Good day. I have a series of commands that I wanted to execute via a function so that I could get the exit code and perform console output accordingly. With that being said, I have two issues here:
1) I can't seem to direct stderr to /dev/null.
2) The first echo line is not displayed until the $1 is executed. It's not really noticeable until I run commands that take a while to process, such as searching the hard drive for a file. Additionally, it's obvious that this is the case, because the output looks like:
sh-3.2# ./runScript.sh
sh-3.2# com.apple.auditd: Already loaded
sh-3.2# Attempting... Enable Security Auditing ...Success
In other words, the stderr was displayed before "Attempting... $2"
Here is the function I am trying to use:
#!/bin/bash
function saveChange {
echo -ne "Attempting... $2"
exec $1
if [ "$?" -ne 0 ]; then
echo -ne " ...Failure\n\r"
else
echo -ne " ...Success\n\r"
fi
}
saveChange "$(launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist)" "Enable Security Auditing"
Any help or advice is appreciated.
this is how you redirect stderr to /dev/null
command 2> /dev/null
e.g.
ls -l 2> /dev/null
Your second part (i.e. ordering of echo) -- It may be because of this you have while invoking the script. $(launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist)
The first echo line is displayed later because it is being execute second. $(...) will execute the code. Try the following:
#!/bin/bash
function saveChange {
echo -ne "Attempting... $2"
err=$($1 2>&1)
if [ -z "$err" ]; then
echo -ne " ...Success\n\r"
else
echo -ne " ...Failured\n\r"
exit 1
fi
}
saveChange "launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist" "Enable Security Auditing"
EDIT: Noticed that launchctl does not actually set $? on failure so capturing the STDERR to detect the error instead.
Related
I want my Vagrant provision script to run some checks that will require user action if they're not satisfied. As easy as:
if [ ! -f /some/required/file ]; then
echo "[Error] Please do required stuff before provisioning"
exit
fi
But, as long as this is not a real error, I got the echo printed in green. I'd like my output to be red (or, a different color at least) to alert the user.
I tried:
echo "\033[31m[Error] Blah blah blah"
that works locally, but on Vagrant output the color code gets escaped and I got it echoed in green instead.
Is that possible?
This is happening because some tools write some of their messages to stderr, which Vagrant then interprets as an error and prints in red.
Not all terminals support ANSI colour codes and Vagrant don't take care of that. Said that, I won't suggest colorizing a word by sending it to stderr unless it is an error.
To achieve that you can simply:
echo "Your error message" > /dev/stderr
You need to use keep_color true then it works as intended;
config.vm.provision "shell", keep_color: true, inline: $echoes
$echoes = <<-ECHOES
echo "\e[32mPROVISIONING DONE\e[0m"
ECHOES
From https://www.vagrantup.com/docs/provisioning/shell.html
keep_color (boolean) - Vagrant automatically colors output in green and red depending on whether the output is from stdout or stderr. If this is true, Vagrant will not do this, allowing the native colors from the script to be outputted.
Vagrant commands runs by default with --no-color option. You could try to set color on with --color. The environmental variables for Vagrant are documented here.
Here is a bash script test.sh which should demonstrate how to output to stderr or stdout conditionally. This form is good for a command like [ / test or touch that does not return any stdout or stderr normally. This form is checking the exit status code of the command which is stored in $?.
test -f $1
if [ $? -eq 0 ]; then
echo "File exists: $1"
else
echo "File not found: $1"
fi
You can alternatively hard code your file path like your question shows:
file="/some/required/file"
test -f $file
if [ $? -eq 0 ]; then
echo "File exists: $file"
else
echo "File not found: $file"
fi
If you have output of the command, but its being sent to stderr rather than stdout and ending up in red in the Vagrant output, you can use the following forms to redirect the output to where you would expect it to be. This is good for commands like update-grub or wget.
wget
url='https://example.com/file'
out=$(wget --no-verbose $url 2>&1)
if [ $? -ne 0 ]; then
echo "$out" > /dev/stderr
else
echo "$out"
fi
update-grub
out=$(update-grub 2>&1)
if [ $? -ne 0 ]; then
echo "$out" > /dev/stderr
else
echo "$out"
fi
One Liners
wget
url='https://example.com/file'
out=$(wget --no-verbose $url 2>&1) && echo "$out" || echo "$out" > /dev/stderr
update-grub
out=$(update-grub 2>&1) && echo "$out" || echo "$out" > /dev/stderr
Here I am again. Today I wrote a little script that is supposed to start an application silently in my debian env.
