从管道将值读入 shell 变量 [英] Read values into a shell variable from a pipe

问题描述

我试图让 bash 处理来自 stdin 的数据，这些数据通过管道输入，但没有运气.我的意思不是以下工作:

I am trying to get bash to process data from stdin that gets piped into, but no luck. What I mean is none of the following work:

echo "hello world" | test=($(< /dev/stdin)); echo test=$test
test=

echo "hello world" | read test; echo test=$test
test=

echo "hello world" | test=`cat`; echo test=$test
test=

我希望输出是 test=hello world.我试过在 "$test" 周围加上 "" 引号也不起作用.

where I want the output to be test=hello world. I've tried putting "" quotes around "$test" that doesn't work either.

推荐答案

使用

IFS= read var << EOF
$(foo)
EOF

你可以欺骗read从管道接受这样的:

You can trick read into accepting from a pipe like this:

echo "hello world" | { read test; echo test=$test; }

甚至写一个这样的函数:

or even write a function like this:

read_from_pipe() { read "$@" <&0; }

但是没有意义——你的变量赋值可能不会持续！管道可能会产生一个子外壳，其中的环境是通过值继承的，而不是通过引用继承的.这就是为什么 read 不关心来自管道的输入 - 它是未定义的.

But there's no point - your variable assignments may not last! A pipeline may spawn a subshell, where the environment is inherited by value, not by reference. This is why read doesn't bother with input from a pipe - it's undefined.

仅供参考，http://www.etalabs.net/sh_tricks.html 很不错收集必要的 cruft 以对抗 bourne shell 的奇怪和不兼容，sh.

FYI, http://www.etalabs.net/sh_tricks.html is a nifty collection of the cruft necessary to fight the oddities and incompatibilities of bourne shells, sh.

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