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How to pick only the latest files based on the timestamp?

 
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# 36  
Old 06-06-2011
Code:
ls -lrt | awk ' {print $6$7" "$9 }' |grep `date '+%b''%d' |sed 's/0\([0-9]\)$/\1/'` | awk ' {print $2} '

# 37  
Old 06-06-2011
Code:
ls -lrt | nawk -v D="$(date +'%b%e:'| sed 's/ //g')" 'D==$6$7":"{print $NF}'

Code:
# ls -lrt
[...]
-rw-r--r--   1 admin    admin         30 Jun  3 21:48 f2
-rw-r--r--   1 admin    admin         12 Jun  3 21:48 f3
-rw-r--r--   1 admin    admin          4 Jun  3 22:39 f4
-rw-r--r--   1 admin    admin       1810 Jun  4 00:08 bigone
-rw-r--r--   1 admin    admin       1568 Jun  6 14:22 tst
-rw-r--r--   1 admin    admin          0 Jun  6 21:35 toto
-rw-r--r--   1 admin    admin          0 Jun  6 21:35 titi
# date
Mon Jun  6 21:36:00 CEST 2011
# date +'%b%e:'| sed 's/ //g'
Jun6:
# ls -lrt | nawk -v D="$(date +'%b%e:'| sed 's/ //g')" 'D==$6$7":"{print $NF}'
tst
toto
titi
#

I added the colon : stuff just in the case you want to match files from let's say for example :

the Jun 1 without matching those from the Jun 10
Last edited by ctsgnb; 06-06-2011 at 04:44 PM..
# 38  
Old 06-06-2011
touch 06060000 /tmp/mark.$$ ; find * -type f -newer /tmp/mark.$$
# 39  
Old 06-06-2011
... to secure a bit the output in the case some of your files contains space in their name :
Code:
ls -lrt | nawk -v D="$(date +'%b%e:'| sed 's/ //g')" 'D==$6$7":"{sub(".*"$9,$9);print}'

# 40  
Old 06-06-2011
the above codes mentioned works fine but i would like to know if the script would work fine for double digit dates as well(10-30/31)....


I added the colon : stuff just in the case you want to match files from let's say for example :

the Jun 1 without matching those from the Jun 10


Dint understand what you meant..could u pls elaborate more on this?


Thanks to both of you
win4luv
# 41  
Old 06-07-2011
What i meant is when you
Code:
ls -lrt | awk '{print $1$2,$9}'

imagine you have some file from 1 and 10 of june,

if you simply grep "Jun1"

you will get both
Jun1 and Jun10 since "Jun1" is a subpattern contained in Jun10
instead of only getting file from the date of Jun1 :

Code:
# echo "Jun  1\nJun 10" | awk '{print $1$2}' | grep 'Jun1'
Jun1
Jun10
# echo "Jun  1\nJun 10" | awk '{print $1$2":"}' | grep 'Jun1:'
Jun1:
# echo "Jun  1\nJun 10" | awk '{print $1$2":"}' | grep 'Jun10:'
Jun10:

# 42  
Old 06-07-2011
Yes, whenever grepping a variable area for an exact value, you need a separator to the right to avoid matching prefixes, and to the left to avoid matching suffixes, but Jun fif that. If you are not anchored to the trailing separator, and if the trailing separator and string can occur in the 'payload' as well as the 'key', you need to anchor your regex to one end of the line. I guess your file names are 'well behaved', but life is not always so.
 
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