extracting date from a filename - UNIX for Dummies Questions & Answers

laiko · #1 (**permalink**) 04-23-2008

Hi,
I am a beginner in Unix so please bear with me...
I have a directory which has files in format: RECF-YYYY-MM-DD-input. For example, RECF-2008-02-25-input. I need to extract the YYYYY-MM-DD substring from this filename and convert that into date and compare it with a date. How do I do that?
For instance, I have a directory with the following files:
RECF-2008-02-21-input
RECF-2008-02-22-input
RECF-2008-02-23-input
RECF-2008-02-24-input

I need to get the files whose date substrings are greater than 2008-02-22 (Feb 22, 2008) which are RECF-2008-02-23-input and RECF-2008-02-24-input in the examples above. Please help

Thanks in advance.

radoulov · #2 (**permalink**) 04-23-2008

Code:

$ ls -1
RECF-2008-02-21-input
RECF-2008-02-22-input
RECF-2008-02-23-input
RECF-2008-02-24-input
$ printf "%s\n" *|awk -F- '$2FS$3FS$4>"2008-02-22"'
RECF-2008-02-23-input
RECF-2008-02-24-input

Just re-read your post, you should use date +% ... to print the date in the desired format

Code:

printf "%s\n" *|awk -F-  '$2FS$3FS$4>d' d="$(date +%Y-%m-%d)"

Or:

Code:

$ printf "%s\n" *|awk '$0>d' d="RECF-$(date +%Y-%m-%d)-input"
$ printf "%s\n" *|awk '$0>d' d="RECF-2008-02-22-input"

era · #3 (**permalink**) 04-23-2008

This requires a version of date which can do -d "day" and +%s which -- with my luck -- probably excludes yours. But anyway, here goes. This converts the date in the file name into a canonical date stamp (seconds since Jan 1, 1970, also known as "Unix epoch") which is then easy to compare by whatever you have which can compare two numbers.

Code:

vnix$ ls RECF-*-*-*-input |
cut -d - -f2-4 |
xargs -i date -d '{}' +'%s RECF-{}-input'
1203544800 RECF-2008-02-21-input
1203631200 RECF-2008-02-22-input
1203717600 RECF-2008-02-23-input
1203804000 RECF-2008-02-24-input

The cut command extracts the date; xargs arranges to have those results passed to date one by one, constructing and executing a command like date -d "2008-04-24" +"%s RECF-2008-02-24-input". The %s format specifier is "seconds since epoch"

Edit: fergeddit, radoulov's solution is much better, as usual /-:

bakunin · #4 (**permalink**) 04-23-2008

Quote:

Originally Posted by era

This requires a version of date which can do -d "day" and +%s which -- with my luck -- probably excludes yours.

You can do date arithmetics even if the date command doesn't support it - at least on the day level - quite easily using the TZ variable (i suppose in the following example that you are in the GMT timezone. You will have to introduce a corrective factor if not.):

Code:

# print - "yesterdays date: $(TZ=GMT+24; date)"
# print - "one week ago: $(TZ=GMT+$(( 24*7 )); date)"

bakunin

denn · #5 (**permalink**) 04-23-2008

print - "one week ago: $(TZ=GMT+$(( 24*7 )); date)"
Note: This doesn't work on HP-UX, this method can be used +-24 hrs max for that OS.
Other variants of UNIX that I tried this on previously, this works as expected.

laiko · #6 (**permalink**) 05-09-2008

Thanks a lot guys. The AWK works.
Now, I have a new problem. I need to do it in an FTP box. It does not accept 'AWK' commands.
I need to find the files in the FTP box whose filename has dates greater than say 2008-02-22 (Feb 22, 2008). Example:

Code:

ftp -pin <<EOF>>$mylog
   open inbox.mi.com
   quote user $user
   quote pass $pass
   lcd $destination_dir
   cd SOURCE
   -- here, get all the files SOURC-YYYY-MM-DD-bin
   -- whose date (the YYYY-MM-DD substr) is greater than '2008-02-22'
   bye
EOF

So, in my FTP box, I have files like:
SOURC-2008-03-21-bin
SOURC-2008-01-23-bin
SOURC-2008-04-17-bin
SOURC-2008-04-08-bin
SOURC-2008-02-19-bin

I only want to get the files whose date component is greater than 2008-02-22 which, in this example, are:
SOURC-2008-03-21-bin
SOURC-2008-04-17-bin
SOURC-2008-04-08-bin

Thanks in advance....

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