i have uploaded a test script remote.sh to a remote webserver like this :
#!/usr/bin/bash
echo "input var is : $1"
and i have a local script local.sh like this :
#!/usr/bin/bash
curl -sS https://remote_host/remote.sh | bash
then i run the local script with some inline parameter :
./local.sh "some input here."
but the remote script i grabbed doesn't seem to see the local inline parameter. how can this be done ?
Your code is starting a second copy of bash, and not passing the arguments retrieved to it.
I would generally suggest not starting a second copy of bash at all:
#!/usr/bin/env bash
eval "$(curl -sS https://remote_host/remote.sh)"
...but you could proceed to do so and pass them through. The following passes that code on the command line, leaving stdin free (so the new copy of bash being started can use it to prompt the user):
#!/bin/sh
code=$(curl -sS https://remote_host/remote.sh) || exit
exec bash -c "$code" bash "$#"
...or, to continue using stdin to pass code, bash -s can be used:
#!/bin/sh
curl -sS https://remote_host/remote.sh | bash -s -- "$#"
By the way -- everywhere I use /bin/sh above you could substitute /bin/bash or any other POSIX-compliant shell; the point being made is that the code given above does not depend on behaviors that are unspecified in the POSIX.2 standard.
Say I have a file at the URL http://mywebsite.example/myscript.txt that contains a script:
#!/bin/bash
echo "Hello, world!"
read -p "What is your name? " name
echo "Hello, ${name}!"
And I'd like to run this script without first saving it to a file. How do I do this?
Now, I've seen the syntax:
bash < <(curl -s http://mywebsite.example/myscript.txt)
But this doesn't seem to work like it would if I saved to a file and then executed. For example readline doesn't work, and the output is just:
$ bash < <(curl -s http://mywebsite.example/myscript.txt)
Hello, world!
Similarly, I've tried:
curl -s http://mywebsite.example/myscript.txt | bash -s --
With the same results.
Originally I had a solution like:
timestamp=`date +%Y%m%d%H%M%S`
curl -s http://mywebsite.example/myscript.txt -o /tmp/.myscript.${timestamp}.tmp
bash /tmp/.myscript.${timestamp}.tmp
rm -f /tmp/.myscript.${timestamp}.tmp
But this seems sloppy, and I'd like a more elegant solution.
I'm aware of the security issues regarding running a shell script from a URL, but let's ignore all of that for right now.
source <(curl -s http://mywebsite.example/myscript.txt)
ought to do it. Alternately, leave off the initial redirection on yours, which is redirecting standard input; bash takes a filename to execute just fine without redirection, and <(command) syntax provides a path.
bash <(curl -s http://mywebsite.example/myscript.txt)
It may be clearer if you look at the output of echo <(cat /dev/null)
This is the way to execute remote script with passing to it some arguments (arg1 arg2):
curl -s http://server/path/script.sh | bash /dev/stdin arg1 arg2
For bash, Bourne shell and fish:
curl -s http://server/path/script.sh | bash -s arg1 arg2
Flag "-s" makes shell read from stdin.
Use:
curl -s -L URL_TO_SCRIPT_HERE | bash
For example:
curl -s -L http://bitly/10hA8iC | bash
Using wget, which is usually part of default system installation:
bash <(wget -qO- http://mywebsite.example/myscript.txt)
You can also do this:
wget -O - https://raw.github.com/luismartingil/commands/master/101_remote2local_wireshark.sh | bash
The best way to do it is
curl http://domain/path/to/script.sh | bash -s arg1 arg2
which is a slight change of answer by #user77115
You can use curl and send it to bash like this:
bash <(curl -s http://mywebsite.example/myscript.txt)
I often using the following is enough
curl -s http://mywebsite.example/myscript.txt | sh
But in a old system( kernel2.4 ), it encounter problems, and do the following can solve it, I tried many others, only the following works
curl -s http://mywebsite.example/myscript.txt -o a.sh && sh a.sh && rm -f a.sh
Examples
$ curl -s someurl | sh
Starting to insert crontab
sh: _name}.sh: command not found
sh: line 208: syntax error near unexpected token `then'
sh: line 208: ` -eq 0 ]]; then'
$
The problem may cause by network slow, or bash version too old that can't handle network slow gracefully
However, the following solves the problem
$ curl -s someurl -o a.sh && sh a.sh && rm -f a.sh
Starting to insert crontab
Insert crontab entry is ok.
Insert crontab is done.
okay
$
Also:
curl -sL https://.... | sudo bash -
Just combining amra and user77115's answers:
wget -qO- https://raw.githubusercontent.com/lingtalfi/TheScientist/master/_bb_autoload/bbstart.sh | bash -s -- -v -v
It executes the bbstart.sh distant script passing it the -v -v options.
Is some unattended scripts I use the following command:
sh -c "$(curl -fsSL <URL>)"
I recommend to avoid executing scripts directly from URLs. You should be sure the URL is safe and check the content of the script before executing, you can use a SHA256 checksum to validate the file before executing.
instead of executing the script directly, first download it and then execute
SOURCE='https://gist.githubusercontent.com/cci-emciftci/123123/raw/123123/sample.sh'
curl $SOURCE -o ./my_sample.sh
chmod +x my_sample.sh
./my_sample.sh
This way is good and conventional:
17:04:59#itqx|~
qx>source <(curl -Ls http://192.168.80.154/cent74/just4Test) Lord Jesus Loves YOU
Remote script test...
