I am trying to install zfs through shell script while installation I am getting some error so, to automate it fully, I want to get the version to be installed from error itself. For all other commands I am getting the error in one variable but for one command its not coming at all. I tried every solution possible.
I need output of this command
sploutput=$(sudo dkms install -m spl -v $version)
echo $sploutput
echo $sploutput # This is giving nothing.
I tried wrapping it around string also like "sploutput=$(sudo dkms install -m spl -v $version)"
echo "{sploutput}"
Nothing seems to work.
sploutput=$(sudo dkms install -m spl -v $version)
echo $sploutput
it might be because the dkms install outputs on STDERR and not on STDOUT, and with the command you're using, you're only getting STDOUT output in the variable. To take both, you can try:
sploutput=$(sudo dkms install -m spl -v $version 2>&1)
to redirect STDERR into STDOUT.
As per ZMO's answer: try running sudo dkms install -m spl -v $version
See, what is it returning? : STDERR or STDOUT.
In case, if it fails, it won't show up anything in sploutput.
It write only STDOUT to the variable.
Use 2>&1(to write Standard error to standard output).
You can refer IO redirection
You can use following:
sploutput=$(sudo dkms install -m spl -v $version 2>&1)
echo $sploutput
Related
I install Pip for a user (not system wide) and I would like to check that pip is installed for that user in my script that I run with sudo: sudo ./script.sh
I know to check for a command with command -v pip3 and that works when I enter it in the shell as the user.
But how can I check it in my script?
command -v pip3 exit code is 1 because I am root (because of sudo).
su -c "command -v pip3" "$SUDO_USER" has exit code 1.
sudo -u "$SUDO_USER" command -v pip3 says "command: command not found"
The simplest is
sudo -u "$SUDO_USER" -i command -v pip3
The -i option causes sudo to pass the supplied command line to the user's configured shell using its -c option, instead of trying to execute the command directly. That's necessary because command is a shell built-in; it doesn't exist as a stand-alone executable. (The -i options runs a "login" shell. There is also the -s option which runs a non-login shell. See below.)
If you want to specify a shell explicitly you could do so instead:
sudo -u "$SUDO_USER" /bin/sh -lc "command -v pip3"
Again, a login shell is forced, here by using the -l option.
As a safety feature, sudo normally resets the $PATH to a "safe" value before executing the shell (or the single command). That value will not have any of the modifications made in the /etc/profile and ~/.profile startup scripts, and without those modifications -- which add one or more user-specific directories to the path -- the shell will not find software such as pip3 which has been installed for individual users.
use following command by replacing $USER with the specific user name.
sudo -H -u $USER bash -c 'command -v pip3'
similarly, you can run any command as another user
syntax : sudo -H -u $USER bash -c 'INSERT_COMMAND_HERE'
I have been trying to run the usr/bin/time command in my terminal (Bash) with the verbose flag --verbose or -v but have repeatedly been getting this error:
/usr/bin/time: illegal option -- v
usage: time [-lp] command.
The command I have been running looks like basically like this:
/usr/bin/time -v python practice.py
Any ideas how to get this to work properly on a Mac? (I have OS X Yosemite)?
If you have homebrew, you can get GNU time by installing the gnu-time package:
brew install gnu-time
After that, it’s available as the gtime command:
$ gtime
Usage: gtime [-apvV] [-f format] [-o file] [--append] [--verbose]
[--portability] [--format=format] [--output=file] [--version]
[--help] command [arg...]
The case is similar for a lot of other homebrew-packaged GNU utilities for OSX; e.g., you can get the GNU df command with gdf, du with gdu, readlink with greadlink, etc.
The homebrew package that has most of those is coreutils, which installs about a hundred different GNU-flavored commands. Other useful packages: findutils, gnu-sed, gnu-tar.
If you don’t have homebrew installed yet, you can get it with just a single command:
Command to download and install homebrew
ruby -e "$(curl -fsSL\
https://raw.githubusercontent.com/Homebrew/install/master/install)"
I think looking at the man page the verbose flag is GNU only. Unfortunately, OSX implementation simply differs.
This question is similar to this one: https://serverfault.com/questions/342697/prevent-sudo-apt-get-etc-from-swallowing-pasted-input-to-stdin but the answer is not satisfying (appending && to each line of bash script is not elegant) and does not explain why some users can paste/execute multiple subsequent apt-get install -y commands and others can't because stdout is swollen by the next command.
I have a script my_script.sh:
sudo apt-get install -y graphicsmagick
sudo apt-get install -y libgraphicsmagick++1-dev
...
It can have only two lines or more of sudo apt-get install stuff. The libraries (graphicsmagick, etc.) doesn't matter, it can be any library.
