combined use of bash options for error handling (e.g. pipefail, errexit, inherit_errexit) - bash

Consider the following Bash script:
#!/usr/bin/env bash
set -o errexit pipefail
shopt -s inherit_errexit
( echo hello ; exit 1 ) | cat
echo world
Running with version 5.0.17, the output is as follows:
hello
world
However, the subshell responsible for printing hello would have failed, with a nonzero exit status. Part of a pipe, with enabling the optionpipefail, the entire pipe should similarly fail, with the same status (the subsequent item in the pipe of course returning a zero status by itself). Thus, the pipe should evaluate to a nonzero status, which due to errexit (if not inherit_errexit as well) would prompt the immediate termination of the script, without reaching the final statement printing world.
As seen, the prediction is not accurate that the final print statement is not reached.
Why?

You're only setting the errexit setting, not pipefail. You can't put multiple options after -o, you need to repeat the -o option. Everything else is being put into the positional arguments.
So change
set -o errexit pipefail
to
set -o errexit -o pipefail

Related

bash set -eo pipefail not immediately exiting

#!/usr/bin/env bash
set -eo pipefail
sha256sum \
Dockerfile-ci \
frontend/pnpm-lock.yaml \
| sha256sum
If frontend/pnpm-lock.yaml does not exist, and the script above is run
sha256sum: frontend/pnpm-lock.yaml: No such file or directory
e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855
It correctly logs that the file doesn't exist, but it continues piping that into the next sha256sum. Shouldn't set -eo pipefail immediately exit the script on the first sha256sum command and not pipe into the second sha256sum?
pipefail doesn't cause the pipeline to abort early if a command fails. The pipeline still runs to completion, until all the commands have exited. That's true with or without pipefail.
What pipefail does do is ensure the return status is a failure if any of the commands fail. Without pipefail the pipeline fails only if the final command fails.
From the bash manual (emphasis added):
The exit status of a pipeline is the exit status of the last command in the pipeline, unless the pipefail option is enabled (see The Set Builtin). If pipefail is enabled, the pipeline’s return status is the value of the last (rightmost) command to exit with a non-zero status, or zero if all commands exit successfully. 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.

Difference of -e, -u and -o pipefail in bash exit status

Being trying to properly guard for non-zero exits on a bash script.
What is the difference between -e, -u and -o pipefail?
-o pipefail is not sufficient to exit with an error code?
set -e: Exit immediately if a command exits with a non-zero status.
set -u: If you try to access an undefined variable, that is an error.
set -o pipefail: If any command in a pipeline returns a non-zero exit code, the return code of the entire pipeline is the exit code of the last failed command.

`grep` cause bash script stop [duplicate]

