I want to execute a long running command in Bash, and both capture its exit status, and tee its output.
So I do this:
command | tee out.txt
ST=$?
The problem is that the variable ST captures the exit status of tee and not of command. How can I solve this?
Note that command is long running and redirecting the output to a file to view it later is not a good solution for me.
There is an internal Bash variable called $PIPESTATUS; it’s an array that holds the exit status of each command in your last foreground pipeline of commands.
<command> | tee out.txt ; test ${PIPESTATUS[0]} -eq 0
Or another alternative which also works with other shells (like zsh) would be to enable pipefail:
set -o pipefail
...
The first option does not work with zsh due to a little bit different syntax.
Dumb solution: Connecting them through a named pipe (mkfifo). Then the command can be run second.
mkfifo pipe
tee out.txt < pipe &
command > pipe
echo $?
using bash's set -o pipefail is helpful
pipefail: the return value of a pipeline is the status of
the last command to exit with a non-zero status,
or zero if no command exited with a non-zero status
There's an array that gives you the exit status of each command in a pipe.
$ cat x| sed 's///'
cat: x: No such file or directory
$ echo $?
0
$ cat x| sed 's///'
cat: x: No such file or directory
$ echo ${PIPESTATUS[*]}
1 0
$ touch x
$ cat x| sed 's'
sed: 1: "s": substitute pattern can not be delimited by newline or backslash
$ echo ${PIPESTATUS[*]}
0 1
This solution works without using bash specific features or temporary files. Bonus: in the end the exit status is actually an exit status and not some string in a file.
Situation:
someprog | filter
you want the exit status from someprog and the output from filter.
Here is my solution:
((((someprog; echo $? >&3) | filter >&4) 3>&1) | (read xs; exit $xs)) 4>&1
echo $?
See my answer for the same question on unix.stackexchange.com for a detailed explanation and an alternative without subshells and some caveats.
By combining PIPESTATUS[0] and the result of executing the exit command in a subshell, you can directly access the return value of your initial command:
command | tee ; ( exit ${PIPESTATUS[0]} )
Here's an example:
# the "false" shell built-in command returns 1
false | tee ; ( exit ${PIPESTATUS[0]} )
echo "return value: $?"
will give you:
return value: 1
So I wanted to contribute an answer like lesmana's, but I think mine is perhaps a little simpler and slightly more advantageous pure-Bourne-shell solution:
# You want to pipe command1 through command2:
exec 4>&1
exitstatus=`{ { command1; printf $? 1>&3; } | command2 1>&4; } 3>&1`
# $exitstatus now has command1's exit status.
I think this is best explained from the inside out - command1 will execute and print its regular output on stdout (file descriptor 1), then once it's done, printf will execute and print icommand1's exit code on its stdout, but that stdout is redirected to file descriptor 3.
While command1 is running, its stdout is being piped to command2 (printf's output never makes it to command2 because we send it to file descriptor 3 instead of 1, which is what the pipe reads). Then we redirect command2's output to file descriptor 4, so that it also stays out of file descriptor 1 - because we want file descriptor 1 free for a little bit later, because we will bring the printf output on file descriptor 3 back down into file descriptor 1 - because that's what the command substitution (the backticks), will capture and that's what will get placed into the variable.
The final bit of magic is that first exec 4>&1 we did as a separate command - it opens file descriptor 4 as a copy of the external shell's stdout. Command substitution will capture whatever is written on standard out from the perspective of the commands inside it - but since command2's output is going to file descriptor 4 as far as the command substitution is concerned, the command substitution doesn't capture it - however once it gets "out" of the command substitution it is effectively still going to the script's overall file descriptor 1.
(The exec 4>&1 has to be a separate command because many common shells don't like it when you try to write to a file descriptor inside a command substitution, that is opened in the "external" command that is using the substitution. So this is the simplest portable way to do it.)
You can look at it in a less technical and more playful way, as if the outputs of the commands are leapfrogging each other: command1 pipes to command2, then the printf's output jumps over command 2 so that command2 doesn't catch it, and then command 2's output jumps over and out of the command substitution just as printf lands just in time to get captured by the substitution so that it ends up in the variable, and command2's output goes on its merry way being written to the standard output, just as in a normal pipe.
