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Editorial Team
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Asked: 2026-05-21T20:06:30+00:00

i want to get the last part since a given timestamp t0 from a

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i want to get the last part since a given timestamp “t0” from a possible huge logfile (>50..1000mb):

     __________________
    |1 xxx xxx ...     |
    |2 xxx ...         |     uninteresting part
    |4 ...             |
    |...               |
 ___|423 ...           | ___ timestamp t0
    |425 xxx ...       |
    |437 ...           |
    |...               |     <-- i want this part ( from t0 to EOF)
    |__________________|

and an additional constraint is that i want to do this using simple bash commands. a simple solution may be:

awk '$1 > 423' file.log

but this scans the whole file with all the unintresting lines. there’s the command tail but i just can give him the number of last lines i want which i don’t know – i just know the timestamp. is there a way “awking” from behind and stop processing when the first timestamp doesn’t match?

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1 Answer

  1. Editorial Team

    tac is your friend here:

    tac file.log | awk '{ if ($1 >= 423) print; else exit; }' | tac

    tac will dump each line of a file starting with the last line, then working to the beginning of the file. do it once to get the lines you want, then do it again to fix their order.

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