The file looks like this:

xxxxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxx   FILE_ABC_200107017_220065.log

What I need to extract is the 2200 and format it to 22:00 .  Yes this
is the time for the log and it will change for every file.  I'm new to
Unix and not sure what or how to use  awk, sed, or grep ???

Much thanks in advance  !!!!

RS

I use gawk and assume:
- you have always 5 columns separated by whitespace;
- the last field is always separated by three "_"
- the time ist always in HHMM format

Then you could write the following script "extract_time.awk":

{  split($5, split_array, "_")
   raw_time = split_array[4]
   formatted_time = substr(raw_time, 1, 2) ":" substr(raw_time, 3, 2)
   print $0 " " formatted_time

The next question is: where do you want the formatted output. The
script appends the formatted time at the end of each line. Change
the line with "print" according to your needs.

Then call the script with:
gawk -f extract_time.awk logfile > outfile

--
hans juergen mueller

YES
Thanks again

On Tue, 17 Jul 2001 21:41:27 +0900 (KST), H...@chonnam.ac.kr (Hans

    sed 's/.*\(..\)\(..\)....../\1:\2/' logfile
assuming the hour part is always 2 digits.

Which should (untested) do what you're after, so long as the line in your
file always ends in hhmmss.log

If you want an explanation I suggest you read the regular expressions
section in "Unix Power Tools"

Hope this helps,

Matt

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