Also, as I understand it, $? will still contain the return code of the second command in the pipe, because variable assignments, command substitutions, and compound commands are all effectively transparent to the return code of the command inside them, so the return status of command2 should get propagated out - this, and not having to define an additional function, is why I think this might be a somewhat better solution than the one proposed by lesmana.
Per the caveats lesmana mentions, it's possible that command1 will at some point end up using file descriptors 3 or 4, so to be more robust, you would do:
exec 4>&1
exitstatus=`{ { command1 3>&-; printf $? 1>&3; } 4>&- | command2 1>&4; } 3>&1`
exec 4>&-
Note that I use compound commands in my example, but subshells (using ( ) instead of { } will also work, though may perhaps be less efficient.)
Commands inherit file descriptors from the process that launches them, so the entire second line will inherit file descriptor four, and the compound command followed by 3>&1 will inherit the file descriptor three. So the 4>&- makes sure that the inner compound command will not inherit file descriptor four, and the 3>&- will not inherit file descriptor three, so command1 gets a 'cleaner', more standard environment. You could also move the inner 4>&- next to the 3>&-, but I figure why not just limit its scope as much as possible.
I'm not sure how often things use file descriptor three and four directly - I think most of the time programs use syscalls that return not-used-at-the-moment file descriptors, but sometimes code writes to file descriptor 3 directly, I guess (I could imagine a program checking a file descriptor to see if it's open, and using it if it is, or behaving differently accordingly if it's not). So the latter is probably best to keep in mind and use for general-purpose cases.
(command | tee out.txt; exit ${PIPESTATUS[0]})
Unlike #cODAR's answer this returns the original exit code of the first command and not only 0 for success and 127 for failure. But as #Chaoran pointed out you can just call ${PIPESTATUS[0]}. It is important however that all is put into brackets.
In Ubuntu and Debian, you can apt-get install moreutils. This contains a utility called mispipe that returns the exit status of the first command in the pipe.
Outside of bash, you can do:
bash -o pipefail -c "command1 | tee output"
This is useful for example in ninja scripts where the shell is expected to be /bin/sh.
The simplest way to do this in plain bash is to use process substitution instead of a pipeline. There are several differences, but they probably don't matter very much for your use case:
When running a pipeline, bash waits until all processes complete.
Sending Ctrl-C to bash makes it kill all the processes of a pipeline, not just the main one.
The pipefail option and the PIPESTATUS variable are irrelevant to process substitution.
Possibly more
With process substitution, bash just starts the process and forgets about it, it's not even visible in jobs.
Mentioned differences aside, consumer < <(producer) and producer | consumer are essentially equivalent.
If you want to flip which one is the "main" process, you just flip the commands and the direction of the substitution to producer > >(consumer). In your case:
command > >(tee out.txt)
Example:
$ { echo "hello world"; false; } > >(tee out.txt)
hello world
$ echo $?
1
$ cat out.txt
hello world
$ echo "hello world" > >(tee out.txt)
hello world
$ echo $?
0
$ cat out.txt
hello world
As I said, there are differences from the pipe expression. The process may never stop running, unless it is sensitive to the pipe closing. In particular, it may keep writing things to your stdout, which may be confusing.
PIPESTATUS[#] must be copied to an array immediately after the pipe command returns.
Any reads of PIPESTATUS[#] will erase the contents.
Copy it to another array if you plan on checking the status of all pipe commands.
"$?" is the same value as the last element of "${PIPESTATUS[#]}",
and reading it seems to destroy "${PIPESTATUS[#]}", but I haven't absolutely verified this.
declare -a PSA
cmd1 | cmd2 | cmd3
PSA=( "${PIPESTATUS[#]}" )
This will not work if the pipe is in a sub-shell. For a solution to that problem,
see bash pipestatus in backticked command?