I'm studying the content of this preinst file that the script executes before that package is unpacked from its Debian archive (.deb) file.
The script has the following code:
#!/bin/bash
set -e
# Automatically added by dh_installinit
if [ "$1" = install ]; then
if [ -d /usr/share/MyApplicationName ]; then
echo "MyApplicationName is just installed"
return 1
fi
rm -Rf $HOME/.config/nautilus-actions/nautilus-actions.conf
rm -Rf $HOME/.local/share/file-manager/actions/*
fi
# End automatically added section
My first query is about the line:
set -e
I think that the rest of the script is pretty simple: It checks whether the Debian/Ubuntu package manager is executing an install operation. If it is, it checks whether my application has just been installed on the system. If it has, the script prints the message "MyApplicationName is just installed" and ends (return 1 mean that ends with an “error”, doesn’t it?).
If the user is asking the Debian/Ubuntu package system to install my package, the script also deletes two directories.
Is this right or am I missing something?
From help set :
-e Exit immediately if a command exits with a non-zero status.
But it's considered bad practice by some (bash FAQ and irc freenode #bash FAQ authors). It's recommended to use:
trap 'do_something' ERR
to run do_something function when errors occur.
See http://mywiki.wooledge.org/BashFAQ/105
set -e stops the execution of a script if a command or pipeline has an error - which is the opposite of the default shell behaviour, which is to ignore errors in scripts. Type help set in a terminal to see the documentation for this built-in command.
I found this post while trying to figure out what the exit status was for a script that was aborted due to a set -e. The answer didn't appear obvious to me; hence this answer. Basically, set -e aborts the execution of a command (e.g. a shell script) and returns the exit status code of the command that failed (i.e. the inner script, not the outer script).
For example, suppose I have the shell script outer-test.sh:
#!/bin/sh
set -e
./inner-test.sh
exit 62;
The code for inner-test.sh is:
#!/bin/sh
exit 26;
When I run outer-script.sh from the command line, my outer script terminates with the exit code of the inner script:
$ ./outer-test.sh
$ echo $?
26
As per bash - The Set Builtin manual, if -e/errexit is set, the shell exits immediately if a pipeline consisting of a single simple command, a list or a compound command returns a non-zero status.
By default, the exit status of a pipeline is the exit status of the last command in the pipeline, unless the pipefail option is enabled (it's disabled by default).
If so, the pipeline's return status of the last (rightmost) command to exit with a non-zero status, or zero if all commands exit successfully.
If you'd like to execute something on exit, try defining trap, for example:
trap onexit EXIT
where onexit is your function to do something on exit, like below which is printing the simple stack trace:
onexit(){ while caller $((n++)); do :; done; }
There is similar option -E/errtrace which would trap on ERR instead, e.g.:
trap onerr ERR
Examples
Zero status example:
$ true; echo $?
0
Non-zero status example:
$ false; echo $?
1
Negating status examples:
$ ! false; echo $?
0
$ false || true; echo $?
0
Test with pipefail being disabled:
$ bash -c 'set +o pipefail -e; true | true | true; echo success'; echo $?
success
0
$ bash -c 'set +o pipefail -e; false | false | true; echo success'; echo $?
success
0
$ bash -c 'set +o pipefail -e; true | true | false; echo success'; echo $?
1
Test with pipefail being enabled:
$ bash -c 'set -o pipefail -e; true | false | true; echo success'; echo $?
1
This is an old question, but none of the answers here discuss the use of set -e aka set -o errexit in Debian package handling scripts. The use of this option is mandatory in these scripts, per Debian policy; the intent is apparently to avoid any possibility of an unhandled error condition.
What this means in practice is that you have to understand under what conditions the commands you run could return an error, and handle each of those errors explicitly.
Common gotchas are e.g. diff (returns an error when there is a difference) and grep (returns an error when there is no match). You can avoid the errors with explicit handling:
diff this that ||
echo "$0: there was a difference" >&2
grep cat food ||
echo "$0: no cat in the food" >&2
(Notice also how we take care to include the current script's name in the message, and writing diagnostic messages to standard error instead of standard output.)
If no explicit handling is really necessary or useful, explicitly do nothing:
diff this that || true
grep cat food || :
(The use of the shell's : no-op command is slightly obscure, but fairly commonly seen.)
Just to reiterate,
something || other
is shorthand for
if something; then
: nothing
else
other
fi
i.e. we explicitly say other should be run if and only if something fails. The longhand if (and other shell flow control statements like while, until) is also a valid way to handle an error (indeed, if it weren't, shell scripts with set -e could never contain flow control statements!)
And also, just to be explicit, in the absence of a handler like this, set -e would cause the entire script to immediately fail with an error if diff found a difference, or if grep didn't find a match.
On the other hand, some commands don't produce an error exit status when you'd want them to. Commonly problematic commands are find (exit status does not reflect whether files were actually found) and sed (exit status won't reveal whether the script received any input or actually performed any commands successfully). A simple guard in some scenarios is to pipe to a command which does scream if there is no output:
find things | grep .
sed -e 's/o/me/' stuff | grep ^
It should be noted that the exit status of a pipeline is the exit status of the last command in that pipeline. So the above commands actually completely mask the status of find and sed, and only tell you whether grep finally succeeded.
(Bash, of course, has set -o pipefail; but Debian package scripts cannot use Bash features. The policy firmly dictates the use of POSIX sh for these scripts, though this was not always the case.)
In many situations, this is something to separately watch out for when coding defensively. Sometimes you have to e.g. go through a temporary file so you can see whether the command which produced that output finished successfully, even when idiom and convenience would otherwise direct you to use a shell pipeline.
I believe the intention is for the script in question to fail fast.
To test this yourself, simply type set -e at a bash prompt. Now, try running ls. You'll get a directory listing. Now, type lsd. That command is not recognized and will return an error code, and so your bash prompt will close (due to set -e).
Now, to understand this in the context of a 'script', use this simple script:
#!/bin/bash
# set -e
lsd
ls
If you run it as is, you'll get the directory listing from the ls on the last line. If you uncomment the set -e and run again, you won't see the directory listing as bash stops processing once it encounters the error from lsd.
set -e The set -e option instructs bash to immediately exit if any command [1] has a non-zero exit status. You wouldn't want to set this for your command-line shell, but in a script it's massively helpful. In all widely used general-purpose programming languages, an unhandled runtime error - whether that's a thrown exception in Java, or a segmentation fault in C, or a syntax error in Python - immediately halts execution of the program; subsequent lines are not executed.
By default, bash does not do this. This default behavior is exactly what you want if you are using bash on the command line
you don't want a typo to log you out! But in a script, you really want the opposite.
If one line in a script fails, but the last line succeeds, the whole script has a successful exit code. That makes it very easy to miss the error.
Again, what you want when using bash as your command-line shell and using it in scripts are at odds here. Being intolerant of errors is a lot better in scripts, and that's what set -e gives you.
copied from : https://gist.github.com/mohanpedala/1e2ff5661761d3abd0385e8223e16425
this may help you .
Script 1: without setting -e
#!/bin/bash
decho "hi"
echo "hello"
This will throw error in decho and program continuous to next line
Script 2: With setting -e
#!/bin/bash
set -e
decho "hi"
echo "hello"
# Up to decho "hi" shell will process and program exit, it will not proceed further
It stops execution of a script if a command fails.
A notable exception is an if statement. eg:
set -e
false
echo never executed
set -e
if false; then
echo never executed
fi
echo executed
false
echo never executed
cat a.sh
#! /bin/bash
#going forward report subshell or command exit value if errors
#set -e
(cat b.txt)
echo "hi"
./a.sh; echo $?
cat: b.txt: No such file or directory
hi
0
with set -e commented out we see that echo "hi" exit status being reported and hi is printed.
cat a.sh
#! /bin/bash
#going forward report subshell or command exit value if errors
set -e
(cat b.txt)
echo "hi"
./a.sh; echo $?
cat: b.txt: No such file or directory
1
Now we see b.txt error being reported instead and no hi printed.
So default behaviour of shell script is to ignore command errors and continue processing and report exit status of last command. If you want to exit on error and report its status we can use -e option.

