bash 输出shell变量时如何保留空格？
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How to preserve spaces when outputting a shell variable?
提问by gluuke
For string matching purposes I need to define a bash variable with leading spaces. I need to define this starting from an integer, like:
出于字符串匹配的目的，我需要定义一个带有前导空格的 bash 变量。我需要从一个整数开始定义它，例如：
jj=5
printf seems to me a good idea, so if I want to fill spaces up to 6 character:
printf 在我看来是个好主意，所以如果我想填充最多 6 个字符的空格：
jpat=`printf " %6i" $jj`
but unluckly when I am trying to recall the variable:
但不幸的是，当我试图回忆这个变量时：
echo $jpat
the leading whitespaces are removed and I only get the $jjinteger as it was.
前导空格被删除，我只得到$jj原来的整数。
Any solution to keep such spaces?
有什么办法可以保留这样的空间吗？
(This is equivalent to this: v=' val'; echo $v$v. Why aren't there leading and multiple spaces in output?)
（这相当于：v=' val'; echo $v$v。为什么输出中没有前导和多个空格？）
回答by l0b0
Use More Quotes!echo "$jpat"will do what you want.
使用更多报价！echo "$jpat"会做你想做的。
There is another issue with what you're doing: Command substitutions will removetrailing newlines. It's not an issue in the printfcommand you're using, but for example assigning jpat=$(printf " %6i\n" "$jj")would give you exactly the same result as your command.
您正在做的事情还有另一个问题：命令替换将删除尾随的换行符。这在printf您使用的命令中不是问题，但例如分配jpat=$(printf " %6i\n" "$jj")会给您与您的命令完全相同的结果。
