Copy stdout and stderr to a log file and leave them on the console within the script itself

Question

Using bash, how do I copy stderr and stdout to a log file and also leave them displayed on the console?

I would like to do this within the script itself using an exec.

I tried with

exec &>> log.out

echo "This is stdout"
echo "This is stderr" >&2

But the above prints nothing on the console. How can I achieve this in bash?

There is a highly upvoted answer to a similar question on StackOverflow which answers this question quite thoroughly. https://stackoverflow.com/a/692407/208257 — Dan Burton, Nov 12 '18 at 15:50

H.-Dirk Schmitt · Answer 1 · 2013-03-12T12:35:20.050

23

You are looking for tee.

See man tee for details.

To combine it with exec, you have to use process substitution. (See man bash for details.)

exec &> >(tee  log.out)
echo "This is stdout"
echo "This is stderr" >&2

edited Mar 12 '13 at 12:35

answered Mar 12 '13 at 11:49

H.-Dirk Schmitt

2,009

I have looked at it. Doing exec 2>&1 | tee -a log.out only prints on the console, nothing in the log file. Maybe I am not using the right syntax? – adarshr Mar 12 '13 at 11:58
@adarshr Please double-check your code. I have writen about tee in combination with process substitution. – H.-Dirk Schmitt Mar 12 '13 at 12:34
3

That merges stdout and stderr though and can have the same kind of side-effect as mentioned in my answer. – Stéphane Chazelas Mar 12 '13 at 13:05
1

@StephaneChazelas Look at the example code in the question - that was the intent from the OP. – H.-Dirk Schmitt Mar 12 '13 at 13:46
What I mean by merge is that now the script's errors go to stdout. So if someone does the-script | wc -l for instance, that will also count the error lines and you won't see the errors. – Stéphane Chazelas Mar 12 '13 at 14:18
As Stephane says, this sends stderr to where stdout should be going. – richvdh Dec 04 '14 at 12:13
1

how can we undo what we have done with exec? – Iresh Dissanayaka Apr 24 '22 at 04:22

user2624583 · Answer 2 · 2017-08-02T15:04:53.953

11

I know this is an old post, but why not just do this?

echo "hi" >> log.txt #stdout -> log
echo "hi" | tee -a log.txt #stdout -> log & stdout
echo "hi" &>> log.txt #stdout & stderr -> log
echo "hi" |& tee -a log.txt #stdout & stderr -> log & stdout

And of course, if you want stdout you can just print regularly.

You can do this with any combination of streams you wish, just using those two basic commands.

I know I came here and did not get an easy to understand/implement answer, hopefully this will be help to someone else who is struggling.

By the way, for noobs out there like my previous self, all the tee command does is output the stdin input to both stdout and the file(s) specified as subsequent arguments. -a stands for append, so you don't overwrite the file with every use of the command. If you have further questions, I find this to be a very helpful resource for quickly learning bash.

edited Aug 02 '17 at 15:04

answered Aug 01 '17 at 20:01

user2624583

139

2

Thanks for contributing. I on the other hand, have totally forgotten what I was trying to do :-) – adarshr Aug 01 '17 at 20:02
1

+1 for giving multiple, nice and simple, tee examples. – Tim May 31 '18 at 10:55
2

Why? Because you have to add tee to every single command in the file. – Dan Dascalescu Jun 01 '20 at 10:00
Also, piping individual commands through tee loses their exit status. – fwyzard Jan 05 '23 at 09:56

Stéphane Chazelas · Answer 3 · 2014-12-04T12:30:20.520

You can do:

: > log # empty log file if necessary
{ { {

  ...the script

} 3>&- | tee -a log >&3 3>&-
exit "${PIPESTATUS[0]}"
} 2>&1 | tee -a log >&2 3>&-
} 3>&1
exit "${PIPESTATUS[0]}"

You could also write it as:

: > log # empty log file if necessary
exec 2> >(tee -a log >&2) > >(tee -a log)

...the script

But because bash is not waiting for those processes started with >(...), that has the nasty effect of sometimes outputting something to the terminal after the command has returned which can have even nastier effects (like silently discarding that output) if the terminal "tostop" attribute is on.

In any case, by making stdout a pipe in both solutions, and because two commands independently output the output and error messages, this will affect output buffering and the order the output and error messages are displayed.

I think your second tee is missing a redirect to stderr. It should be: tee -a log >&2 3>&- — richvdh, Dec 04 '14 at 12:26

score 5 · Answer 4 · answered Oct 24 '14 at 14:50

5

One more way of doing it is using redirections within functions.

#!/bin/bash

function1 () {
    echo 'STDOUT from function 1'
    echo 'STDERR from function 1' >&2
}

function2 () {
    echo 'STDOUT from function 2'
    echo 'STDERR from function 2' >&2
}


function3 () {
    echo 'STDOUT from function 3'
    echo 'STDERR from function 3' >&2
}

main() {
    function1
    function2
    function3
}

main 2>&1 |tee log.txt

Here we have a main function which invokes all other functions. Now redirecting STDOUT and STDERR of main function to tee.

answered Oct 24 '14 at 14:50

Kannan Mohan

3,231

2

IMHO this is the cleanest way of doing this, at least for simple cases. Thanks. – ACK_stoverflow Dec 03 '16 at 21:56
1

While this is clean, it may cause commands that emit progress output (e.g. 7z) to not output anything. 7z displays progress on the same line, using Carriage Return (\r), and if run piped, it won't output any progress (maybe it detects the lack of an interactive terminal?). Output can be preserved though by using script – Dan Dascalescu Jun 01 '20 at 10:33

Copy stdout and stderr to a log file and leave them on the console within the script itself

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