Base on #brian-s-wilson 's answer; this bash helper function:
pipestatus() {
local S=("${PIPESTATUS[#]}")
if test -n "$*"
then test "$*" = "${S[*]}"
else ! [[ "${S[#]}" =~ [^0\ ] ]]
fi
}
used thus:
1: get_bad_things must succeed, but it should produce no output; but we want to see output that it does produce
get_bad_things | grep '^'
pipeinfo 0 1 || return
2: all pipeline must succeed
thing | something -q | thingy
pipeinfo || return
Pure shell solution:
% rm -f error.flag; echo hello world \
| (cat || echo "First command failed: $?" >> error.flag) \
| (cat || echo "Second command failed: $?" >> error.flag) \
| (cat || echo "Third command failed: $?" >> error.flag) \
; test -s error.flag && (echo Some command failed: ; cat error.flag)
hello world
And now with the second cat replaced by false:
% rm -f error.flag; echo hello world \
| (cat || echo "First command failed: $?" >> error.flag) \
| (false || echo "Second command failed: $?" >> error.flag) \
| (cat || echo "Third command failed: $?" >> error.flag) \
; test -s error.flag && (echo Some command failed: ; cat error.flag)
Some command failed:
Second command failed: 1
First command failed: 141
Please note the first cat fails as well, because it's stdout gets closed on it. The order of the failed commands in the log is correct in this example, but don't rely on it.
This method allows for capturing stdout and stderr for the individual commands so you can then dump that as well into a log file if an error occurs, or just delete it if no error (like the output of dd).
It may sometimes be simpler and clearer to use an external command, rather than digging into the details of bash. pipeline, from the minimal process scripting language execline, exits with the return code of the second command*, just like a sh pipeline does, but unlike sh, it allows reversing the direction of the pipe, so that we can capture the return code of the producer process (the below is all on the sh command line, but with execline installed):
$ # using the full execline grammar with the execlineb parser:
$ execlineb -c 'pipeline { echo "hello world" } tee out.txt'
hello world
$ cat out.txt
hello world
$ # for these simple examples, one can forego the parser and just use "" as a separator
$ # traditional order
$ pipeline echo "hello world" "" tee out.txt
hello world
$ # "write" order (second command writes rather than reads)
$ pipeline -w tee out.txt "" echo "hello world"
hello world
$ # pipeline execs into the second command, so that's the RC we get
$ pipeline -w tee out.txt "" false; echo $?
1
$ pipeline -w tee out.txt "" true; echo $?
0
$ # output and exit status
$ pipeline -w tee out.txt "" sh -c "echo 'hello world'; exit 42"; echo "RC: $?"
hello world
RC: 42
$ cat out.txt
hello world
Using pipeline has the same differences to native bash pipelines as the bash process substitution used in answer #43972501.
* Actually pipeline doesn't exit at all unless there is an error. It executes into the second command, so it's the second command that does the returning.
Why not use stderr? Like so:
(
# Our long-running process that exits abnormally
( for i in {1..100} ; do echo ploop ; sleep 0.5 ; done ; exit 5 )
echo $? 1>&2 # We pass the exit status of our long-running process to stderr (fd 2).
) | tee ploop.out
So ploop.out receives the stdout. stderr receives the exit status of the long running process. This has the benefit of being completely POSIX-compatible.
(Well, with the exception of the range expression in the example long-running process, but that's not really relevant.)
Here's what this looks like:
...
ploop
ploop
ploop
ploop
ploop
ploop
ploop
ploop
ploop
ploop
5
Note that the return code 5 does not get output to the file ploop.out.
What is the purpose of a shell command (part of a shell script) starting with an exclamation mark?
Concrete example:
In foo.sh:
#!/usr/bin/env bash
set -e
! docker stop foo
! docker rm -f foo
# ... other stuff
I know that without the space the exclamation mark is used for history replacements and ! <expression> according to the man page can be used to evaluate "True if expr is false". But in the example context that does not make sense to me.
TL;DR: This is just by-passing the set -e flag in the specific line where you are using it.
Adding add to hek2mgl's correct and useful answer.
You have:
set -e
! command
Bash Reference Manual → Pipelines describes:
Each command in a pipeline is executed in its own subshell. The exit status of a pipeline is the exit status of the last command in the pipeline (...). If the reserved word ‘!’ precedes the pipeline, the exit status is the logical negation of the exit status as described above. The shell waits for all commands in the pipeline to terminate before returning a value.
This means that ! preceding a command is negating the exit status of it:
$ echo 23
23
$ echo $?
0
# But
$ ! echo 23
23
$ echo $?
1
Or:
$ echo 23 && echo "true" || echo "fail"
23
true
$ ! echo 23 && echo "true" || echo "fail"
23
fail
The exit status is useful in many ways. In your script, used together with set -e makes the script exit whenever a command returns a non-zero status.
Thus, when you have:
set -e
command1
command2
If command1 returns a non-zero status, the script will finish and won't proceed to command2.
