list oldest file in directories in a loop

Question

I have a set of files in a structure like so;

regions ├── ap-northeast-1 │   └── sg-66497903 │   ├── sg-66497903-2017-10-03-Tue-12.39.json │   ├── sg-66497903-2017-10-03-Tue-12.42.json │   ├── sg-66497903-2017-10-03-Tue-12.49.json │   ├── sg-66497903-2017-10-03-Tue-12.53.json │   └── sg-66497903-2017-10-03-Tue-13.12.json ├── ap-northeast-2 │   └── sg-824282eb │   ├── sg-824282eb-2017-10-03-Tue-12.39.json │   ├── sg-824282eb-2017-10-03-Tue-12.42.json │   ├── sg-824282eb-2017-10-03-Tue-12.49.json │   ├── sg-824282eb-2017-10-03-Tue-12.53.json │   └── sg-824282eb-2017-10-03-Tue-13.12.json ├── ap-south-1 │   └── sg-4fec0526 │   ├── sg-4fec0526-2017-10-03-Tue-12.39.json │   ├── sg-4fec0526-2017-10-03-Tue-12.42.json │   ├── sg-4fec0526-2017-10-03-Tue-12.49.json │   ├── sg-4fec0526-2017-10-03-Tue-12.53.json │   └── sg-4fec0526-2017-10-03-Tue-13.12.json 

The list is longer but you get the idea. I am trying to find the oldest json file in each directory to use as at the standard to set in an array and diff the other files in the individual directories.

I have this so far, but it's not right.

#!/bin/bash mapfile -t awsReg < <(ls ~/regions) for awsrg in "${awsReg[@]}" do mapfile -t awsSG < <(ls regions/"$awsrg") for sg in "${awsSG[@]}" do find "$sg" -mindepth 2 -type f -printf '%T+ %p\n' | sort | head -n 1 diff -q ##oldest file## ##all other files### done done 

For example I would like in regions -> ap-northeast-1 -> sg-66497903 to find the file sg-66497903-2017-10-03-Tue-12.39.json, set it as the file to diff the other files in the directory with a diff -q. Move on to next directory, find the oldest in that directory....etc

Answer 1

It's a lot easier in zsh:

for sg in ~/region/ap-*/sg-*(/); do files=($sg/sg-*.json(N.Om)) # list of json regular files sorted # from oldest to newest if (($#files >= 2)); then oldest=$files[1] files[1]=() for file in $files; do cmp -s $oldest $file || printf '"%s" differs from "%s"\n' $file $oldest done fi done 

With ksh93 or bash and GNU ls, you could translate that to:

for sg in ~/region/ap-*/sg-*/; do eval "files=($(ls --quoting-style=shell-always -rtd -- "$sg"sg-*.json))" if ((${#files[@]} >= 2)); then oldest=${files[1]} files=("${files[@]:1}") for file in "${files[@]}"; do cmp -s "$oldest" "$file" || printf '"%s" differs from "%s"\n' "$file" "$oldest" done fi done 

Answer 2

Well you have the file very well named - from most significant value to least so even simple text sort will do it In each directly you can do in bash

myfile=$(ls -1 *.json| sort | head -1) 

this will set your variable $myfile to what you need and diff-away as much as you need

to get the other files all but your myfile

ls -1 *.json | grep -v $myfile 

Answer 3

The proper way would be to use something like:

 #!/usr/bin/env bash unset -v oldest oldest() { local files f old; files=("${1:-.}"/*) old="${files[0]}" for f in "${files[@]}" do [[ $f -ot $old ]] && old=$f ;done printf '%s\n' "$old" } 

Call the script as in: ./oldest regions/*