Sort logs by date field in bash

Question

let\'s have

126 Mar  8 07:45:09 nod1 /sbin/ccccilio[12712]: INFO: sadasdasdas
  2 Mar  9 08:16:22 nod1 /sbin/zzzzo[12712]: sadsdasdas
  1 Mar  8 17:20:01 no         


        

Answer 1

For GNU sort: sort -k2M -k3n -k4

• -k2M sorts by second column by month (this way "March" comes before "April")
• -k3n sorts by third column in numeric mode (so that " 9" comes before "10")
• -k4 sorts by the fourth column.

See more details in the manual.

Answer 2

little off-topic - but anyway. only useful when working within filetrees

ls -l -r --sort=time

from this you could create a one-liner which for example deletes the oldest backup in town.

ls -l -r --sort=time | grep backup | head -n1 | while read line; do oldbackup=\`echo $line | awk '{print$8}'\`; rm $oldbackup; done;

Answer 3

You can use the sort command:

cat $logfile | sort -M -k 2

That means: Sort by month (-M) beginning from second column (-k 2).

Answer 4

days need numeric (not lexical) sort, so it should be sort -s -k 2M -k 3n -k 4,4

See more details here.