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Re: Sorting a logfile by date to print to htmlby cjensen (Sexton) |
on Oct 09, 2001 at 21:50 UTC ( [id://117816]=note: print w/replies, xml ) | Need Help?? |
Your date format isn't perfect for sorting, but you can pipe your file through the unix sort command using ~ as your separator, on the fourth (0, 1, 2, 3) column, ignoring blank spaces: sort -b -t '~' +3 That's assuming your OS has a compatible 'sort'. You can reverse it for descending order: sort -b -r -t '~' +3 It would be better if date was in a fully sortable format like yyyymmdd, as referenced above, or even yy/mm/dd How about this mess: cat <log file> | \ perl -p -e 's|(\d\d)/(\d\d)/(\d\d)|$3/$1/$2|' | \ sort -r -b -t '~' +3 | \ perl -p -e 's|(\d\d)/(\d\d)/(\d\d)|$2/$3/$1|' Note: You can do a lot on the command-line, but it's not always the best way to do things. If you want to parse a few decades of historical data and aren't worried about your script being around in 50 years or so, you can go way overboard: cat <log file> | \ perl -p -e 's|(\d\d)/(\d\d)/(\d\d)|($3 > 50 ? "19$3" : "20$3")."$1$2"|e' | \ sort -r -b -t '~' +3 | \ perl -p -e 's|(\d\d)(\d\d)(\d\d)(\d\d)|$3/$4/$2|'
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