) # Example with an operand
filename="$1"
shift
;;
--*)
echo "Invalid option: '$opt'" >&2
exit 1
;;
*)
# end of long options
break;
;;
esac
done

You can use the 'getopts' builtin, like so:
#!/bin/bash
# Parse arguments
usage() {
echo "Usage: $0 [-h] [-v] [-f FILE]"
echo " -h Help. Display this message and quit.
echo " -v Version. Print version number and quit.
echo " -f Specify configuration file FILE."
exit
}
optspec="hvf:"
while getopts "$optspec" optchar
do
case "${optchar}" in
h)
usage
;;
v)
version
;;
f)
file=${OPTARG}
;;
*)
usage
;;
esac
done
This only works for single character options, not for long options like -help or --help. In practice, I've never found that this is a significant restriction; any script which is complex enough to require long options is probably something that I would write in a different language.

There is probably a better way to do this, but here is what I find useful:
Each argument is represented by a variable in BASH. The first argument is $1. The second is $2, and so on. Match an option string with the first argument, and if it matches run some code accordingly.
Example:
#!/bin/bash
if [ $1 == "--hello" ]
then
echo "Hello"
else
echo "Goodbye"
fi
If you code in C or C++, then use the **argv variable. **argv is a double pointer that holds a list of all arguments passed to the program (with argv[0] being the program name).

Related

how to pass other arguments besides flags

I am trying to execute my file by passing in an absolute path as the first argument ($1). I also want to add flags from that absolute path onward, but i do not know how to tell optargs to start counting from $2 forward since if i pass in the absolute path as the $1 it seems to break the getopts loop.
I'm gussing i have to implement a shift for the first argument in the following code:
while getopts :lq flag; do
case $flag in
l) echo "executing -l flag"
;;
q) echo "executing -q flag"
;;
esac
done
I'm not sure how to approach this. Any tips are welcome, thank you.
getopts does, indeed, stop processing the arguments when it sees the first non-option argument. For what you want, you can explicitly shift the first argument if it is not an option. Something like
if [[ $1 != -* ]]; then
path=$1
shift
fi
while getopts :lq flag; do
...
done
Keep the options before file argument (i.e. absolute path).
Many standard bash commands follow the same practice.
Example :
wc -wl ~/sample.txt
ls -lR ~/sample_dir
So if you follow the above practice, your code goes like this.
This code works even if options are not provided.
In general, that is the desired behavior with options.
# Consider last argument as file path
INPUT_FILEPATH=${*: -1}
echo $INPUT_FILEPATH
# Process options
while getopts :lq flag
do
case $flag in
l) echo "executing -l flag"
;;
q) echo "executing -q flag"
;;
esac
done
Sample execution :
bash sample.sh /home/username/try.txt
/home/username/try.txt
bash sample.sh -lq /home/username/try.txt
/home/username/try.txt
executing -l flag
executing -q flag

Saving argument that comes before the flag

In bash shell scripting, I'm trying to take the argument that comes before the flag.
When the argument comes after the flag, I know that I could use getopts and have the case smth like echo "there's an -g flag! Argument: $OPTARG
However I have no clue how to take an argument that comes before the flag. Let's say I would like to process this command: ./filename 2345 -g.
And the argument is a PID that the flag is trying to take argument as.
Thanks in advance!
Assuming Best Practices
Let's say your -g stands for global, and that you support passing -g either before or after the number whose meaning it changes. A mostly-conventional parser (not compliant with baseline POSIX conventions only inasmuch as the latter require all options to come before any positional arguments) might look a bit like the following:
#!/usr/bin/env bash
args=( )
global=0
while (( $# )); do
case $1 in
-g) global=1 ;;
--) shift; args+=( "$#" ); break ;;
-*) echo "Unrecognized argument $1" >&2; exit 1 ;;
*) args+=( "$1" ) ;;
esac
shift
done
if (( global )); then
echo "Doing something with global PID ${args[0]}"
fi
That is to say: Store your positional arguments in a separate location (in this case, the args array), and refer back to them as-needed.
Real, Literal (Awful) Answer
If you really want to store your last argument in a variable and refer back to that variable when you see -g, of course, you can do that:
#!/usr/bin/env bash
last_arg=
while (( $# )); do
case $1 in
-g) global=$last_arg ;;
esac
last_args=$1
shift
done
if [[ $global ]]; then
echo "Global value is $global"
fi
...however: Don't. This violates both POSIX and GNU command-line utility conventions, and thus will be surprising to any of your users who are long-time UNIX users.

