using $ arguments on the command line and providing values for them at the end

Question

rather than write scripts for every single process I need to do, I like to just save commands in a text file and copy and paste them on the command line. Of course, some of the specifics of the commands change each time I paste and run it. Then I need to arrow thru the command and change the important parts. So, I would like to do something like this.

For a simple example:

chmod $1 /home/$2/.bashrc 740 bobsa

Can that be done some way? Is xargs something that can be used for this? I know I can set ENV variables before running the command to attain this but I was hoping from something else I can tac on the end thanks!

Answer 1

You can use the bash command with -c flag as follows:

bash -c 'chmod "$1" /home/"$2"/.bashrc' bash 740 bobsa

Remember to use single quotes for the bash command.

Answer 2

If you have commands you use often, write them as functions, and store them in the shell's startup files, e.g. .bashrc. That one could be defined like this (though you'd need a better name):

f() {
    chmod "$1" "/home/$2/.bashrc"
}

and called as

f 740 bobsa

That doesn't work too well if you need to edit the command often on the command line. But even then, you could just drop the function definition on the same command line and immediately run it:

f() { chmod "$1" "/home/$2/.bashrc"; }; f 740 bobsa

With {} you need the spaces and semicolons, but you could also use () instead. They start a subshell but you probably don't mind:

f() (chmod "$1" "/home/$2/.bashrc"); f 740 bobsa