输出shell变量时如何保留空格? [英] How to preserve spaces when outputting a shell variable?

问题描述

出于字符串匹配的目的，我需要定义一个带前导空格的bash变量. 我需要从一个整数开始定义它，例如:

For string matching purposes I need to define a bash variable with leading spaces. I need to define this starting from an integer, like:

jj=5

printf在我看来是个好主意，因此，如果我想填充最多6个字符的空格:

printf seems to me a good idea, so if I want to fill spaces up to 6 character:

jpat=`printf "  %6i"  $jj`

但是当我尝试调出变量时却很不幸:

but unluckly when I am trying to recall the variable:

echo $jpat

前导空格被删除，我只得到$jj整数.

the leading whitespaces are removed and I only get the $jj integer as it was.

有任何解决方案来保留这样的空间吗?

Any solution to keep such spaces?

(这等效于:v=' val'; echo $v$v.为什么输出中没有前导空格和多个空格?)

(This is equivalent to this: v=' val'; echo $v$v. Why aren't there leading and multiple spaces in output?)

推荐答案

使用更多引号！ 会做您想要的.

Use More Quotes! echo "$jpat" will do what you want.

您正在执行的操作还有另一个问题:命令替换将删除尾随的换行符.在您使用的printf命令中这不是问题，但是例如分配jpat=$(printf " %6i\n" "$jj")将为您提供与命令完全相同的结果.

There is another issue with what you're doing: Command substitutions will remove trailing newlines. It's not an issue in the printf command you're using, but for example assigning jpat=$(printf " %6i\n" "$jj") would give you exactly the same result as your command.

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