CODEWITHSUNDEEP
rubylinuxbash
The Mechanic
The Mechanic

Reputation: 145

Count files in directory on a time

I'm using macOSx, and i want to count all files in directory then print the result sorted by last Modified time. like this:

1241 2014-12-08 07:00:00 +0700
4123 2014-12-08 08:00:00 +0700
2241 2014-12-08 09:00:00 +0700

etc.. i tried with ruby, but i can only puts count of files with this:

dir = 'original'
puts Dir[File.join(dir, '**', '*')].count { |file| File.file?(file) }

or puts all file and sorts it by time like this:

Dir[File.join(dir, '**', '*')].sort_by{ |f| File.mtime(f) }

So how can i do this? may be with bash script, or ruby...

Upvotes: 0

Views: 439

Answers (6)

The Mechanic
The Mechanic

Reputation: 145

This is my ruby code:

require "Date"
dir = 'original'

for date in 1..7 do
  for hour in 0..23 do
    d = DateTime.new(2014, 1, 1, hour, 0, 0)
    d1 = DateTime.new(2014, 1, 1, hour+1, 0, 0)
    time = d.strftime("%H:%M")
    time1 = d1.strftime("%H:%M")

    puts ("Files on time 12/0#{date}/2014 #{time} (from #{time} -- #{time1}) :")
    puts Dir[File.join(dir, '**', '*')].count {|file|
      if ("12/0#{date}/2014 #{time}" < File.mtime(file).strftime('%m/%d/%Y %I:%M')) && (File.mtime(file).strftime('%m/%d/%Y %I:%M') < "12/0#{date}/2014 #{time1}")
        File.file?(file)
      end
    }
  end
  puts ("Files on time 12/0#{date}/2014 23:00 (from 23:00 -- 00:00 next day) :")
  puts Dir[File.join(dir, '**', '*')].count {|file|
    if ("12/0#{date}/2014 23:00" < File.mtime(file).strftime('%m/%d/%Y %I:%M')) && (File.mtime(file).strftime('%m/%d/%Y %I:%M') < "12/0#{date+1}/2014 #{time1}")
      File.file?(file)
    end
  }
end

Upvotes: 0

Yevgeniy Anfilofyev
Yevgeniy Anfilofyev

Reputation: 4847

You could use tap to count files, like this:

require 'fileutils'
dir = '.'
p Dir[File.join(dir, '**', '*')]
    .tap{|el| p el.length }
    .sort_by{ |f| File.mtime(f) }

Which prints count and sorted array of files.

Upvotes: 0

Ananth Srinivasan
Ananth Srinivasan

Reputation: 136

Here is the bash script to go into folder and execute command from there. 

#!/bin/bash
curdir=$(pwd)
for f in $curdir/<parent folder name>/*
  do
     [ -d $f ] && cd "$f" && echo Entering into $f && wc -l
  done;

Upvotes: 0

Fuelen
Fuelen

Reputation: 41

If I understood correctly

Dir["Other/**"].sort_by{ |f| File.mtime(f) }.each do |item|
  if File.directory? item  
    print "#{Dir[File.join(item, '**', '*')].count} #{File.mtime(item)}\n"
  end
end

Upvotes: 0

shivam
shivam

Reputation: 16506

arr = []
count = 0
Dir.glob("/path/**/*").each { |file| 
  if  File.file?(file)
    count+=1
    arr << File.mtime(file)
  end
}
puts count.to_s + " " + arr.sort.last.to_s
#=> 2241 2014-12-08 09:00:00 +0700

This will not consider the sub directory files though

Upvotes: 0

BMW
BMW

Reputation: 45293

Not sure if I understand the request. do you have subdirectories or need sort all files in subdirectories as well? if not, I use this command

ls -lctr

explanation

 -l      (The lowercase letter ``ell''.)  List in long format.  (See below.)  If the output is to a terminal, a total sum
         for all the file sizes is output on a line before the long listing.
 -c      Use time when file status was last changed for sorting (-t) or long printing (-l).
 -t      Sort by time modified (most recently modified first) before sorting the operands by lexicographical order.
 -r      Reverse the order of the sort to get reverse lexicographical order or the oldest entries first (or largest files
         last, if combined with sort by size

If you need count the files, use this:

find . -type f |wc -l

Upvotes: 2

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