Bash command as variable

Question

I am trying to store the start of a sed command inside a variable like this:

sedcmd=\"sed -i \'\' \"

Later I then execute a command like so:

Answer 1

It works when the command is only one word long:

$ LS=ls
$ $LS

But in your case, the shell is trying the execute the program sed -i '', which does not exist.

The workaround is to use $SHELL -c:

$ $SHELL -c "$LS"
total 0

(Instead of $SHELL, you could also say bash, but that's not entirely reliable when there are multiple Bash installations, the shell isn't actually Bash, etc.)

However, in most cases, I'd actually use a shell function:

sedcmd () {
    sed -i '' "$@"
}