sort logfile by timestamp on linux command line

Question

I have a logfile with entries like:

...    
freeswitch.log:2011-09-08 12:21:07.282236 [ERR] ftdm_queue.c:136 Failed to enqueue obj 0x7f2cda3525c0 in queue 0x         


        

Answer 1

Crude but effective technique: Prefix each line with a numeric representation of the date, sort numerically, then remove the extra info.

Oneliner:

while IFS=' ' read -r name_date trailing ; do date=$(cut -d: -f2 <<<"$name_date") ; printf '%s:%s\n' $(date -d "$date" +%s) "$name_date $trailing" ; done < freeswitch.log | sort -k1 -t: | cut -d: -f2-

Shell script:

#!/usr/bin/env bash

logfile="$1"

if [ -f "$logfile" ] ; then
    while IFS=' ' read -r name_date trailing ; do
            date=$(cut -d: -f2 <<<"$name_date")
        printf '%s:%s\n' $(date -d "$date" +%s) "$name_date $trailing"
    done < "$logfile" | sort -k1 -t: | cut -d: -f2-
fi

Note: Requires GNU date.

If the output at this point is the reverse of what you want it is simple to pipe through tac or to modify the script to also pass -r to sort.

EDIT: I missed the part where the filename was literally on each line. Updated version will now actually work.