Easy as
silent "npm search 1234556"
This works but not at all.
As you can see, I commented the section where I have some troubles.
This line:
$($cmdLine) &
doesn't hide application output but this one
$($1 >/dev/null 2>/dev/null) &
works perfectly. What am I missing? Many thanks.
#!/bin/sh
# Daniele Brugnara
# October, 2013
# Silently exec a command line passed as argument
errorsRedirect=""
if [ -z "$1" ]; then
echo "Please, don't joke me..."
exit 1
fi
cmdLine="$1 >/dev/null"
# if passed a second parameter, errors will be hidden
if [ -n "$2" ]; then
cmdLine="$cmdLine 2>/dev/null"
fi
# not working
$($cmdLine) &
# works perfectly
#$($1 >/dev/null 2>/dev/null) &
With the use of evil eval following script will work:
#!/bin/sh
# Silently exec a command line passed as argument
errorsRedirect=""
if [ -z "$1" ]; then
echo "Please, don't joke me..."
exit 1
fi
cmdLine="$1 >/dev/null"
# if passed a second parameter, errors will be hidden
if [ -n "$2" ]; then
cmdLine="$cmdLine 2>&1"
fi
eval "$cmdLine &"
Rather than building up a command with redirection tacked on the end, you can incrementally apply it:
#!/bin/sh
if [ -z "$1" ]; then
exit
fi
exec >/dev/null
if [ -n "$2" ]; then
exec 2>&1
fi
exec $1
This first redirects stdout of the shell script to /dev/null. If the second argument is given, it redirects stderr of the shell script too. Then it runs the command which will inherit stdout and stderr from the script.
I removed the ampersand (&) since being silent has nothing to do with running in the background. You can add it back (and remove the exec on the last line) if it is what you want.
I added exec at the end as it is slightly more efficient. Since it is the end of the shell script, there is nothing left to do, so you may as well be done with it, hence exec.
& means that you're doing sort of multitask whereas
1 >/dev/null 2>/dev/null
means that you redirect the output to a sort of garbage and that's why you don't see anything.
Furthermore cmdLine="$1 >/dev/null" is incorrect, you should use ' instead of " :
cmdLine='$1 >/dev/null'
you can build your command line in a var and run a bash with it in background:
bash -c "$cmdLine"&
Note that it might be useful to store the output (out/err) of the program, instead of trow them in null.
In addition, why do you need errorsRedirect??
You can even add a wait at the end, just to be safe...if you want...
#!/bin/sh
# Daniele Brugnara
# October, 2013
# Silently exec a command line passed as argument
[ ! $1 ] && echo "Please, don't joke me..." && exit 1
cmdLine="$1>/dev/null"
# if passed a second parameter, errors will be hidden
[ $2 ] && cmdLine+=" 2>/dev/null"
# not working
echo "Running \"$cmdLine\""
bash -c "$cmdLine" &
wait
I implemented the following in POSIX shell (not bash):
fail.sh:
#!/bin/sh
echo something useful
echo warning 1 >&2
echo warning 2 >&2
echo an error message >&2
exit 100
The command prints something I want to use on stdout, some warnings on stderr and an error message on stderr as well before failing with exit code 100.
success.sh:
#!/bin/sh
echo something useful
echo warning 1 >&2
echo warning 2 >&2
exit 0
This command prints something to stdout and some warnings to stderr but finishes successfully with exit code 0.
test.sh:
#!/bin/sh -e
script=$1
rm -f success
msg=$({ $script > useful; touch success; } 2>&1 | tail -1;)
if [ -f success ]; then
echo success
else
echo failure
echo last error was: $msg
fi
In this script I want to run either of those two scripts and provide the following functionality:
the output of the scripts must be redirected to a file
the last line of stderr must be saved to a variable so that I can print that last line later in case the command didnt exit successfully
I want to detect whether or not the command exited successfully by checking its exit status
My script test.sh achieves all of that but it uses an external file. Since I use -e the touch will only be executed if $script executed successfully. Can I capture the exit code of $script without this technique?
