bash 不传递参数时如何读取标准输入？
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How to read stdin when no arguments are passed?
提问by aron23
Script doesn't work when I want to use standard input when there are no arguments (files) passed. Is there any way how to use stdin instead of a file in this code?
当我想在没有传递参数（文件）的情况下使用标准输入时，脚本不起作用。有什么办法可以在这段代码中使用stdin而不是文件吗？
I tried this:
我试过这个：
if [ ! -n ] # check if argument exists
then
=$(</dev/stdin) # if not use stdin as an argument
fi
var=""
while read line
do
... # find the longest line
done <"$var"
采纳答案by pilcrow
Just substitute bash's specially interpreted /dev/stdinas the filename:
只需替换特别解释/dev/stdin为文件名的bash ：
VAR=
while read blah; do
...
done < "${VAR:-/dev/stdin}"
(Note that bash will actually use that special file/dev/stdinif built for an OS that offers it, but since bash 2.04 will work around that file's absence on systems that do not support it.)
（请注意，如果为提供它的操作系统构建，bash 实际上将使用该特殊文件，但由于 bash 2.04 将在不支持它的系统上解决该文件缺失的问题。）/dev/stdin
回答by Ryan
For a general case of wanting to read a value from stdin when a parameter is missing, this will work.
对于在缺少参数时想要从 stdin 读取值的一般情况，这将起作用。
$ echo param | script.sh
$ script.sh param
script.sh
脚本文件
#!/bin/bash
set -- "${1:-$(</dev/stdin)}" "${@:2}"
echo
回答by mklement0
pilcrow's answerprovides an elegant solution; this is an explanation of why the OP's approach didn't work.
pilcrow 的回答提供了一个优雅的解决方案；这是对为什么 OP 的方法不起作用的解释。
The main problem with the OP's approachwas the attempt to assign topositional parameter $1with $1=..., which won't work.
主要与OP的做法的问题是要尝试分配给位置参数$1有$1=...，这是行不通的。
The LHS is expanded by the shell to the valueof $1, and the resultis interpreted as the name of the variable to assign to - clearly, not the intent.
LHS由shell来扩展价值的$1，并且结果被解释为变量的名称，分配到-显然，并不是意图。
The only way to assign to $1in bash is via the setbuiltin.
The caveat is that setinvariably sets allpositional parameters, so you have to include the other ones as well, if any.
在 bash 中分配给$1set的唯一方法是通过内置的. 需要注意的是，set总是设置所有位置参数，因此您必须包括其他参数（如果有）。
set -- "${1:-/dev/stdin}" "${@:2}" # "${@:2}" expands to all remaining parameters
(If you expect only at most 1argument, set -- "${1:-/dev/stdin}"will do.)
（如果您期望最多只有1 个参数，set -- "${1:-/dev/stdin}"则可以。）
The above also corrects a secondary problem with the OP's approach: the attempt to store the contentsrather than the filenameof stdin in $1, since <is used.
以上还纠正了 OP 方法的一个次要问题：尝试在 中存储内容而不是stdin的文件名$1，因为<使用了。
${1:-/dev/stdin}is an application of bash parameter expansionthat says: return the value of $1, unless $1is undefined (no argument was passed) or its value is the empty string (""or ''was passed). The variation ${1-/dev/stdin}(no :) would only return /dev/stdinif $1is undefined(if it contains anyvalue, even the empty string, it would be returned).
${1:-/dev/stdin}是 bash参数扩展的应用程序，它说：返回 , 的值$1，除非$1未定义（未传递参数）或其值为空字符串（""或''已传递）。变化${1-/dev/stdin}（无:）将只返回/dev/stdin如果$1是不确定的（如果它包含任何值，甚至空字符串，它会被退回）。
If we put it all together:
如果我们把它们放在一起：
# Default to filename '/dev/stdin' (stdin), if none was specified.
set -- "${1:-/dev/stdin}" "${@:2}"
while read -r line; do
... # find the longest line
done < ""
But, of course, the much simpler approach would be to use ${1:-/dev/stdin}as the filename directly:
但是，当然，更简单的方法是${1:-/dev/stdin}直接用作文件名：
while read -r line; do
... # find the longest line
done < "${1:-/dev/stdin}"
or, via an intermediate variable:
或者，通过中间变量：
filename=${1:-/dev/stdin}
while read -r line; do
... # find the longest line
done < "$filename"
回答by kenorb
Here is my version of script:
这是我的脚本版本：
#!/bin/bash
file=${1--} # POSIX-compliant; ${1:--} can be used either.
while IFS= read -r line; do
printf '%s\n' "$line"
done < <(cat -- "$file")
If file is not present in the argument, read the from standard input.
如果参数中不存在文件，则从标准输入中读取。
See more examples: How to read from file or stdin in bash?at stackoverflow SE
查看更多示例：如何在 bash 中读取文件或标准输入？在 stackoverflow SE
回答by Oswald
Variables are assigned a value by Var=Valueand that variable is used by e.g. echo $Var. In your case, that would amount to
变量被赋值，Var=Value并且该变量被例如使用echo $Var。在你的情况下，这相当于
1=$(</dev/stdin)
when assigning the standard input. However, I do not think that variable names are allowed to start with a digit character. See the question bash read from file or stdinfor ways to solve this.
分配标准输入时。但是，我认为不允许变量名以数字字符开头。有关解决此问题的方法，请参阅问题bash 从文件或标准输入读取。