How use local in Bash in conjunction with set -o errexit?

I'm new to Bash and I had a hard time to figure out why when I was using set -o errexit and a command fail, the script was not exiting.
It seems because I declared a local variable!
Please tell me how I can use local variable and set -o errexit at the same time.
Example:
#!/bin/bash
set -o errexit
set -o nounset
set -o pipefail
function test {
local output=$(ls --badoption)
echo "error code is $?"
echo "output=$output"
}
test
Result:
./test.sh
ls: illegal option -- -
usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]
error code is 0
output=
But:
#!/bin/bash
set -o errexit
set -o nounset
set -o pipefail
function test {
output=$(ls --badoption)
echo "error code is $?"
echo "output=$output"
}
test
Result:
ls: illegal option -- -
usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]
It's because set -o errexit (or running bash with -e) only affects simple commands (as defined in the bash man page). cmd in local output=$(cmd) isn't considered a simple command, so -e doesn't have any affect (and $? doesn't have any relation to how cmd exited). A simple workaround is to split the line by replacing:
local output=$(ls --badoption)
with:
local output
output=$(ls --badoption)
In my testing that does everything you want, it will exit immediately, and if you replace the ls --badoption with just ls, it works, and setting output before calling test then echoing it afterwords shows output really is local to test.
The documentation of local says:
The return status is zero unless local is used outside a function, an invalid name is supplied, or name is a readonly variable.
So the return status of the local command is not affected by whether the command executed in the command substitution succeeded.
As blm indicated in his answer, you can get the effect you want by separating the declaration and assignment of the local variable. The return status of an ordinary assignment that uses command substitution is the return status of the subshell that executes the command.

