I am trying to run script a from script b, I didn't create script a and I am not sure if i want to add script as location to PATH.
Making things even more hard to do is the fact I want to run script a with sudo, but not script b so I have whiptail in script b to ask for password and pass along with echo $pass | sudo -S.
my current code:
#!/bin/bash
input=$(whiptail --passwordbox "Enter password" 10 50 3>&1 1>&2 2>&3 )
echo $input | sudo -S bash /home/<username>/multibootusb/makeUSB.sh -e -b "$1" "$2"
Related
There is a simple cron job:
#reboot /home/user/scripts/run.sh > /dev/null 2>&1
run.sh starts a binary (simple web server):
#!/usr/bin/env bash
NPID=/home/user/server/websrv
if [ ! -f $NPID ]
then
echo "Not started"
echo "Starting"
nohup home/user/server/websrv &> my_script.out &
else
NUM=$(ps ax | grep $(cat $NPID) | grep -v grep | wc -l)
if [ $NUM -lt 1 ]
then
echo "Not working"
echo "Starting"
nohup home/user/server/websrv &> my_script.out &
else
ps ax | grep $(cat $NPID) | grep -v grep
echo "All Ok"
fi
fi
websrv gets JSON from user, and runs work.sh script itselves.
The problem is that sh script, which is invoked by websrv, "does not see" commands and stops with exit 1.
The script work.sh is like this:
#!/bin/sh -e
if [ "$#" -ne 1 ]; then
echo "Usage: $0 INPUT"
exit 1
fi
cd $(dirname $0) #good!
pwd #good!
IN="$1"
echo $IN #good!
KEYFORGIT="/some/path"
eval `ssh-agent -s` #good!
which ssh-add #good! (returns /usr/bin/ssh-add)
ssh-add $KEYFORGIT/openssh #error: exit 1!
git pull #error: exit 1!
cd $(dirname $0) #good!
rm -f somefile #error: exit 1!
#############==========Etc.==============
Usage of the full paths does not help.
If the script has been executed itself, it works.
If run.sh manually, it also works.
If I run the command nohup home/user/server/websrv & if works as well.
However, if all this chain of tools is started by cron on boot, work.sh is not able to perform any command except of cp, pwd, which, etc. But invoke of ssh-add, git, cp, rm, make etc., forces exit 1 status of the script. Why it "does not see" the commands? Unfortunately, I also cannot get any extended log which might explain the particular errors.
Try adding the path from the session that runs the script correctly to the cron entry (or inside the script)
Get the current path (where the script runs fine) with echo $PATH and add that to the crontab: replacing the string below with the output -> <REPLACE_WITH_OUTPUT_FROM_ABOVE>
#reboot export PATH=$PATH:<REPLACE_WITH_OUTPUT_FROM_ABOVE>; /home/user/scripts/run.sh > /dev/null 2>&1
You can compare paths with a cron entry like this to see what cron's PATH is:
* * * * * echo $PATH > /tmp/crons_path
Then cat /tmp/crons_path to see what it says.
Example output:
$ crontab -l | grep -v \#
* * * * * echo $PATH >> /tmp/crons_path
# wait a minute or so...
$ cat /tmp/crons_path
/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games:/snap/bin
$ echo $PATH
/home/ubuntu/.local/bin:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games:/snap/bin
As the commenter above mentioned, crontab doesn't always use the same path as user so likely something is missing.
Be sure to remove the temp cron entry after testing (crontab -e, etc.)...
I have a script (myscript.sh) which runs a few commands which need elevated privileges (i.e. needs to run with sudo).
Script is quite complex, but to demonstrate it is like below:
#!/bin/bash
echo "hello"
command1_which_needs_sudo
echo "hello2"
command2_which_needs_sudo
echo "hello3"
...
If I run it as a normal user without the required privileges:
$ ./myscript.sh
hello
must be super-user to perform this action
However if I run it with the correct privileges, it will work fine:
$ sudo ./myscript.sh
hello
hello2
hello3
Can I somehow achieve to run myscript.sh without sudo, and make the script requesting the elevated privileges only once in the beginning (and pass it back once it has finished)?
So obviously, sudo command1_which_needs_sudo will not be good, as command2 also need privileges.
How can I do this if I don't want to create another file, and due to script complexity I also don't want to do this with heredoc syntax?
If your main concern is code clarity, using wrapper functions can do a lot of good.
# call any named bash function under sudo with arbitrary arguments
run_escalated_function() {
local function_name args_q
function_name=$1; shift || return
printf -v args_q '%q ' "$#"
sudo bash -c "$(declare -f "$function_name"); $function_name $args_q"
}
privileged_bits() {
command1_which_needs_sudo
echo "hello2"
command2_which_needs_sudo
}
echo "hello"
run_escalated_function privileged_bits
echo "hello3"
If you want to run the script with root privileges without having to type sudo in the terminal nor having to type the password more than once then you can use:
#!/bin/bash
if [ "$EUID" -ne 0 ]
then
exec sudo -s "$0" "$#"
fi
echo "hello"
command1_which_needs_sudo
echo "hello2"
command2_which_needs_sudo
echo "hello3"
# ...
sudo -k
Update:
If your goal is to execute one part of the script with sudo rights then using a quoted here‑document is probably the easiest solution; there won't be any syntax issues because the current shell won't expand anything in it.
