CODEWITHSUNDEEP
linuxbashawksedgrep
meallhour
meallhour

Reputation: 15581

Using sed to fetch date

I have a file which contains two values for abc... keyword. I want to grab the latest date for matching abc... string. After getting the date I also need to format the date by replacing / with -

    ---other data
2018/01/15 01:56:14.944+0000 INFO  newagent.bridge BridgeTLSAssetector::setupACBContext() - abc...

2018/02/14 01:56:14.944+0000 INFO  newagent.bridge BridgeTLSAssetector::setupACBContext() - abc...

    ---other data

In the above example, my output should be 2018-02-14. Here, I am fetching the line which contains abc... value and only getting the line with latest date value. Then, I need to strip out the remaining string and fetch only the date value.

I am using the following sed but it is not working

grep -iF "abc..." file.txt | tail -n 1 | sed -e 's/^[^|]*|[^|]*| *\([^ ]*\) .*/\1/' -e 's%/%-%g'

Upvotes: 0

Views: 916

Answers (5)

gautam patel
gautam patel

Reputation: 28

Easy and more simple try this.

cat filename.txt | grep 'abc' | awk -F' ' '{print $1}'

As pattern abc always fix as per the given logs. So this will be more easier way to get desire output.

Upvotes: 0

mjuarez
mjuarez

Reputation: 16834

This does what you want:

grep -iF "abc..." file.txt | tail -n 1 | awk '{print $1}' | sed 's#/#-#g'

Outputs this:

2018-02-14

Upvotes: 1

steffen
steffen

Reputation: 16948

With awk:

$ awk '/abc\.\.\./{d=$1} END{gsub("/", "-", d); print d}' file.txt
2018-2-14

Something with sed:

tac file.txt | grep -Fi 'abc...' | sed 's/ .*//;s~/~-~g;q'

Upvotes: 1

Walter A
Walter A

Reputation: 20002

When you use sed for matching a part of the date, you can have it match year. month, date and abc... in one command.

sed -rn 's#([0-9]{4})/([0-9]{2})/([0-9]{2}).*abc[.]{3}.*#\1-\2-\3#p' file.txt | tail -1

Upvotes: 0

Paul Hodges
Paul Hodges

Reputation: 15293

Since you asked for sed -

$: sed -nE ' / abc[.]{3}/x; $ { x; s! .*!!; s!/([0-9])/!/0\1/!g; s!/([0-9])$!/0\1!g; s!/!-!g; p; }' in
2018-02-14

arguments

  • -n says don't print by default
  • -E says use extended regexes

the script

  • / abc[.]{3}/x; say on each line with abc... swap the line for the buffer
  • $ { x; s! .*!!; s!/([0-9])/!/0\1/!g; s!/([0-9])$!/0\1!g; s!/!-!g; p; } says on the LAST line($) do the set of commands inside the {}.
    • x swaps the buffer to get the last saved record back.
    • s! .*!!; deletes everything from the first space (after the date)
    • s!/([0-9])/!/0\1/!g; adds a zero to the month if needed
    • s!/([0-9])$!/0\1!g; adds a zero to the day if needed
    • s!/!-!g; converts the /'s to dashes
    • p prints the resulting record.

Upvotes: 0

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