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.
I'm stuck passing a parameter(URL to download) to a script.
My goal is to create a script for deployment that downloads and installs an app.
The script I run:
curl url_GitHub | bash -s url_download_app
The script on GitHub:
#! /bin/sh
url="$2"
filename=$(basename "$url")
workpath=$(dirname $(readlink -f $0))
curl $url -o $workpath/$filename -s
sudo dpkg --install $workpath/$filename
As I understood it doesn't pass the URL to download the app to the URL="$2" variable.
If I run the GitHub script locally, and pass the URL to download the app, it executes successfully.
Smth like:
bash install.sh -s url_download_app
Please help=)
-s appears to be an option intended for the downloaded script. However, it is also an option accepted by bash, so what I think you want is
curl url_GitHub | bash -s -- -s url_download_app
As the script on GitHub use $2, we should pass it as second argument :
curl url_GitHub | bash -s _ url_download_app
_ url_download_app will be passed to the script on GitHub.
What about the following (using process substitution):
bash <(curl -Ss url_GitHub) url_download_app
I did a proof of concept with the following script:
$ cat /tmp/test.sh
#!/bin/bash
echo "I got '$1'"
exit 0
and when you run it you get:
$ bash <(cat /tmp/test.sh) "test input argument"
I got 'test input argument'
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
I'm testing a Bash script I created on GitHub for behavioral correctness (e.g. that it parses options correctly). I want to do this without having to clone the repository locally, so here is how I'm doing it:
curl -sSL https://github.com/jamesqo/gid/raw/master/gid | xargs -0 bash -c
My question is, how can I pass arguments to the script in question? I tried bash -c --help, but that didn't work since it got interpreted as part of the script.
Thanks!
You’re actually over-complicating things by using xargs with Bash’s -c option.
Download the script directly
You don’t need to clone the repository to run the script. Just download it directly:
curl -o gid https://raw.githubusercontent.com/jamesqo/gid/master/gid
Now that it’s downloaded as gid, you can run it as a Bash script, e.g.,
bash gid --help
You can also make the downloaded script executable in order to run it as a regular Unix script file (using its shebang, #!/bin/bash):
chmod +x gid
./gid --help
Use process substitution
If you wanted to run the script without actually saving it to a file, you could use Bash process substitution:
bash <(curl -sSL https://github.com/jamesqo/gid/raw/master/gid) --help
I'll echo Anthony's comments - it makes a lot more sense to download the script and execute it directly, but if you're really set on using the -c option for bash, it's a little bit complicated, the problem is that when you do:
something | xargs -0 bash -c
there's no opportunity to pass any arguments. They all get swallowed as the argument to -c - it essentially gets turned into:
bash -c "$(something)"
so if you place something after the -c in the xargs, it gets before the something. There is no opportunity to put anything after something, as xargs doesn't let you.
If you want to pass arguments, you have to use the substitution position option for xargs, which allows you to place where the argument goes, The option is -J <item>, and the next thing to realize is that the first argument will be $0, so you have to do:
something | xargs -0 -I # bash -c # something <arg1> <arg2>…
I can emulate this with:
echo 'echo hi: ~$0~ ~$1~ ~$2~ ~$3~' | xargs -0 -I # bash -c # something one two three four
which yields:
hi: ~something~ ~one~ ~two~ ~three~