#!/bin/bash
echo "hello"
sudo -s var="hello2" <<'END_OF_SUDO'
command1_which_needs_sudo
echo "$var"
command2_which_needs_sudo
END_OF_SUDO
sudo -k
echo "hello3"
#...
remark: take notice that you can use external values in the here-document script by setting varname=value in the sudo command.
In reference to https://stackoverflow.com/a/11886837/1996022 (also shamelessly stole the title) where the question is how to capture the script's output I would like to know how I can additionally capture the scripts input. Mainly so scripts that also have user input produce complete logs.
I tried things like
exec 3< <(tee -ia foo.log <&3)
exec <&3 <(tee -ia foo.log <&3)
But nothing seems to work. I'm probably just missing something.
Maybe it'd be easier to use the script command? You could either have your users run the script with script directly, or do something kind of funky like this:
#!/bin/bash
main() {
read -r -p "Input string: "
echo "User input: $REPLY"
}
if [ "$1" = "--log" ]; then
# If the first argument is "--log", shift the arg
# out and run main
shift
main "$#"
else
# If run without log, re-run this script within a
# script command so all script I/O is logged
script -q -c "$0 --log $*" test.log
fi
Unfortunately, you can't pass a function to script -c which is why the double-call is necessary in this method.
If it's acceptable to have two scripts, you could also have a user-facing script that just calls the non-user-facing script with script:
script_for_users.sh
--------------------
#!/bin/sh
script -q -c "/path/to/real_script.sh" <log path>
real_script.sh
---------------
#!/bin/sh
<Normal business logic>
It's simpler:
#! /bin/bash
tee ~/log | your_script
The wonderful thing is your_script can be a function, command or a {} command block!
I am trying to write a bash script which simply acts as an emulator. It takes input from the user and executes the command while forwarding the command along with the result onto a file. I am unable to handle inputs which have either a | or a > in them.
The only option I could find was segregating the commands based on the | into an array and run them individually. However, this does not allow > redirects.
Thanking in advance.
$cmd is a command taken as input from the user
I used the command
$cmd 2>&1 | tee -a $flname
but this does not work if there is a | or a > in $cmd
/bin/bash -c "$cmd 2>&1 | tee -a $flname" does not run/store the command either
Try this:
#!/bin/bash
read -r -p "Insert command to execute"$'\n' cmd
echo "Executing '$cmd'"
/bin/bash -c "$cmd"
# or eval "$cmd"
Example of execution:
$ ./script.sh
Insert command to execute
printf '1\n2\n3\n4\n' | grep '1\|3'
Executing 'printf '1\n2\n3\n4\n' | grep '1\|3''
1
3
In a script.sh,
source a.sh
source b.sh
CMD1
CMD2
CMD3
how can I replace the source *.sh with their content (without executing the commands)?
I would like to see what the bash interpreter executes after sourcing the files and expanding all variables.
I know I can use set -n -v or run bash -n -v script.sh 2>output.sh, but that would not replace the source commands (and even less if a.sh or b.sh contain variables).
I thought of using a subshell, but that still doesn't expand the source lines. I tried a combination of set +n +v and set -n -v before and after the source lines, but that still does not work.
I'm going to send that output to a remote machine using ssh.
I could use <<output.sh to pipe the content into the ssh command, but I can't log as root onto the remote machine, but I am however a sudoer.
Therefore, I thought I could create the script and send it as a base64-encoded string (using that clever trick )
base64 script | ssh remotehost 'base64 -d | sudo bash'
Is there a solution?
Or do you have a better idea?
You can do something like this:
inline.sh:
#!/usr/bin/env bash
while read line; do
if [[ "$line" =~ (\.|source)\s+.+ ]]; then
file="$(echo $line | cut -d' ' -f2)"
echo "$(cat $file)"
else
echo "$line"
fi
done < "$1"
Note this assumes the sourced files exist, and doesn't handle errors. You should also handle possible hashbangs. If the sourced files contain themselves source, you need to apply the script recursively, e.g. something like (not tested):
while egrep -q '^(source|\.)' main.sh; do
bash inline.sh main.sh > main.sh
done
Let's test it
main.sh:
source a.sh
. b.sh
echo cc
echo "$var_a $var_b"
a.sh:
echo aa
var_a="stack"
b.sh:
echo bb
var_b="overflow"
Result:
bash inline.sh main.sh
echo aa
var_a="stack"
echo bb
var_b="overflow"
echo cc
echo "$var_a $var_b"
bash inline.sh main.sh | bash
aa
bb
cc
stack overflow
BTW, if you just want to see what bash executes, you can run
bash -x [script]
or remotely
ssh user#host -t "bash -x [script]"