Saving argument that comes before the flag - bash

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.

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

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.

How to write a command line tool using 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).

Multiple compound comparison in bash

I have a bash script which is getting options (using getopts). Let say they are '-n' , '-d' , '-u' . I only want to have one of the option being chosen, if not, it will prompt the user error.
The code is like this:
while getopts ":dun" name; do
case $name in
d )
DELETE='YES'
;;
u )
UPDATE='YES'
;;
n )
NEW='YES'
;;
esac
done
I can only have $DELETE or $UPDATE or $NEW being 'YES' in one time, meaning the user cannot specific '-n' and '-d' in the same time or '-u' and '-n' in the same time, how do I achieve that in a IF statement ?
I have been looking for this in stackoverflow, but can't find any. Thanks for the help, mate!
You can increment a counter every time getopts() senses one of the valid commandline options. Then, after the loop test the counter's value. If it is greater than one, then multiple options were specified.
This is a complete hack, and depends on the option variables being either unset or "YES" (and in the form I've written it, is bash-only):
if [[ "$DELETE$UPDATE$NEW" == YESYES* ]]; then
echo "Please only use one of -d, -u, and -n" >&2
exit 1
fi
(If you were using the brand-x shell instead of bash, it'd be something like if [ "$DELETE$UPDATE$NEW" = "YESYES" -o "$DELETE$UPDATE$NEW" = "YESYESYES" ]; then)
You're better off setting your variables to true or false, so you can simply write
if $DELETE; then
echo "You've already specified UPDATE!"
fi
and similar.
However, options are called that because they are supposed to be optional, and usually orthogonal. (There are a few instances where options of common utilities exclude each other, but the vast majority don't). What you want is really a mandatory mode of operation, so you shouldn't receive it as an option at all, but simply as the first command-line argument.
You can use some getopt-kludge to manage having only one of the option being chosen.
#!/bin/bash
while [ $# -gt 0 ]; do
case "$1" in
-u)
echo "got -u"
break
;;
-d)
echo "got -d"
break
;;
-n)
echo "got -n"
break
;;
*)
echo "some error"
break
;;
esac
done

Using getopts within user-defined-function in bourne shell

Is it possible to pass command line arguments into a function from within a bourne script, in order to allow getopts to process them.
The rest of my script is nicely packed into functions, but it's starting to look like I'll have to move the argument processing into the main logic.
The following is how it's written now, but it doesn't work:
processArgs()
{
while getopts j:f: arg
do
echo "${arg} -- ${OPTARG}"
case "${arg}" in
j) if [ -z "${filename}" ]; then
job_number=$OPTARG
else
echo "Filename ${filename} already set."
echo "Job number ${OPTARG} will be ignored.
fi;;
f) if [ -z "${job_number}" ]; then
filename=$OPTARG
else
echo "Job number ${job_number} already set."
echo "Filename ${OPTARG} will be ignored."
fi;;
esac
done
}
doStuff1
processArgs
doStuff2
Is it possible to maybe define the function in a way that it can read the scripts args? Can this be done some other way? I like the functionality of getopts, but it looks like in this case I'm going to have to sacrifice the beauty of the code to get it.
You can provide args to getopts after the variable. The default is $#, but that's also what shell functions use to represent their arguments. Solution is to pass "$#" — representing all the script's command-line arguments as individual strings — to processArgs:
processArgs "$#"
Adding that to your script (and fixing the quoting in line 11), and trying out some gibberish test args:
$ ./try -j asdf -f fooo -fasdfasdf -j424pyagnasd
j -- asdf
f -- fooo
Job number asdf already set.
Filename fooo will be ignored.
f -- asdfasdf
Job number asdf already set.
Filename asdfasdf will be ignored.
j -- 424pyagnasd

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