About: extracting *.gz files and move a original file to other folder

Question

I am almost new on shell script but don't know some commands. I am trying to write below shell script , please give some direction. 1. Read *.gz files from specific directory 2. Extract it to other folder 3. Move a original file to another folder. i can do it three separate shell scripts but i want it include one shell script. Then this script will be cronjob and will run every 5 minutes. i was trying to start like below but somehow i am bit confused how to get filelist. I can do here another script but want to include in one script."

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#!/bin/bash

while IFS= read file; do
    gzip -c "$file" > "zipdir/$(basename "$file").gz"
done < filelist
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PS: Files are created in every 5 minutes.

Answer 1

There are several ways to implement what you're looking for (I would consider notify). Anyhow... this is a very simple implementation:

$ source=~/tmp/source   # directory where .gz files will be created
$ target=~/tmp/target   # target directory for uncompressed files
$ archive=~/tmp/archive # archive dir for .gz files
$ shopt -s nullglob     # avoid retiring unexpanded paths
$ for gz in ${source}/*.gz ; do gzip -dc "$gz" > ${target}/$(basename "$gz" .gz) ; mv "$gz" ${archive}/ ; done
$ shopt -u nullglob     # reset nullglob

If you know for sure "source" directory will always contain .gz files you can avoid shopt.

Another solution (not requiring shopt) is this:

find ${source} -name '*.gz' -print0 | while read -d '' -r gz; do
    gzip -dc "$gz" > ${target}/$(basename "$gz" .gz)
    mv "$gz" ${archive}/
done

The first line looks a little bit complicated because it manages source file names containing spaces...