An easy way to diff log files, ignoring the time stamps?

Solution 1:

Depending on the shell you are using, you can turn the approach @Blair suggested into a 1-liner

diff <(cut -b13- file1) <(cut -b13- file2)

(+1 to @Blair for the original suggestion :-)

Solution 2:

@EbGreen said

I would just take the log files and strip the timestamps off the start of each line then save the file out to different files. Then diff those files.

That's probably the best bet, unless your diffing tool has special powers. For example, you could

cut -b13- file1 > trimmed_file1
cut -b13- file2 > trimmed_file2
diff trimmed_file1 trimmed_file2

See @toolkit's response for an optimization that makes this a one-liner and obviates the need for extra files. If your shell supports it. Bash 3.2.39 at least seems to...

Solution 3:

Answers using cut are fine but sometimes keeping timestamps within the diff output is appreciable. As the OP's question is about ignoring the time stamps (not removing them), I share here my tricky command line:

diff -I '^#' <(sed -r 's/^((.){12})/#\1\n/' 1.log) <(sed -r 's/^((.){12})/#\1\n/' 2.log)

• sed isolates the timestamps (# before and \n after) within a process substitution
• diff -I '^#' ignores lines having these timestamps (lines beginning by #)

example

Two log files having same content but different timestamps:

$> for ((i=1;i<11;i++)) do echo "09:0${i::1}:00.000 data $i"; done > 1.log
$> for ((i=1;i<11;i++)) do echo "11:00:0${i::1}.000 data $i"; done > 2.log

Basic diff command line says all lines are different:

$> diff 1.log 2.log
1,10c1,10
< 09:01:00.000 data 1
< 09:02:00.000 data 2
< 09:03:00.000 data 3
< 09:04:00.000 data 4
< 09:05:00.000 data 5
< 09:06:00.000 data 6
< 09:07:00.000 data 7
< 09:08:00.000 data 8
< 09:09:00.000 data 9
< 09:01:00.000 data 10
---
> 11:00:01.000 data 1
> 11:00:02.000 data 2
> 11:00:03.000 data 3
> 11:00:04.000 data 4
> 11:00:05.000 data 5
> 11:00:06.000 data 6
> 11:00:07.000 data 7
> 11:00:08.000 data 8
> 11:00:09.000 data 9
> 11:00:01.000 data 10

Our tricky diff -I '^#' does not display any difference (timestamps ignored):

$> diff -I '^#' <(sed -r 's/^((.){12})/#\1\n/' 1.log) <(sed -r 's/^((.){12})/#\1\n/' 2.log)
$>

Change 2.log (replace data by foo on the 6th line) and check again:

$> sed '6s/data/foo/' -i 2.log
$> diff -I '^#' <(sed -r 's/^((.){12})/#\1\n/' 1.log) <(sed -r 's/^((.){12})/#\1\n/' 2.log)
11,13c11,13
11,13c11,13
< #09:06:00.000
<  data 6
< #09:07:00.000
---
> #11:00:06.000
>  foo 6
> #11:00:07.000

=> timestamps are kept in the diffoutput!

You can also use the side by side feature using -y or --side-by-side option:

$> diff -y -I '^#' <(sed -r 's/^((.){12})/#\1\n/' 1.log) <(sed -r 's/^((.){12})/#\1\n/' 2.log)
#09:01:00.000                   #11:00:01.000
 data 1                          data 1
#09:02:00.000                   #11:00:02.000
 data 2                          data 2
#09:03:00.000                   #11:00:03.000
 data 3                          data 3
#09:04:00.000                   #11:00:04.000
 data 4                          data 4
#09:05:00.000                   #11:00:05.000
 data 5                          data 5
#09:06:00.000                 | #11:00:06.000
 data 6                       |  foo 6
#09:07:00.000                 | #11:00:07.000
 data 7                          data 7
#09:08:00.000                   #11:00:08.000
 data 8                          data 8
#09:09:00.000                   #11:00:09.000
 data 9                          data 9
#09:01:00.000                   #11:00:01.000
 data 10                         data 10

old sed

If your sed implementation does not support the -r option, you may have to count the twelve dots <(sed 's/^\(............\)/#\1\n/' 1.log) or use another pattern of your choice ;)