Automatic exit from Bash shell script on error [duplicate]

This question already has answers here:
Aborting a shell script if any command returns a non-zero value
(10 answers)
Closed 3 years ago.
I've been writing some shell script and I would find it useful if there was the ability to halt the execution of said shell script if any of the commands failed. See below for an example:
#!/bin/bash
cd some_dir
./configure --some-flags
make
make install
So in this case, if the script can't change to the indicated directory, then it would certainly not want to do a ./configure afterwards if it fails.
Now I'm well aware that I could have an if check for each command (which I think is a hopeless solution), but is there a global setting to make the script exit if one of the commands fails?
Use the set -e builtin:
#!/bin/bash
set -e
# Any subsequent(*) commands which fail will cause the shell script to exit immediately
Alternatively, you can pass -e on the command line:
bash -e my_script.sh
You can also disable this behavior with set +e.
You may also want to employ all or some of the the -e -u -x and -o pipefail options like so:
set -euxo pipefail
-e exits on error, -u errors on undefined variables, -x prints commands before execution, and -o (for option) pipefail exits on command pipe failures. Some gotchas and workarounds are documented well here.
(*) Note:
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 following the if or elif reserved words, 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 value is being inverted with
!
(from man bash)
To exit the script as soon as one of the commands failed, add this at the beginning:
set -e
This causes the script to exit immediately when some command that is not part of some test (like in a if [ ... ] condition or a && construct) exits with a non-zero exit code.
Use it in conjunction with pipefail.
set -e
set -o pipefail
-e (errexit): Abort the script at the first error, when a command exits with non-zero status (except in until or while loops, if-tests, and list constructs)
-o pipefail: Causes a pipeline to return the exit status of the last command in the pipe that returned a non-zero return value.
Chapter 33. Options
Here is how to do it:
#!/bin/sh
abort()
{
echo >&2 '
***************
*** ABORTED ***
***************
'
echo "An error occurred. Exiting..." >&2
exit 1
}
trap 'abort' 0
set -e
# Add your script below....
# If an error occurs, the abort() function will be called.
#----------------------------------------------------------
# ===> Your script goes here
# Done!
trap : 0
echo >&2 '
************
*** DONE ***
************
'
An alternative to the accepted answer that fits in the first line:
#!/bin/bash -e
cd some_dir
./configure --some-flags
make
make install
One idiom is:
cd some_dir && ./configure --some-flags && make && make install
I realize that can get long, but for larger scripts you could break it into logical functions.
I think that what you are looking for is the trap command:
trap command signal [signal ...]
For more information, see this page.
Another option is to use the set -e command at the top of your script - it will make the script exit if any program / command returns a non true value.
One point missed in the existing answers is show how to inherit the error traps. The bash shell provides one such option for that using set
-E
If set, any trap on ERR is inherited by shell functions, command substitutions, and commands executed in a subshell environment. The ERR trap is normally not inherited in such cases.
Adam Rosenfield's answer recommendation to use set -e is right in certain cases but it has its own potential pitfalls. See GreyCat's BashFAQ - 105 - Why doesn't set -e (or set -o errexit, or trap ERR) do what I expected?
According to the manual, set -e exits
if a simple commandexits with 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 a if statement, part of an && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command's return value is being inverted via !".
which means, set -e does not work under the following simple cases (detailed explanations can be found on the wiki)
Using the arithmetic operator let or $((..)) ( bash 4.1 onwards) to increment a variable value as
#!/usr/bin/env bash
set -e
i=0
let i++ # or ((i++)) on bash 4.1 or later
echo "i is $i"
If the offending command is not part of the last command executed via && or ||. For e.g. the below trap wouldn't fire when its expected to
#!/usr/bin/env bash
set -e
test -d nosuchdir && echo no dir
echo survived
When used incorrectly in an if statement as, the exit code of the if statement is the exit code of the last executed command. In the example below the last executed command was echo which wouldn't fire the trap, even though the test -d failed
#!/usr/bin/env bash
set -e
f() { if test -d nosuchdir; then echo no dir; fi; }
f
echo survived
When used with command-substitution, they are ignored, unless inherit_errexit is set with bash 4.4
#!/usr/bin/env bash
set -e
foo=$(expr 1-1; true)
echo survived
when you use commands that look like assignments but aren't, such as export, declare, typeset or local. Here the function call to f will not exit as local has swept the error code that was set previously.
set -e
f() { local var=$(somecommand that fails); }
g() { local var; var=$(somecommand that fails); }
When used in a pipeline, and the offending command is not part of the last command. For e.g. the below command would still go through. One options is to enable pipefail by returning the exit code of the first failed process:
set -e
somecommand that fails | cat -
echo survived
The ideal recommendation is to not use set -e and implement an own version of error checking instead. More information on implementing custom error handling on one of my answers to Raise error in a Bash script

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