Command substitution interpreting spaces within quotes as argument separators

Question

I'd like an alias that additionally appends itself to ~/.bashrc, e.g.

function tailias
{
    $(echo "alias $1='${*:2}'" | tee -a ~/.bashrc)
}

I'm using tee to split the command to ~/.bashrc while leaving the stdout intact to be run by $().

This works fine for aliasing one-word commands, e.g.

[/home/acheong87] tailias f g
[/home/acheong87] alias
alias f='g'

and I see it appended to my ~/.bashrc. But for commands with spaces, I get

[/home/acheong87] tailias ll 'ls -al'
-bash: alias: -al: not found

It looks correct in ~/.bashrc—

alias ll='ls -al'

—but in the alias list I get this crazy thing:

[/home/acheong87] alias
alias ll=''\''ls'

I've tried putting the quotes in different places, to no avail. I've reduced the function to just the echo, and like in the ~/.bashrc, it looks correct—so I can't figure out what hidden step is averting my attention.

Answer 1

You can eval to achieve what you want:

function tailias
{
    eval $(echo "alias $1='${*:2}'" | tee -a ~/.bashrc)
}

This is better expained in this question (where an alternative solution is given)

https://stackoverflow.com/questions/11079342/execute-command-containing-quotes-from-shell-variable

or directly in the bash FAQ:

http://mywiki.wooledge.org/BashFAQ/050

Answer 2

Instead of using eval and tee to execute the command twice to add both a temporary alias and permanent alias in ~/.bashrc, you can simply reload the ~/.bashrc file after it is appended, which will update the alias list.

function tailias
{
    echo "alias $1='${*:2}'" >> ~/.bashrc && source ~/.bashrc
}