bash read is being skipped when run from curl pipe

问题

I'm building a bootstrap for a github project and would like it to be a simple one-liner. The script requires a password input.

This works and stops the script to wait for an input:

curl -s https://raw.github.com/willfarrell/.vhosts/master/setup.sh -o setup.sh
bash setup.sh

This does not, and just skips over the input request:

curl -s https://raw.github.com/willfarrell/.vhosts/master/setup.sh | bash

setup.sh contains code is something like:

# code before
read -p "Password:" -s password
# code after

Is it possible to have a clean one-liner? If so, how might one do it?

Workaround:

Use three commands instead of piping output.

curl -s https://raw.github.com/willfarrell/.vhosts/master/setup.sh -o vhosts.sh && bash vhosts.sh && rm vhosts.sh

回答1:

With the pipe, the read reads from standard input (the pipe), but the shell already read all the standard input so there isn't anything for the read to read.

来源：https://stackoverflow.com/questions/16854041/bash-read-is-being-skipped-when-run-from-curl-pipe