Boolean cli flag using getopts in bash?

Is it possible to implement a boolean cli option using getopts in bash? Basically I want to do one thing if -x is specified and another if it is not.
Of course it is possible. #JonathanLeffler already pretty much gave the answer in the comments to the question, so all I'm going to do here is add an example of the implementation and a few niceties to consider:
#!/usr/bin/env bash
# Initialise option flag with a false value
OPT_X='false'
# Process all options supplied on the command line
while getopts ':x' 'OPTKEY'; do
case ${OPTKEY} in
'x')
# Update the value of the option x flag we defined above
OPT_X='true'
;;
'?')
echo "INVALID OPTION -- ${OPTARG}" >&2
exit 1
;;
':')
echo "MISSING ARGUMENT for option -- ${OPTARG}" >&2
exit 1
;;
*)
echo "UNIMPLEMENTED OPTION -- ${OPTKEY}" >&2
exit 1
;;
esac
done
# [optional] Remove all options processed by getopts.
shift $(( OPTIND - 1 ))
[[ "${1}" == "--" ]] && shift
# "do one thing if -x is specified and another if it is not"
if ${OPT_X}; then
echo "Option x was supplied on the command line"
else
echo "Option x was not supplied on the command line"
fi
A few notes about the above example:
true and false are used as option x indicators because both are valid UNIX commands. This makes the test for the option presence more readable, in my opinion.
getopts is configured to run in silent error reporting mode because it suppressed default error messages and allows for a more precise error handling.
the example includes fragments of code for dealing with missing option arguments and post-getopts command line arguments. These are not part of the OP's question.
They are added for the sake of completeness as this code will be required in any reasonably complex script.
For more information about getopts see Bash Hackers Wiki: Small getopts tutorial

Bug in parsing args with getopts in bash

I was trying to modify the bd script to use getopts. I am a newbie at bash scripting
my script is
while getopts ":hvis:d:" opt
do
...
done
...
echo $somedirpath
cd "$somedirpath"
this runs fine when doing
$ ./bd -v -i -s search
or
$ ./bd -is search -d dir
But when running it like this
$ . ./bd -s search
getopts doesn't read the arguments at all. And all the variables I set in the while loop according to the arguments are all not set, so the script no longer works. Please help!
Setting OPTIND=1 before invoking getopts works fine.
The problem is that getopts relies on OPTIND to loop through the arguments provided, and after sourcing the script, it will be set to some value greater than 1 by getopts according to how many arguments you pass. This value gets carried over even after the script ends(because its being sourced). So the next time its sourced, getopts will pick up from that OPTIND, rather than starting from 1!
This might cause strange behaviour with other scripts, and I don't know how safe this is. But it works!
For a better workaround, I think what #tripleee suggests looks safe and robust.
When you source a script, the arguments parsed by getopts are those of the current shell, not the parameters on the source command line.
The common workaround is to have your script merely print the path, and invoke it like cd "$(bd)" instead (perhaps indirectly through a function or alias).
Setting OPTIND=1 may not work reliably on zsh. Try to use something different than getopts:
while [ "$#" -gt 0 ]
do
case "$1" in
-h|--help)
help
return 0
;;
-o|--option)
option
return 0
;;
-*)
echo "Invalid option '$1'. Use -h|--help to see the valid options" >&2
return 1
;;
*)
echo "Invalid option '$1'. Use -h|--help to see the valid options" >&2
return 1
;;
esac
shift
done

