Bash script re-source a file

Question

i am just wondering is there anyway to source a file and then down in track source the file again?

I am using https://github.com/renatosilva/easyoptions on my bash script, i am sourcing easyoption.sh on the main script and which works fine. but when i have other scripts which gets loaded later on from the main script, i want easyoptions.sh to be sourced and --help should work on the last loaded file.

example:

test.sh

#!/bin/bash

## EasyOptions Sub Test
## Copyright (C) Someone
## Licensed under XYZ
##     -h, --help              All client scripts have this, it can be omitted.

script_dir=$(dirname "$BASH_SOURCE")
# source "${script_dir}/../easyoptions" || exit # Ruby implementation
source "${script_dir}/easyoptions.sh" || exit # Bash implementation, slower

main.sh

#!/bin/bash

## EasyOptions Main
## Copyright (C) Someone
## Licensed under XYZ
## Options:
##     -h, --help              All client scripts have this, it can be omitted.
##         --test      This loads test.sh.

script_dir=$(dirname "$BASH_SOURCE")
 # source "${script_dir}/../easyoptions" || exit # Ruby implementation
 source "${script_dir}/easyoptions.sh" || exit # Bash implementation, slower

if [[ -n "$test" ]];then
    source "${script_dir}/test.sh"
fi

Now when i try ./main.sh --help it displays

  EasyOptions Main
      Copyright (C) Someone
      Licensed under XYZ
      Options:
          -h, --help              All client scripts have this, it can be omitted.
>              --test      This loads test.sh.

Now i want below to work ./main.sh --test --help and it should output

  EasyOptions Sub Test
      Copyright (C) Someone
      Licensed under XYZ
          -h, --help              All client scripts have this, it can be omitted.

But instead it always displays main.sh help

pynexj · Accepted Answer · 2017-10-12 06:18:37Z

1

In main.sh when you source easyoptions.sh it will parse all the command line options (including both --help and --test). Later when you source test.sh the easyoptions would have nothing to parse (i.e. it would not see --help). You can verify this by adding echo "$@" before source test.sh.

answered Oct 12, 2017 at 6:18

pynexj

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peter__barnes · Accepted Answer · 2017-10-12 06:53:19Z

1

as @pynexj said, "when you source easyoptions.sh it will parse all the command line options " so you need following steps:

1.you need to check arguments in main process:

1.1 if the 1st argument is --help(1st argument means $1,not $0 "file name"),then show main help,

1.2 if the 1st argument is --test, loads test.sh and pass other arguments to child.

if the child process get argument --help, it show child's help.

here is an simple example, passing main.sh's argument to child(proc.sh).

main.sh:

echo "main:"
echo $1
echo $2
source ./proc.sh $2

proc.sh:

echo "proc:"
echo $1

when you run cmd

 ./main.sh test help

output：

main:
test
help
proc:
help

you can see,the second argument of main.sh is passed to child

answered Oct 12, 2017 at 6:53

peter__barnes

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1 Comment

Abhishek Salian Over a year ago

i want to use github.com/renatosilva/easyoptions but thanks after seeing your solution i got an idea to skip showing help when --help is not the first option. Which works but now issue is i cant re-source easyoptions second time, so i have to find a work around for that.

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