Sort datetime string in 12-hour format I have a log file with contents like this: 11-12-2014 - 03:03:59AM lat = 41.990516; lon = -93.430704<br> 11-12-2014 - 03:05:15AM lat = 41.001546; lon = -93.443352<br> 11-12-2014 - 03:11:50AM lat = 42.039054; lon = -93.442001<br> 11-12-2014 - 12:08:03AM lat = 41.937911; lon = -93.369249<br> 11-12-2014 - 12:11:29AM lat = 41.949656; lon = -93.329133<br> 11-12-2014 - 12:23:02AM lat = 42.025385; lon = -93.347026<br> 11-12-2014 - 12:29:10AM lat = 41.033341; lon = -93.380586<br> 11-12-2014 - 12:38:08AM lat = 41.036720; lon = -93.436851<br> 11-12-2014 - 12:45:20AM lat = 41.998129; lon = -93.400943<br> 11-12-2014 - 12:53:36AM lat = 41.961489; lon = -93.414624<br> How can I convert this to 24-hour time and sort it correctly?

With `perl`:


$ perl -MTime::Piece -anle '
$F[2] = Time::Piece->strptime($F[2],"%r")->strftime("%H:%M:%S");
push @out, [$F[2]."-".join("-", reverse(split("-",$F[0]))), join(" ",@F)];
END {
print for map { $_->[1] }
sort { $a->[0] cmp $b->[0] } @out;
}' file
11-12-2014 - 00:08:03 lat = 41.937911; lon = -93.369249

11-12-2014 - 00:11:29 lat = 41.949656; lon = -93.329133

11-12-2014 - 00:23:02 lat = 42.025385; lon = -93.347026

11-12-2014 - 00:29:10 lat = 41.033341; lon = -93.380586

11-12-2014 - 00:38:08 lat = 41.036720; lon = -93.436851

11-12-2014 - 00:45:20 lat = 41.998129; lon = -93.400943

11-12-2014 - 00:53:36 lat = 41.961489; lon = -93.414624

11-12-2014 - 03:03:59 lat = 41.990516; lon = -93.430704

11-12-2014 - 03:05:15 lat = 41.001546; lon = -93.443352

11-12-2014 - 03:11:50 lat = 42.039054; lon = -93.442001