When I copy this script and paste it's contents to bash or just execute it like this:
cat my_script.sh | sudo -i bash
then for some reason only the first line (graphicsmagick) gets executed and the rest is just printed to the console. It happens only with sudo apt-get install -y, other scripts, which doesn't contain this command behave normally.
If I change bash to sh (which is dash) I get expected behaviour:
cat my_script.sh | sudo -i sh
Can you explain why this happens?
When answering, can you please avoid this questions/comments:
Why are you doing it this way?
Piping to your bash is not safe
Some other aspects are not safe or hackish
I just want to know why bash doesn't work as I would expect and sh does.
PS. I'm using Ubuntu 14.04, sh is dash as you can see here:
vagrant#vagrant-ubuntu-trusty-64:/tmp$ ls -l /bin/sh
lrwxrwxrwx 1 root root 4 Feb 19 2014 /bin/sh -> dash
Bash and dash simply behave different when using -i flag.
Bash always goes to interactive mode even when stdin is not a terminal.
Dash on the other hand will not go into interactive mode, even with -i flag.
Probably need the -s option
If the -s option is present, or if no arguments remain after option
processing, then commands are read from the standard input. This option allows
the positional parameters to be set when invoking an interactive shell.
Bash man page
curl -s http://foo.com/bar.sh | sudo -i bash -s
Example
I'm creating a BASH scrip which requires a couple of applications to be installed. ffmpeg and sox
To ensure they are in place when my script runs I first check for the installation of Homebrew with :
#!/bin/bash
which -s brew
if [[ $? != 0 ]] ; then
# Install Homebrew
/usr/bin/ruby -e "$(curl -fsSL https://raw.github.com/Homebrew/homebrew/go/install)"
fi
Then I check that sox and ffmpeg are installed with :
echo "---- checking for sox ----"
which -s sox || /usr/local/bin/brew install sox
echo "---- checking for ffmpeg ----"
which -s ffmpeg || /usr/local/bin/brew install ffmpeg
The problem I am facing is when Homebrew is installed but in a non-standard location.
I have to use the full path to Homebrew because this script is being run within Playtypus.
So the question is : How can I reliably get the installed path of Homebrew in a BASH script?
Answering my own question...
You can test the output of which brew and deal with things accordingly. To gracefully deal with the case where Homebrew is not installed you can use if which brew 2> /dev/null which redirects stderr to /dev/null.
brew --prefix is also useful here as it give the path to where Homebrew installed applications are symlinked to, rather than their actual install path.
A script which works and shows this working :
#!/bin/bash
if which brew 2> /dev/null; then
brewLocation=`which brew`
appLocation=`brew --prefix`
echo "Homebrew is installed in $brewLocation"
echo "Homebrew apps are run from $appLocation"
else
echo "Can't find Homebrew"
echo "To install it open a Terminal window and type :"
echo /usr/bin/ruby -e \"\$\(curl\ \-fsSL\ https\:\/\/raw\.github\.com\/Homebrew\/homebrew\/go\/install\)\"
fi
Thanks to Allendar for the pointers.
Just to add to this, Homebrew's --prefix mode has been enhanced here in the far-flung future of 2020 (or maybe it was always this way), so that it now takes a package name as an argument. Meaning locating those "keg-only" packages which aren't linked into standard paths is as easy as:
$ brew --prefix ffmpeg
/usr/local/opt/ffmpeg
Goal: when the user types 'make packages', automatically search for the package libx11-dev (required for my program to compile) and, if not found, install it. Here's a stripped-down version of my makefile:
PACKAGES = $(shell if [ -z $(dpkg -l | grep libx11-dev) ]; then sudo apt-get install libx11-dev; fi)
[other definitions and targets]
packages: $(PACKAGES)
When I type 'make packages', I'm prompted for the super-user password. If entered correctly, it then hangs indefinitely.
Is what I'm trying to do even possible from within the makefile? If so, how?
Thanks so much.
The problem is that the shell function acts like backticks in the shell: it takes the output to stdout and returns it as the value of the function. So, apt-get is not hanging, it's waiting for you to enter a response to some question. But you cannot see the question because make has taken the output.
The way you're doing this is not going to work. Why are you using shell instead of just writing it as a rule?
packages:
[ -z `dpkg -l | grep libx11-dev` ] && sudo apt-get install libx11-dev
.PHONY: packages
I figured out a better way, which avoids the problem of having unexpected arguments to the if statement:
if ! dpkg -l | grep libx11-dev -c >>/dev/null; then sudo apt-get install libx11-dev; fi
The -c flag on grep makes it return the number of lines in dpkg -l which contain the string libx11-dev, which will either be 0 (if uninstalled) or 1 (if installed), allowing
dpkg -l | grep libx11-dev -c
to be treated like an ordinary boolean variable.