I'm running a program that takes variable numbers of arguments with the same flag. For example

myprogram -args 'var1' 'var2' 'var3' 'var4'
myprogram -args 'var5' 'var6'

I have to launch this program several times with different sets of arguments provided in a test.txt file.

arg1 arg2 arg3
arg5 arg6
arg7
arg8 arg9 arg9 arg10

The program must be inside its own script to request resources in our HPCC.

while read p; do
  launchmyprogram.sh "$p"
done < test.txt

I know I can use var1=$1 syntax inside launchmyprogram.sh to collect and allocate the variables, but this cannot handle variable number of arguments, and I'd have to create a script for each line. Is there a way to create a script in bash that takes variable numbers of arguments?

CodePudding user response：

Use arrays to store dynamically sized sequences of strings. Bash has two ways of reading input into an array:

• readarray -t somearray turns lines of an entire input file into array elements.
• read -a somearray turns tokens of a single line of input into array elements.

In this case you can use the latter. Here’s a runnable MWE:

myprogram() {
  local -i i
  echo "Got ${#} arguments."
  for ((i = 1; i <= $#;   i)); do
    echo "Argument No. ${i} is '${!i}'."
  done
}

while read -ra args; do
  myprogram -args "${args[@]}"
done <<-INPUT
    arg1 arg2 arg3
    arg5 arg6
    arg7
    arg8 arg9 arg9 arg10
INPUT

That way the arguments from each line are kept separate, as the output suggests:

Got 4 arguments.
Argument No. 1 is '-args'.
Argument No. 2 is 'arg1'.
Argument No. 3 is 'arg2'.
Argument No. 4 is 'arg3'.
Got 3 arguments.
Argument No. 1 is '-args'.
Argument No. 2 is 'arg5'.
Argument No. 3 is 'arg6'.
Got 2 arguments.
Argument No. 1 is '-args'.
Argument No. 2 is 'arg7'.
Got 5 arguments.
Argument No. 1 is '-args'.
Argument No. 2 is 'arg8'.
Argument No. 3 is 'arg9'.
Argument No. 4 is 'arg9'.
Argument No. 5 is 'arg10'.

CodePudding user response：

You can use $# to query the number of arguments passed to launchmyprogram.sh. Something like

if [ $# -eq 1 ]; then
  echo "one argument"
elif [ $# -eq 2 ]; then
  echo "two arguments"
elif [ $# -eq 3 ]; then
  echo "three arguments"
else
    echo "too many arguments"
    exit 1
fi

CodePudding user response：

All bash scripts take a variable number of arguments, your question is about how to access them. The simplest method is:

for arg; do
    my-cmd "$arg"
done

This repeats my-cmd with each argument, individually. You can use this loop in launchmyprogram.sh. You can also put the relevant code in a function, and use just the function inside the loop.

Parsing arguments from a file is more complicated. If the arguments aren't quoted for the shell, and don't contain spaces or wildcard characters ([]?*), you could just unquote $p in your example. It will be split on white space in to multiple arguments.

In this case you could also just parse the whole file in launchmyprogram.txt:

for arg in $(<test.txt); do
    my-cmd "$arg"
done