Param size: 4
---------
17:19:31#node7|/var/www/html/cent74
arch>cat just4Test
echo Remote script test...
echo Param size: $#
If you want the script run using the current shell, regardless of what it is, use:
${SHELL:-sh} -c "$(wget -qO - http://mywebsite.example/myscript.txt)"
if you have wget, or:
${SHELL:-sh} -c "$(curl -Ls http://mywebsite.example/myscript.txt)"
if you have curl.
This command will still work if the script is interactive, i.e., it asks the user for input.
Note: OpenWRT has a wget clone but not curl, by default.
bash | curl http://your.url.here/script.txt
actual example:
juan#juan-MS-7808:~$ bash | curl https://raw.githubusercontent.com/JPHACKER2k18/markwe/master/testapp.sh
Oh, wow im alive
juan#juan-MS-7808:~$
As the title says, within linux how can I feed input to the bash when I do sudo bash
Lets say I have a bash script that reads the name.
The way I execute the script is through sudo using:
cat read-my-name-script.sh | sudo bash
Lets just say this is how I execute the script throught the network.
Now I want to fill the name automatically, is there a way to feed the input. I tried doing this: cat read-my-name-script.sh < name-input-file | sudo bash where the name-input-file is a file for the input that the user will be using to feed the script.
I am new to linux and learning to automate the input and wanted to create a file for input where the user can fill it and feed it to my script.
This is convoluted, but might do what you want.
sudo bash -c "$(cat read-my-name.sh)" <name-input-file
The -c says the next quoted argument are the commands to run (so, read the script as a string on the command line, instead of from a file), and the calling shell interpolates the contents of the file inside the double quotes before the sudo command gets evaluated. So if read-my-name.sh contains
#!/bin/bash
read -p "I want your name please"
then the command gets expanded into
sudo bash -c '#!/bin/bash
read -p "I want your name please"' <name-input-file
(where of course at this time the shell has actually removed the outer double quotes altogether; I put in single quotes in their place instead to show how this would look as actually executable, syntactically valid code).
I think you need that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done <name-input-file
So each line of name-input-file will be passed as argument to sudo bash read-my-name-script.sh
If your argslist located on http server, you can do that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done < <(wget -q -O- http://some/address/in/internet/name-input-file)
UPD
add [[ -f name-input-file ]] && readarray -t args <name-input-file
to read-my-name-script.sh
and use "${args[#]}" as arguments of command in the script.
For example echo "${args[#]}" or cmd "${args[0]}" "${args[1]}" ... "${args[100]}" in any order.
In this case you can use
wget -q -O- http://some/address/in/internet/read-my-name-script.sh | bash
for run your script with arguments from name-input-file whitout saving script to the local machine
This question already has answers here:
Usage of expect command within a heredoc
(1 answer)
Pass commands as input to another command (su, ssh, sh, etc)
(3 answers)
Closed 4 years ago.
I have a script that I need to execute through ssh as another use, is there a way to pass whole script like this:
ssh -t user#server.com sudo -u user2 sh -c << EOF
cd /home
ls
dir=$(pwd)
echo "$dir"
echo "hello"
....
EOF
Returns: sh: -c: option requires an argument
ssh'ing and sudo'ing separately is not an option and putting .sh file directly on the machine is not possible.
sh -c requires a command string as the argument. Since you are reading the commands from standard input (through heredoc), you need to use sh -s option:
ssh -t user#server.com sudo -u user2 sh -s << 'EOF'
cd /home
ls
dir=$(pwd)
echo "$dir"
echo "hello"
...
EOF
From man sh:
-c
string If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.
-s
If the -s option is present, or if no arguments remain after option processing, then commands are read from the standard input. This option allows the positional parameters to be set when invoking an interactive shell.
Need to quote the heredoc marker to prevent the parent shell from interpreting the content.
I need your help in understanding this behaviour of sudo.
sudo -s -- 'ls -l' this command works but sudo 'ls -l' throws error saying
sudo: ls -l: command not found I realize it treats the entire string within quote as single command (including the spaces) but what I don't get is how does it work fine with -s flag but fails when -s is not there.
Without -s, the first argument is the name of the command to execute. With -s, the first argument is a string passed to the -c option of whatever shell ($SHELL or your system shell) is used to execute the argument.
That is, assuming $SHELL is sh, the following are equivalent:
sudo -s -- 'ls -l'
sudo -- sh -c 'ls -l'
From the sudo man page:
-s [command]
The -s (shell) option runs the shell specified by the SHELL environment variable if it is set or the shell as specified in
the password database. If a command is specified, it is passed to the
shell for execution via the shell's -c option. If no command is
specified, an interactive shell is executed.
It behaves like it does because a new shell is spawned which breaks up the words in your "quoted command" like shells do.