Find specific argument and pass it to command

Question

I have a script called test_script that needs to run a file called run_command as follows:
./run_command <input path>

If I supply the script with the following arguments:
./test_script argument1=sometext argument2=othertext inputpath=/folder1/folder2/file.txt argument4=moretext

How can I make the script find the argument inputpath=/folder1/folder2/file.txt and pass /folder1/folder2/file.txt to run_command? Keep in mind that inputpath=/folder1/folder2/file.txt won't necessarily always be in third place and the amount of arguments supplied may vary.

Stéphane Chazelas · Answer

With zsh:
#! /bin/zsh -
inputpath=(${${(M)argv:#inputpath=*}#*=})
(($#inputpath > 0)) && ./run-command $inputpath

Would extract all the paths in inputpath=path arguments to the script and store the non-empty ones in the $inputpath array. Then we run ./run-command with those inputpath as arguments if any were found.
POSIXly, you could do something like:
#! /bin/sh -
run_with_extracted_keyword() (
  cmd="${1?}" keyword="${2?}"
  shift 2
  for arg do
    case $arg in
      ("$keyword="*) set -- "$@" "${arg#*=}"
    esac
    shift
  done
  [ "$#" -gt 0 ] && exec "$cmd" "$@"
)

run_with_extracted_keyword ./run-command inputpath "$@"

GNUly, you could do:
#! /bin/bash -
set -o pipefail
printf '%s' "$@" |
  LC_ALL=C grep -zPo '^inputpath=K(?s:.*)' |
  xargs -r0 ./run-command

Gilles Quenot · Answer

What I would do, if you control the input, and can restrict the script access to prevent shell injection as stated by Stéphane in comments:
#!/bin/bash
  
for arg; do
    declare "$arg"
done

echo "$argument1"

sometext

Usually, to access positional parameters, you can do echo $1, $2, $3...
But if you use this var=value form, this code will fit better

Find specific argument and pass it to command

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