Bash：在命令行上使用空格输入保留字符串？
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Bash: preserve string with spaces input on command line?
提问by asking
I'd like to allow a string to be captured with spaces, so that:
我想允许用空格捕获字符串，以便：
echo -n "Enter description: "
read input
echo $input
Would produce:
会产生：
> Enter description: My wonderful description!
> My wonderful description!
Possible?
可能的？
回答by Gordon Davisson
The main thing to worry about is that when you refer to a variable without enclosing it in double-quotes, the shell does word splitting (splits it into multiple words wherever there's a space or other whitespace character), as well as wildcard expansion. Solution: use double-quotes whenever you refer to a variable (e.g. echo "$input").
需要担心的主要问题是，当您引用一个变量而不用双引号将其括起来时，shell 会进行分词（在有空格或其他空白字符的地方将其拆分为多个词）以及通配符扩展。解决方案：无论何时引用变量（例如echo "$input"），都使用双引号。
Second, readwill trim leading and trailing whitespace (i.e. spaces at the beginning and/or end of the input). If you care about this, use IFS= read(this essentially wipes out its definition of whitespace, so nothing gets trimmed). You might also want to use read's -r("raw") option, so it doesn't try to interpret backslash at the end of a line as a continuation character.
其次，read将修剪前导和尾随空格（即输入开头和/或结尾的空格）。如果您关心这一点，请使用IFS= read（这基本上消除了其对空白的定义，因此没有任何内容被修剪）。您可能还想使用read's -r("raw") 选项，因此它不会尝试将行尾的反斜杠解释为连续字符。
Finally, I'd recommend using read's -poption to supply the prompt (instead of echo -n).
最后，我建议使用read's-p选项来提供提示（而不是echo -n）。
With all of these changes, here's what your script looks like:
通过所有这些更改，您的脚本如下所示：
IFS= read -r -p "Enter description: " input
echo "$input"
