If I have a text file with a separate command on each line how would I make terminal run each line as a command? I just don't want to have to copy and paste 1 line at a time. It doesn't HAVE to be a text file... It can be any kind of file that will work.
example.txt:
sudo command 1
sudo command 2
sudo command 3
you can make a shell script with those commands, and then chmod +x <scriptname.sh>, and then just run it by
./scriptname.sh
Its very simple to write a bash script
Mockup sh file:
#!/bin/sh
sudo command1
sudo command2
.
.
.
sudo commandn
you can also just run it with a shell, for example:
bash example.txt
sh example.txt
Execute
. example.txt
That does exactly what you ask for, without setting an executable flag on the file or running an extra bash instance.
For a detailed explanation see e.g. https://unix.stackexchange.com/questions/43882/what-is-the-difference-between-sourcing-or-source-and-executing-a-file-i
You can use something like this:
for i in `cat foo.txt`
do
sudo $i
done
Though if the commands have arguments (i.e. there is whitespace in the lines) you may have to monkey around with that a bit to protect the whitepace so that the whole string is seen by sudo as a command. But it gives you an idea on how to start.
cat /path/* | bash
OR
cat commands.txt | bash
Related
I am new to linux, and I want to write to a .txt file all of the running processes on my PC that has the word "con" in them.
The script I wrote:
#!/bin/bash
ps -A | grep "con" > con_proc.txt
Why is this not working?
#!/bin/bash
ps -eaf | grep -i "con" > con_proc.txt
If you want to place inside of a script the contents of the script would be the above contents, for example script.sh.
To invoke the script you will need to do the following:
chmod +x script.sh
./script.sh
The first command gives the script execute permissions and the second command invokes the script.
Linux has pgrep to do this.
$ pgrep -a con
...
This is what was trying to do:
$ wget -qO- www.example.com/script.sh | sh
which quietly downloads the script and prints it to stdout which is then piped to sh. This unfortunately doesn't quite work, failing to wait for user input at various points, aswell as a few syntax errors.
This is what actually works:
$ wget -qOscript www.example.com/script.sh && chmod +x ./script && ./script
But what's the difference?
I'm thinking maybe piping the file doesn't execute the file, but rather executes each line individually, but I'm new to this kind of thing so I don't know.
When you pipe to sh , stdin of that shell/script will be the pipe. Thus the script cannot take e.g. user input from the console. When you run the script normally, stdin is the console - where you can enter input.
You might try telling the shell to be interactive:
$ wget -qO- www.example.com/script.sh | sh -i
I had the same issue, and after tinkering and googling this is what worked for me.
wget -O - www.example.com/script.sh | sh
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 know what they do. I was just wondering what kind of command are they. How can you make one using shell scripting.
For example, command like:
ignoreError ls /Home/
ignoreError mkdir /Home/
ignoreError cat
ignoreError randomcommand
Hope you get the idea
The way to do it in a shell script is with the "$#" construct.
"$#" expands to a quoted list of all of the arguments you passed to your shell script. $1 would be the command you want your shell script to run, and $2 $3 etc are the arguments to that command.
The only example I have is from cygwin. Cygwin does not have sudo, but I have this script that emulates it:
#!/usr/bin/bash
cygstart --action=runas "$#"
So when I run a command like
$ sudo ls -l
my sudo script does whatever it needs to do (cygstart --action=runas) and calls the ls command with the -l argument.
Try this script:
#!/bin/sh
"$#"
Call it, for example, run, make it runnable chmod u+x run, and try it:
$ run ls -l #or ./run ls -l
...
output of ls
...
The idea is that the script takes the parameters specified on the command line and use them as a (sub)command... Modify the script this way:
#!/bin/sh
echo "Trying to run $*"
"$#"
and you will see.
I'm trying some staff that is working perfectly when I write it in the regular shell, but when I include it in a bash script file, it doesn't.
First example:
m=`date +%m`
m_1=$((m-1))
echo $m_1
This gives me the value of the last month (actual minus one), but doesn't work if its executed from a script.
Second example:
m=6
m=$m"t"
echo m
This returns "6t" in the shell (concatenates $m with "t"), but just gives me "t" when executing from a script.
I assume all these may be answered easily by an experienced Linux user, but I'm just learning as I go.
Thanks in advance.
Re-check your syntax.
Your first code snippet works either from command line, from bash and from sh since your syntax is valid sh. In my opinion you probably have typos in your script file:
~$ m=`date +%m`; m_1=$((m-1)); echo $m_1
4
~$ cat > foo.sh
m=`date +%m`; m_1=$((m-1)); echo $m_1
^C
~$ bash foo.sh
4
~$ sh foo.sh
4
The same can apply to the other snippet with corrections:
~$ m=6; m=$m"t"; echo $m
6t
~$ cat > foo.sh
m=6; m=$m"t"; echo $m
^C
~$ bash foo.sh
6t
~$ sh foo.sh
6t
Make sure the first line of your script is
#!/bin/bash
rather than
#!/bin/sh
Bash will only enable its extended features if explicitly run as bash. If run as sh, it will operate in POSIX compatibility mode.
First of all, it works fine for me in a script, and on the terminal.
Second of all, your last line, echo m will just output "m". I think you meant "$m"..