I have a simple script :
#!/bin/bash
set -e
trap "echo BOO!" ERR
function func(){
ls /root/
}
func
I would like to trap ERR if my script fails (as it will here b/c I do not have the permissions to look into /root). However, when using set -e it is not trapped. Without set -e ERR is trapped.
According to the bash man page, for set -e :
... A trap on ERR, if set, is executed before the shell exits. ...
Why isn't my trap executed? From the man page it seems like it should.
chepner's answer is the best solution: If you want to combine set -e (same as: set -o errexit) with an ERR trap, also use set -o errtrace (same as: set -E).
In short: use set -eE in lieu of just set -e:
#!/bin/bash
set -eE # same as: `set -o errexit -o errtrace`
trap 'echo BOO!' ERR
function func(){
ls /root/
}
# Thanks to -E / -o errtrace, this still triggers the trap,
# even though the failure occurs *inside the function*.
func
A more sophisticated example trap example that prints the message in red and also prints the exit code:
trap 'printf "\e[31m%s: %s\e[m\n" "BOO!" $?' ERR
man bash says about set -o errtrace / 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.
What I believe is happening:
Without -e: The ls command fails inside your function, and, due to being the last command in the function, the function reports ls's nonzero exit code to the caller, your top-level script scope. In that scope, the ERR trap is in effect, and it is invoked (but note that execution will continue, unless you explicitly call exit from the trap).
With -e (but without -E): The ls command fails inside your function, and because set -e is in effect, Bash instantly exits, directly from the function scope - and since there is no ERR trap in effect there (because it wasn't inherited from the parent scope), your trap is not called.
While the man page is not incorrect, I agree that this behavior is not exactly obvious - you have to infer it.
You need to use set -o errtrace for the function to inherit the trap.
We have these options for debugging:
-e Exit immediately on failure
-E If set, any trap on ERR is inherited by shell functions
-u Exit when there is an unbound variable
-o Give a option-name to set
pipefail The return values of last (rightmost) command (exit code)
-v Print all shell input lines as they are read
-x Print trace of commands
For handling the errors we can catch directory with trap
trap 'echo >&2 "Error - exited with status $? at line $LINENO' ERR
Or a better version ref :
trap 'echo >&2 "Error - exited with status $? at line $LINENO:";
pr -tn $0 | tail -n+$((LINENO - 3)) | head -n7' ERR
Or a function:
function __error_handing__(){
local last_status_code=$1;
local error_line_number=$2;
echo 1>&2 "Error - exited with status $last_status_code at line $error_line_number";
perl -slne 'if($.+5 >= $ln && $.-4 <= $ln){ $_="$. $_"; s/$ln/">" x length($ln)/eg; s/^\D+.*?$/\e[1;31m$&\e[0m/g; print}' -- -ln=$error_line_number $0
}
and call it this way:
trap '__error_handing__ $? $LINENO' ERR
Replace ERR with EXIT and it will work.
The syntax of the trap command is: trap [COMMANDS] [SIGNALS]
For more info, please read http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_12_02.html
Related
I have this script (test.sh) :
#!/bin/bash
set -o errtrace
trap 'echo $BASH_VERSION >&2' ERR
echo <(cat<<EOF
Hello world
EOF
)
Running it, I get :
~/tmp$ bash test.sh
/dev/fd/63
~/tmp$ 5.0.18(1)-release
Two questions :
Why ERR trap get triggerred ?
Why 5.0.18(1)-release is after next prompt, not before it
Your copy of cat exits with a nonzero exit status because it's not able to write its output to stdout (since nothing in your code running in the parent shell actually reads from /dev/fd/63). The subshell spawned to run that copy of cat thus registers an error. As errtrace causes your ERR handler to be used in subshells and other contexts, this thus causes the process substitution's subshell to invoke the handler.
Because this process substitution is running in a subshell, it's asynchronous from the rest of your script; and cat is only given an error after the echo has exited (since it's only at that time that the read end of the FIFO is closed, and thus cat's attempts to write result in a SIGPIPE).
I have a Bash shell script that invokes a number of commands.
I would like to have the shell script automatically exit with a return value of 1 if any of the commands return a non-zero value.
Is this possible without explicitly checking the result of each command?
For example,
dosomething1
if [[ $? -ne 0 ]]; then
exit 1
fi
dosomething2
if [[ $? -ne 0 ]]; then
exit 1
fi
Add this to the beginning of the script:
set -e
This will cause the shell to exit immediately if a simple command exits with a nonzero exit value. A simple command is any command not part of an if, while, or until test, or part of an && or || list.
See the bash manual on the "set" internal command for more details.
It's really annoying to have a script stubbornly continue when something fails in the middle and breaks assumptions for the rest of the script. I personally start almost all portable shell scripts with set -e.
If I'm working with bash specifically, I'll start with
set -Eeuo pipefail
This covers more error handling in a similar fashion. I consider these as sane defaults for new bash programs. Refer to the bash manual for more information on what these options do.
To add to the accepted answer:
Bear in mind that set -e sometimes is not enough, specially if you have pipes.
For example, suppose you have this script
#!/bin/bash
set -e
./configure > configure.log
make
... which works as expected: an error in configure aborts the execution.