The script must be written in POSIX shell and must use -e.
#!/bin/sh -e
script=$1
if msg=$($script 2>&1 >useful); then
echo success
else
echo failure
msg=$(echo "$msg" | tail -1)
echo last error was: $msg
fi
I have a bash script that I want to be quiet when run without attached tty (like from cron).
I now was looking for a way to conditionally redirect output to /dev/null in a single line.
This is an example of what I had in mind, but I will have many more commands that do output in the script
#!/bin/bash
# conditional-redirect.sh
if tty -s; then
REDIRECT=
else
REDIRECT=">& /dev/null"
fi
echo "is this visible?" $REDIRECT
Unfortunately, this does not work:
$ ./conditional-redirect.sh
is this visible?
$ echo "" | ./conditional-redirect.sh
is this visible? >& /dev/null
what I don't want to do is duplicate all commands in a with-redirection or with-no-redirection variant:
if tty -s; then
echo "is this visible?"
else
echo "is this visible?" >& /dev/null
fi
EDIT:
It would be great if the solution would provide me a way to output something in "quiet" mode, e.g. when something is really wrong, I might want to get a notice from cron.
For bash, you can use the line:
exec &>/dev/null
This will direct all stdout and stderr to /dev/null from that point on. It uses the non-argument version of exec.
Normally, something like exec xyzzy would replace the program in the current process with a new program but you can use this non-argument version to simply modify redirections while keeping the current program.
So, in your specific case, you could use something like:
tty -s
if [[ $? -eq 1 ]] ; then
exec &>/dev/null
fi
If you want the majority of output to be discarded but still want to output some stuff, you can create a new file handle to do that. Something like:
tty -s
if [[ $? -eq 1 ]] ; then
exec 3>&1 &>/dev/null
else
exec 3>&1
fi
echo Normal # won't see this.
echo Failure >&3 # will see this.
I found another solution, but I feel it is clumsy, compared to paxdiablo's answer:
if tty -s; then
REDIRECT=/dev/tty
else
REDIRECT=/dev/null
fi
echo "Normal output" &> $REDIRECT
You can use a function:
function the_code {
echo "is this visible?"
# as many code lines as you want
}
if tty -s; then # or other condition
the_code
else
the_code >& /dev/null
fi
This works well for me. If DUMP_FILE is empty things go to stdout otherwise to the file. It does the job without using explicit redirection, but just uses pipes and existing applications.
function stdout_or_file
{
local DUMP_FILE=${1:-}
if [ -z "${DUMP_FILE}" ]; then
cat
else
sed -n "w ${DUMP_FILE}"
fi
}
function foo()
{
local MSG=$1
echo "info: ${MSG}"
}
foo "bar" | stdout_or_file ${DUMP_FILE}
Of course, you can squeeze this also in one line
foo "bar" | if [ -z "${DUMP_FILE}" ]; then cat; else sed -n "w ${DUMP_FILE}"; fi
Besides sed -n "w ${DUMP_FILE}" another command that does the same is dd status=none of=${DUMP_FILE}
The simplest solution is to use eval (a shell builtin), as it will act on the redirection in the expanded variable... and also act on anything else in the command line, so add extra quoting as required (note the extra single quotes added around the echo string below due to the '?' which would otherwise cause shell filename expansion to be attempted).
#!/bin/bash
# conditional-redirect.sh
if tty -s; then
REDIRECT=
else
REDIRECT=">& /dev/null"
fi
eval echo '"is this visible?"' $REDIRECT
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!"