从卷曲管运行bash的时候读被跳过 [英] bash read is being skipped when run from curl pipe

问题描述

我要建一个github上项目的引导，并希望这是一个简单的一行。该脚本需要输入密码。

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.

这工作和停止脚本等待输入：

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

这不，刚跳过输入请求：

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

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

setup.sh包含code是这样的：

setup.sh contains code is something like:

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

是否有可能有一个干净的单行？如果是这样，怎么会做呢？

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

解决方法：

使用，而不是管道输出三个命令。

Use three commands instead of piping output.

卷曲-s https://raw.github.com/willfarrell/.vhosts/master/setup.sh -o vhosts.sh＆放大器;＆安培;庆典vhosts.sh＆放大器;＆安培; RM vhosts.sh

推荐答案

通过管道，在读从标准输入（管道）读取，但已经看过外壳所有的标准输入，所以没有任何的读阅读。

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.

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