Why do the following commands produce different output?
false ; echo $?
output: 1
bash -c "false ; echo $?"
output: 0
Both echo $SHELL and bash -c "echo $SHELL" return /bin/bash so I am not sure why the commands would differ in output.
Since the argument to bash -c is in double quotes, the original shell performs variable substitution in it. So you're actually executing
bash -c "false; echo 0"
Change it to single quotes and you'll get the output you expect.
bash -c 'false; echo $?'
See Difference between single and double quotes in Bash
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.
I'm a bit confused here. My goal here is to have the bash script exit with a non-zero exit code when any of the commands within the script fails. Using the -e flag, I assumed this would be the case, even when using subshells. Below is a simplified example:
#!/bin/bash -e
(false)
echo $?
echo "Line reached!"
Here is the output when ran:
[$]>Tests/Exec/continuous-integration.sh
1
Line reached!
Bash version: 3.2.25 on CentOS
It appears as though this is related to your version of bash. On machines that I have access to, bash version 3.1.17 and 3.2.39 exhibit this behaviour, bash 4.1.5 does not.
Although a bit ugly, a solution that works in both versions could be something like this:
#!/bin/bash -e
(false) || exit $?
echo $?
echo "Line reached!"
There are some notes in the bash source changelog which related to bugs with the set -e option.
I have seen this behavior in bash version 3.2.51 on both SuSE 11.3 and Mac OS prior to El Capitan. Bash 3.2.57 on El Capitan has the "correct" behavior, ie like bash 4.
However, the workaround proposed above, adding "|| exit $?" after the subshell's closing paren, defeats the intent of the -e flag no matter what version of bash. From man bash:
-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 the command list immediately following a while or until keyword, part
of the test in an if statement, part of a && or || list, ...
A subshell followed by "|| exit $?" apparently counts as a command list; and the bash -e flag will not apply to ANY command inside the subshell. Try it:
$ set -e
$ ( echo before the error; false; echo after the error, status $?; ) || echo after the subshell, status $?
before the error
after the error, status 1
$
Because the subshell is followed by ||, that "echo after the error" is run, even with set -e. Not only that, the subshell exits 0, because that "echo" ran. So "|| exit $?" would not even run the "exit". Probably not what we wanted!
So far as I know, the following formula is compatible with bash versions whether they honor bash -e after subshell, or not. It even behaves correctly if the -e flag happens to be reset:
Add the following line immediately after the closing parenthesis of every subshell in the bash script:
case $?/$- in ( 0/* ) ;; ( */*e* ) exit $? ;; esac # honor bash -e flag when subshell returns
The -e option is for current shell, for sub-shell, functions, etc we use -E
form man bash
-E If set, any trap on ERR is inherited by shell functions, command substitutions, and commands executed in a subshell environment.
And for advanced users, a kind of strict mode we have which is:
set -Eeuo pipefail
-E explained above
-e exit immediately
-u exit for unbound variable found
-o set option
pipefail exit if we had failure on a pipe
In shell scripts set -e is often used to make them more robust by stopping the script when some of the commands executed from the script exits with non-zero exit code.
It's usually easy to specify that you don't care about some of the commands succeeding by adding || true at the end.
The problem appears when you actually care about the return value, but don't want the script to stop on non-zero return code, for example:
output=$(possibly-failing-command)
if [ 0 == $? -a -n "$output" ]; then
...
else
...
fi
Here we want to both check the exit code (thus we can't use || true inside of command substitution expression) and get the output. However, if the command in command substitution fails, the whole script stops due to set -e.
Is there a clean way to prevent the script from stopping here without unsetting -e and setting it back afterwards?
Yes, inline the process substitution in the if-statement
#!/bin/bash
set -e
if ! output=$(possibly-failing-command); then
...
else
...
fi
Command Fails
$ ( set -e; if ! output=$(ls -l blah); then echo "command failed"; else echo "output is -->$output<--"; fi )
/bin/ls: cannot access blah: No such file or directory
command failed
Command Works
$ ( set -e; if ! output=$(ls -l core); then echo "command failed"; else echo "output is: $output"; fi )
output is: -rw------- 1 siegex users 139264 2010-12-01 02:02 core
Suppose a shell script (/bin/sh or /bin/bash) contained several commands. How can I cleanly make the script terminate if any of the commands has a failing exit status? Obviously, one can use if blocks and/or callbacks, but is there a cleaner, more concise way? Using && is not really an option either, because the commands can be long, or the script could have non-trivial things like loops and conditionals.
With standard sh and bash, you can
set -e
It will
$ help set
...
-e Exit immediately if a command exits with a non-zero status.
It also works (from what I could gather) with zsh. It also should work for any Bourne shell descendant.
With csh/tcsh, you have to launch your script with #!/bin/csh -e
May be you could use:
$ <any_command> || exit 1
You can check $? to see what the most recent exit code is..
e.g
#!/bin/sh
# A Tidier approach
check_errs()
{
# Function. Parameter 1 is the return code
# Para. 2 is text to display on failure.
if [ "${1}" -ne "0" ]; then
echo "ERROR # ${1} : ${2}"
# as a bonus, make our script exit with the right error code.
exit ${1}
fi
}
### main script starts here ###
grep "^${1}:" /etc/passwd > /dev/null 2>&1
check_errs $? "User ${1} not found in /etc/passwd"
USERNAME=`grep "^${1}:" /etc/passwd|cut -d":" -f1`
check_errs $? "Cut returned an error"
echo "USERNAME: $USERNAME"
check_errs $? "echo returned an error - very strange!"