Source From Standard In (Bash on OSX)

Question

I am trying to do something like this

 ruby test.rb | source /dev/stdin

where test.rb just prints out cd /. There are no errors, but it doesn't do anything either. If I use this:

 ruby test.rb > /tmp/eraseme2352; source /tmp/eraseme2352

it works fine, but I want to avoid the intermediate file.

Edit: The whole point of this is that the changes need to persist when the command is done. Sorry I didn't make that clearer earlier.

Answer 1

How about just

ruby test.rb | bash

Answer 2

Until a more experienced bash hacker comes along to correct me, you could do this:

for c in `ruby test.rb` ; do $c ; done

Caution: This doesn't do what you want. Read the comments!

Answer 3

you can try:

$(ruby test.rb)

$(...) tells bash to execute whatever output is produced by command inside ().

Answer 4

eval `ruby test.rb`

Answer 5

bash (not sh):

while read -a line
do
  "${line[@]}"
done < <(somescript)

Spaces in arguments to commands will need to be backslash-escaped in order to work.