Parsing a flag with a list of values

I'm creating a bash script which involves parsing arguments. The usage would be:
$ ./my_script.sh -a ARG_1 -b ARG_2 [-c LIST_OF_ARGS...]
Using getopts I'm able to parse -a and -b and get their respective values ARG_1 and ARG_2. If and only if user places -c as last argument, then I'm also able to get -c and create a list with all values in LIST_OF_ARGS....
But I would not like to force user to insert -c as the last flag. For instance, it would be great if the script can be invoked this way:
$ ./my_script.sh -b ARG_2 -c V1 V2 V3 -a ARG_1
Here is my current code:
while getopts a:b:c opt
do
case $opt in
a)
A_FLAG=$OPTARG
;;
b)
B_FLAG=$OPTARG
;;
c)
# Handle values as regular expressions
args=("$#")
C_LIST=()
for (( i=$OPTIND-1 ; i <= $#-1 ; i++ ))
do
C_LIST=("${C_LIST[#]}" ${args[$i]})
done
;;
?)
usage
;;
esac
done
You need to separate your detection of the -c flag with the processing associated with it. For example, something like:
while getopts a:b:c opt
do
case $opt in
a)
A_FLAG=$OPTARG
;;
b)
B_FLAG=$OPTARG
;;
c)
C_FLAG=1
;;
?)
usage
;;
esac
done
# discard all of our options.
shift `expr $OPTIND - 1`
if [ "$C_FLAG" = 1 ]; then
# Handle values as regular expressions
args=("$#")
C_LIST=()
for (( i=0 ; i <= $#-1 ; i++ ))
do
C_LIST=("${C_LIST[#]}" ${args[$i]})
done
fi
This script doesn't collect all the non-option arguments until after processing all the command line options.
Here's a question: why have a -c option at all?
If the full usage involves a list of values, why not just have no -c option and allow the -a and -b options only while the rest are regular args as in ./myscript.sh -a ARG_1 -b ARG_2 [argument ...], where any arguments are optional (like the -c option and its arguments are in your usage example?
Then your question becomes "how do I intersperse program options and arguments", to which I would respond: "You shouldn't do this, but to achieve this anyway, parse the command line yourself; getopts won't work the way you want it to otherwise."
Of course, parsing is the hard way. Another possibility involves adding the values after -c to a list, so long as you don't encounter another option or the end of the options:
C_LIST=()
while getopts a:b:c: opt; do
#Skipping code...
c)
C_LIST+="$OPTARG"
shift $(expr $OPTIND - 1)
while [ -n "$1" ] && [ $(printf "%s" "$1" | grep -- '^[^-]') ]; do
C_LIST+="$1"
shift
done
OPTIND=1
;;
The behaviour of getopts is mimicked: even if OPTARG begins with a '-' character, it is still kept, but after OPTARG, any string starting with the '-' character may simply be an invalid option such as -n. I used printf instead of echo because some versions of echo, such as the one that bash has built-in, have a -e option that may or may not allow the loop to continue, which isn't desired. The grep expression should prevent this, but who knows if that version of echo allows for -e'hello', which would cause grep to succeed because it sees "hello"? While possibly unnecessary, why take chances?
Personally, I'd avoid this behaviour if you can, but I also don't understand why you're asking for this behaviour in the first place. If I were to recommend anything, I'd suggest the more common /path/to/script -a ARG_1 -b ARG_2 [argument ...] style above any other possible choice of implementation.
On my system, I haven a /usr/share/doc/util-linux/examples/getopt-parse.bash file. It puts the result of getopt into a variable, and set the positional parameters to that variable. Then uses a switch similar to yours, but uses shift to remove arguments when found.
You could do something similar, but for your -c option use shift until you get an option or run out of arguments.
Or it might be enough for you to use your current solution, but remember to set the OPTIND variable after the loop.

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