如何在 Bash 中使用 read 命令？
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How to use the read command in Bash?
提问by Determinant
When I try to use the readcommand in Bash like this:
当我尝试read在 Bash 中使用该命令时：
echo hello | read str
echo $str
Nothing echoed, while I think strshould contain the string hello. Can anybody please help me understand this behavior?
没有回显，而我认为str应该包含字符串hello. 有人可以帮我理解这种行为吗？
回答by jpalecek
The readin your script command is fine. However, you execute it in the pipeline, which means it is in a subshell, therefore, the variables it reads to are not visible in the parent shell. You can either
该read脚本中的命令是好的。但是，您在管道中执行它，这意味着它在子 shell 中，因此，它读取的变量在父 shell 中不可见。你可以
move the rest of the script in the subshell, too:
echo hello | { read str echo $str }or use command substitution to get the value of the variable out of the subshell
str=$(echo hello) echo $stror a slightly more complicated example (Grabbing the 2nd element of ls)
str=$(ls | { read a; read a; echo $a; }) echo $str
也在子shell中移动脚本的其余部分：
echo hello | { read str echo $str }或者使用命令替换从子shell中获取变量的值
str=$(echo hello) echo $str或者稍微复杂一点的例子（获取 ls 的第二个元素）
str=$(ls | { read a; read a; echo $a; }) echo $str
回答by glenn Hymanman
Other bash alternatives that do not involve a subshell:
其他不涉及子 shell 的 bash 替代方案：
read str <<END # here-doc
hello
END
read str <<< "hello" # here-string
read str < <(echo hello) # process substitution
回答by Shizzmo
Typical usage might look like:
典型用法可能如下所示：
i=0
echo -e "hello1\nhello2\nhello3" | while read str ; do
echo "$((++i)): $str"
done
and output
和输出
1: hello1
2: hello2
3: hello3
回答by l0b0
The value disappears since the read command is run in a separate subshell: Bash FAQ 24
由于 read 命令在单独的子 shell 中运行，该值消失了：Bash FAQ 24
回答by RAKK
To put my two cents here: on KSH, reading as is to a variable willwork, because according to the IBM AIX documentation, KSH's readdoesaffects the current shell environment:
把我的两分钱放在这里：在 KSH 上，read按原样对变量进行操作是可行的，因为根据IBM AIX 文档，KSHread确实会影响当前的 shell 环境：
The setting of shell variables by the read command affects the current shell execution environment.
read命令对shell变量的设置会影响当前的shell执行环境。
This just resulted in me spending a good few minutes figuring out why a one-liner ending with readthat I've used a zillion times before on AIX didn't work on Linux... it's because KSH does saves to the current environment and BASH doesn't!
这只是导致我花了几分钟弄清楚为什么read我之前在 AIX 上使用了无数次的单行结尾在 Linux 上不起作用......这是因为 KSH 确实保存到当前环境和 BASH没有！
回答by HymanDR
I really only use read with "while" and a do loop:
我真的只将 read 与 "while" 和 do 循环一起使用：
echo "This is NOT a test." | while read -r a b c theRest; do
echo "$a" "$b" "$theRest"; done
This is a test.
For what it's worth, I have seen the recommendation to always use -r with the read command in bash.
这是一个测试。
就其价值而言，我已经看到了在 bash 中始终将 -r 与 read 命令一起使用的建议。
回答by jxqz
Another alternative altogether is to use the printf function.
另一种选择是使用 printf 函数。
printf -v str 'hello'
Moreover, this construct, combined with the use of single quotes where appropriate, helps to avoid the multi-escape problems of subshells and other forms of interpolative quoting.
此外，这种结构与在适当情况下使用单引号相结合，有助于避免子外壳和其他形式的插值引用的多重转义问题。
回答by mmrtnt
Do you need the pipe?
你需要管子吗？
echo -ne "$MENU"
read NUMBER
