I am looking into running matlab script in Linux similar to bash/python scripts. I.e., a matlab script that can be run as an application.
You can get a similar effect without your custom mash script by adding the following header to the files you want to be executable:
#/usr/bin/bash
/path/to/matlab -r "$(sed -n -e '4,$p' < "$0")"
exit $?
If you want matlab to terminate after executing the script, as in your example, you could replace the second line with
sed -n -e '4,$p' < "$0" | /path/to/matlab
The idea here is to execute a bash command that simply clips off the header of the script, and passes the rest along to matlab.
Here is the implementation I came up with -
Create /usr/bin/mash script file containing the following lines -
#!/bin/bash
grep -ve '^(#!\|^\s*$)' ${#: -1} | ${#: 1:$#-1}
exit $?
Make mash script executable -
$ chmod +x /usr/bin/mash
Write matlab script file called test.msh
#!/usr/bin/mash /usr/local/MATLAB/R2012a/bin/matlab -nodisplay
format long
a = 2*pi % matlab commands ...
Make test.msh script executable -
$ chmod +x mash
Run test.msh
$ ./test.msh
...
>> >> a =
6.283185307179586
Related
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:
Run bash commands from txt file
(4 answers)
Closed 4 years ago.
I tried to execute commands read it from txt file. But only 1st command is executing, after that script is terminated. My script file name is shellEx.sh is follows:
echo "pwd" > temp.txt
echo "ls" >> temp.txt
exec < temp.txt
while read line
do
exec $line
done
echo "printed"
if I keep echo in the place of exec, just it prints both pwd and ls. But i want to execute pwd and ls one by one.
o/p am getting is:
$ bash shellEx.sh
/c/Users/Aditya Gudipati/Desktop
But after pwd, ls also need to execute for me.
Anyone can please give better solution for this?
exec in bash is meant in the Unix sense where it means "stop running this program and start running another instead". This is why your script exits.
If you want to execute line as a shell command, you can use:
line="find . | wc -l"
eval "$line"
($line by itself will not allow using pipes, quotes, expansions or other shell syntax)
To execute the entire file including multiline commands, use one of:
source ./myfile # keep variables, allow exiting script
bash myfile # discard variables, limit exit to myfile
A file with one valid command per line is itself a shell script. Just use the . command to execute it in the current shell.
$ echo "pwd" > temp.txt
$ echo "ls" >> temp.txt
$ . temp.txt
Consider the following shell script on example.com
#/bin/bash
export HELLO_SCOPE=WORLD
eval $#
Now, I would like to download and then execute this shell script with parameters in the simplest way and be able to launch an interactive bash terminal with the HELLO_SCOPE variable set.
I have tried
curl http://example.com/hello_scope.sh | bash -s bash -i
But it quits the shell immediately. From what I can understand, it's because curls stdout, the script, remains the stdin of the bash, preventing it from starting interactively (as that would require my keyboard to be stdin).
Is there a way to avoid this without going through the extra step of creating a temporary file with the shell script?
You can source it:
# open a shell
. <(curl http://example.com/hello_scope.sh)
# type commands ...
You could just download this script you (using wget for example) and source this script, isn't it ?
script_name="hello_scope.sh"
[[ -f $script_name ]] && rm -rf "$script_name"
wget "http://example.com/$script_name" -O "$script_name" -o /dev/null
&& chmod u+x "$script_name"
&& source "$script_name"
You could use . "$script_name" instead of source "$script_name" if you want (. is POSIX compliant). You could write the previous code in a script and source it to have interactive shell with the setted variable $HELLO_SCOPE.
Finally you could remove the eval line in your remote shell script.
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 $i
done
Though if the commands have arguments (i.e. there is whitespace in the lines) you may have to monkey around with that a bit to protect the whitepace so that the whole string is seen by sudo as a command. But it gives you an idea on how to start.
cat /path/* | bash
OR
cat commands.txt | bash
I'm trying some staff that is working perfectly when I write it in the regular shell, but when I include it in a bash script file, it doesn't.
First example:
m=`date +%m`
m_1=$((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"..