However, there is also an interesting point to mention, described in 4.3.1 The Set Builtin:
-e
Exit immediately if a pipeline (see Pipelines), which may consist of a single simple command (see Simple Commands), a list (see Lists), or a compound command (see Compound Commands) returns a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command’s return status is being inverted with !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits.
Taking all of these into consideration, when you have:
set -e
! command1
command2
What you are doing is to by-pass the set -e flag in the command1. Why?
if command1 runs properly, it will return a zero status. ! will negate it, but set -e won't trigger an exit by the because it comes from a return status inverted with !, as described above.
if command1 fails, it will return a non-zero status. ! will negate it, so the line will end up returning a zero status and the script will continue normally.
If you don't want the script to fail in both cases, error or success of the command, you can also use this alternative:
set -e
docker stop foo || true
The boolean or true makes the pipeline always have 0 as the return value.
Sometimes there are two commands which I often invoke in a row. However the second command only makes sense in case the first command was successful.
I wanted to do something like this:
#!/bin/bash
if [ $? -gt 0 ]
then
echo "WARNING: previous command entered at the shell prompt failed"
fi
But it doesn't work:
t#quad:~$ echo "abc" | grep def
t#quad:~$ ./warnme.sh
Last command succeeded
What I'd like is something a bit like this:
t#quad:~$ echo "abc" | grep def
t#quad:~$ echo ${PIPESTATUS[1]}
1
Where we can clearly see that the last command failed.
The result I'd like to have:
t#quad:~$ echo "abc" | grep def
t#quad:~$ ./warnme.sh
WARNING: previous command entered at the shell prompt failed
I can't find a way to do it.
command1 && command2
does exactly what you want: command 2 is executed only if command1 succeeds. For example you could do:
ls a.out && ./a.out
Then a.out would only be executed if it could be listed. I wikiblogged about this at http://www.linuxintro.org/wiki/%26%26
One option is to put this just before the list of commands you want to execute only if the previous was successful:
set -e
This will exit the script if any of the commands following it return non-zero (usually a fail). You can switch it off again with:
set +e
Or if you'd prefer to switch it off for just one line you can just logical-OR the command with true:
mycommand || true
For a lot of my scripts I have set -e at the top of the script as a safety feature to prevent the script cascading and doing something dangerous.
How about:
echo "abc" | grep def || ./warnme.sh
Where warnme.sh is now only the warning without the test. It's only called if the first command fails.
In other words, it would be the same as:
echo "abc" | grep def || echo "WARNING: That didn't work..."
Since cruise control is full of bugs that have wasted my entire week, I have decided the existing shell scripts I have are simpler and thus better.
Here is what I have so far
svn update /var/www/k12/
#svn log --revision "HEAD" /var/www/code/ | head -2 | tail -1 | awk '{print $1}' > /var/www/path/version.txt
# upload the files
rsync -ar --verbose --stats --progress --delete --exclude=*.svn /var/www/code/ example.com:/home/path
# bring database up to date
ssh example.com 'php /path/tasks/dbrefactor.php'
# notify me
ssh example.com 'php /path/tasks/build.php'
Only thing is the other day I changed the paths and forgot to update the rsync call. As a result the "notify me" step ran several times while I was figuring stuff out.
I know in linux you can do command1 && command2 and if command 1 "fails" command2 will not run, but how can I observe the "failure/success" exit codes for debugging purposes. Some of the scripts I wrote myself and I'm sure I will need to do something special.
The best option, especially for unattended scripts, is to set the -e shell option:
#!/bin/sh -e
or
set -e
This will cause the shell to stop executing if any (untested) command exits with a nonzero error code.
-e Exit immediately if a simple command (see SHELL GRAMMAR
above) exits with a non-zero status. The shell does not
exit if the command that fails is part of an until or
while loop, part of an if statement, part of a && or ||
list, or if the command's return value is being inverted
via !. A trap on ERR, if set, is executed before the
shell exits.
The exit code of a previous process happens to be in $? variable right after its execution. Usually (that's not required, but it's the convention everyone follows) the exit code of a successful command will be equal to 0, and any other value means an error.
