How to listout the files based on group by the date...?

psiva_arul · #1 (**permalink**) 04-18-2008

Hi,

How to listout the files based on group by of date...?

suppose if have list of 10 files which was created by on today(Apr 18) and yesterday(Apr 17 ) in my personal directory like this

if i issue ls -rtl | grep "Apr 18" list out the file which was created/updated only on Apr 18th.

Actual files listing
=============

-rwxrwxrwx 1 geae users 278 Apr 17 06:07 sendmail.ksh
-rw-rw-r-- 1 geae users 15 Apr 17 06:26 P.txt
-rw-rw-r-- 1 geae users 375 Apr 17 07:12 simple.ksh
-rwxrwxrwx 1 geae users 375 Apr 17 10:04 dates.ksh
-rwxrwxrwx 1 geae users 555 Apr 18 01:12 IN.txt
-rwxrwxrwx 1 geae users 400 Apr 18 01:13 date_convert.ksh
-rw-rw-r-- 1 geae users 562 Apr 18 01:13 Updt_IN.txt
-rw-rw-r-- 1 geae users 590 Apr 18 02:43 Dir_File.txt
-rwxrwxrwx 1 geae users 1667 Apr 18 02:45 noofdir.ksh
-rw-rw-r-- 1 geae users 301 Apr 18 03:00 date.txt

Now i want to list out the based on the date and count the no of files basded on the date like this

Expected list of output
=================
[UDate: Apr 17 => No of Files:4[/U]
-rwxrwxrwx 1 geae users 278 Apr 17 06:07 sendmail.ksh
-rw-rw-r-- 1 geae users 15 Apr 17 06:26 P.txt
-rw-rw-r-- 1 geae users 375 Apr 17 07:12 simple.ksh
-rwxrwxrwx 1 geae users 375 Apr 17 10:04 dates.ksh
Apr 18 => No of files: 6
-rwxrwxrwx 1 geae users 555 Apr 18 01:12 IN.txt
-rwxrwxrwx 1 geae users 400 Apr 18 01:13 date_convert.ksh
-rw-rw-r-- 1 geae users 562 Apr 18 01:13 Updt_IN.txt
-rw-rw-r-- 1 geae users 590 Apr 18 02:43 Dir_File.txt
-rwxrwxrwx 1 geae users 1667 Apr 18 02:45 noofdir.ksh
-rw-rw-r-- 1 geae users 301 Apr 18 03:00 date.txt

So, How can list out the files based on the date?

Thanks and Regards,
Siva. P
Bangalore

krishmaths · #2 (**permalink**) 04-18-2008

This works for me:

Code:

ls -ltr | tail +2 > tempfile
token=`head -1 tempfile`
count=1
while read line
do
 dt=`echo $line | awk '{print $6" "$7}'`
 if [ "$dt" != "$token" ]
 then
  echo "$dt => No of files: $count"
  token=$dt
  count=1
 else
  count=`expr $count + 1`
 fi
done<tempfile

krishmaths · #3 (**permalink**) 04-21-2008

Code:

ls -ltr | tail +2 | awk '{print $6" "$7}' | uniq -c

psiva_arul · #4 (**permalink**) 04-21-2008

Hi Krishnamath,

Thanks for your response, your command has retruning correct values but it retruend wrong output for some dates which is in single value date(Dec 5)

i have checked with some ordinary command

ls -rtl | grep "Dec 5" -- it won;t work why because we need to give 5 blank space in between month and date if the value of the date is single digit
instead of that we can use ls -rtl | grep "Dec 5" (its working)

how can we avoid the problem.

we should have to use the double blank spaces if the date of value is single digit.

Thanks and Regards,
Siva.P
Bangalore

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