Tomorrow you make a seemingly trivial change:
#!/bin/bash
set -e
./configure | tee configure.log
make
... and now it does not work. This is explained here, and a workaround (Bash only) is provided:
#!/bin/bash
set -e
set -o pipefail
./configure | tee configure.log
make
The if statements in your example are unnecessary. Just do it like this:
dosomething1 || exit 1
If you take Ville Laurikari's advice and use set -e then for some commands you may need to use this:
dosomething || true
The || true will make the command pipeline have a true return value even if the command fails so the the -e option will not kill the script.
If you have cleanup you need to do on exit, you can also use 'trap' with the pseudo-signal ERR. This works the same way as trapping INT or any other signal; bash throws ERR if any command exits with a nonzero value:
# Create the trap with
# trap COMMAND SIGNAME [SIGNAME2 SIGNAME3...]
trap "rm -f /tmp/$MYTMPFILE; exit 1" ERR INT TERM
command1
command2
command3
# Partially turn off the trap.
trap - ERR
# Now a control-C will still cause cleanup, but
# a nonzero exit code won't:
ps aux | grep blahblahblah
Or, especially if you're using "set -e", you could trap EXIT; your trap will then be executed when the script exits for any reason, including a normal end, interrupts, an exit caused by the -e option, etc.
The $? variable is rarely needed. The pseudo-idiom command; if [ $? -eq 0 ]; then X; fi should always be written as if command; then X; fi.
The cases where $? is required is when it needs to be checked against multiple values:
command
case $? in
(0) X;;
(1) Y;;
(2) Z;;
esac
or when $? needs to be reused or otherwise manipulated:
if command; then
echo "command successful" >&2
else
ret=$?
echo "command failed with exit code $ret" >&2
exit $ret
fi
Run it with -e or set -e at the top.
Also look at set -u.
On error, the below script will print a RED error message and exit.
Put this at the top of your bash script:
# BASH error handling:
# exit on command failure
set -e
# keep track of the last executed command
trap 'LAST_COMMAND=$CURRENT_COMMAND; CURRENT_COMMAND=$BASH_COMMAND' DEBUG
# on error: print the failed command
trap 'ERROR_CODE=$?; FAILED_COMMAND=$LAST_COMMAND; tput setaf 1; echo "ERROR: command \"$FAILED_COMMAND\" failed with exit code $ERROR_CODE"; put sgr0;' ERR INT TERM
An expression like
dosomething1 && dosomething2 && dosomething3
will stop processing when one of the commands returns with a non-zero value. For example, the following command will never print "done":
cat nosuchfile && echo "done"
echo $?
1
#!/bin/bash -e
should suffice.
I am just throwing in another one for reference since there was an additional question to Mark Edgars input and here is an additional example and touches on the topic overall:
[[ `cmd` ]] && echo success_else_silence
Which is the same as cmd || exit errcode as someone showed.
For example, I want to make sure a partition is unmounted if mounted:
[[ `mount | grep /dev/sda1` ]] && umount /dev/sda1
I'm trying to remove some bashisms from a shell script.
I can't figure out how I would replicate 'trap ERR': ERR is not a signal, and is not in the standard, but is a common bashism (see e.g. the LDP guide).
How do I replicate trap ERR in a standards-compliant /bin/sh script?
You can use a combination of set -e and trap ... EXIT.
#!/bin/sh
set -e
err_handler () {
[ $? -eq 0 ] && exit
# Code for non-zero exit status here
}
trap err_handler EXIT
set -e will cause the script to exit whenever an unguarded command has an non-zero exit status. The error handler will be called unconditionally when the script exits, but inside the handler you can simply exit if the current exit status is 0, i.e., we reached it without any errors occurring.
By "unguarded" command I mean a command that isn't run in a context where a non-zero exit status is reasonably expected to occur, such as in the condition for an if statement.
ERR is not really a signal. See: http://en.wikibooks.org/wiki/Bourne_Shell_Scripting/Debugging_and_signal_handling#Err..._ERR.3F at the bottom.
This trap on ERR halts the script. Other examples show that a trap can be used to continue program execution, so why does this halt?
#!/bin/sh -e
trap 'echo error' ERR
echo begin
false
echo end
Returns
$ ./test.sh
begin
$
ERR traps are still affected by the shell's errexit option (set -e). The problem here is that the shebang sets errexit. This will work as expected:
#!/bin/sh
trap 'echo error' ERR
echo begin
false
echo end
returning
$ ./test.sh
begin
error
end
$
errexit is very aggressive. It will instantly kill the script if any statement is non-true.
This will push you as a developer to write very defensively. For example, you cannot ever use $? with errexit because the script itself will not continue if the exit code is non-zero. This "push the developer to get it perfect" approach is useful for some scenarios, e.g. commonly errexit is used to run very early boot scripts when automatically provisioning a system.
If you find yourself in an errexit scenario but needing to handle expected errors, here is the common approach:
#!/bin/sh -e
trap 'echo error' ERR
echo begin
false || { echo "It was false"; }
echo end
Which will output:
begin
It was false
end
Note that the trap is still not fired, because no error happened. As far as I can tell, the "error" condition referred to by ERRtrap and errexit is functionally identical
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