Remember of the caveats! One of them is that after these commands:
svn log --revision "HEAD" /var/www/code/ | head -2 | tail -1 | awk '{print $1}'
echo "$?"
the zero result would most likely be returned, because in the $? the return code of awk is contained. To avoid it, set the pipefail option somewhere above the code:
set -o pipefail 1
The return value of the last-run command is stored in the variable $?. You can use that to determine which command to run next. Overview of special variables.
i think $? contains the last exit code
if [[ -z $? ]]
then
# notify me
ssh example.com 'php /path/tasks/build.php'
fi
I would suggest you can use the exit non zero at the points where the failure is expected and before processing step further you will check
if [ $? -neq 0 ]
then there is a failure.
The $? will always return a non zero number if the last process does not executed successfully.
I have a bash script that checks some log files created by a cron job that have time stamps in the filename (down to the second). It uses the following code:
CRON_LOG=$(ls -1 $LOGS_DIR/fetch_cron_{true,false}_$CRON_DATE*.log 2> /dev/null | sed 's/^[^0-9][^0-9]*\([0-9][0-9]*\).*/\1 &/' | sort -n | cut -d ' ' -f2- | tail -1 )
if [ -f "$CRON_LOG" ]; then
printf "Checking $CRON_LOG for errors\n"
else
printf "\n${txtred}Error: cron log for $CRON_NOW does not exist.${txtrst}\n"
printf "Either the specified date is too old for the log to still be around or there is a problem.\n"
exit 1
fi
CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code")
if [ -z "$CRIT_ERRS" ]; then
printf "%74s[${txtgrn}PASS${txtrst}]\n"
else
printf "%74s[${txtred}FAIL${txtrst}]\n"
printf "Critical errors detected! Outputting to console...\n"
echo $CRIT_ERRS
fi
So this bit of code works fine, but I'm trying to clean up my scripts now and implement set -e at the top of all of them. When i do it to this script, it exits with error code 1. Note that I have errors form the first statement dumping to /dev/null. This is because some days the file has the word "true" and other days "false" in it. Anyway, i don't think this is my problem because the script outputs "Checking xxxxx.log for errors." before exiting when I add set -e to the top.
Note: the $CRON_DATE variable is derived form user input. I can run the exact same statement from command line "$./checkcron.sh 01/06/2010" and it works fine without the set -e statement at the top of the script.
UPDATE: I added "set -x" to my script and narrowed the problem down. The last bit of output is:
Checking /map/etl/tektronix/logs/fetch_cron_false_010710054501.log for errors
++ cat /map/etl/tektronix/logs/fetch_cron_false_010710054501.log
++ grep ERROR
++ grep -v 'Duplicate tracking code'
+ CRIT_ERRS=
[1]+ Exit 1 ./checkLoad.sh...
So it looks like the problem is occurring on this line:
CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code")
Any help is appreciated. :)
Thanks,
Ryan
Adding set -x, which prints a trace of the script's execution, may help you diagnose the source of the error.
Edit:
Your grep is returning an exit code of 1 since it's not finding the "ERROR" string.
Edit 2:
My apologies regarding the colon. I didn't test it.
However, the following works (I tested this one before spouting off) and avoids calling the external cat. Because you're setting a variable using the results of a subshell and set -e looks at the subshell as a whole, you can do this:
CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code"; true)
bash -c 'f=`false`; echo $?'
1
bash -c 'f=`true`; echo $?'
0
bash -e -c 'f=`false`; echo $?'
bash -e -c 'f=`true`; echo $?'
0
Note that backticks (and $()) "return" the error code of the last command they run. Solution:
CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code" | cat)
Redirecting error messages to /dev/null does nothing about the exit status returned by the script. The reason your ls command isn't causing the error is because it's part of a pipeline, and the exit status of the pipeline is the return value of the last command in it (unless pipefail is enabled).
Given your update, it looks like the command that's failing is the last grep in the pipeline. grep only returns 0 if it finds a match; otherwise it returns 1, and if it encounters an error, it returns 2. This is a danger of set -e; things can fail even when you don't expect them to, because commands like grep return non-zero status even if there hasn't been an error. It also fails to exit on errors earlier in a pipeline, and so may miss some error.
The solutions given by geocar or ephemient (piping through cat or using || : to ensure that the last command in the pipe returns successfully) should help you get around this, if you really want to use set -e.
Asking for set -e makes the script exit as soon as a simple command exits with a non-zero exit status. This combines perniciously with your ls command, which exits with a non-zero status when asked to list a non-existent file, which is always the case for you because the true and false variants don't co-exist.