Question

给定一个目录树，我将如何从字符串中删除分隔符的最后一个字段并返回没有该分隔符的字符串，假设我不知道该字符串的结束位置？

例如，给定

/1/2/3/4/5

我知道我可以用

返回5

cut -f 5 -d '/'

如果我知道最后一个字段是第5个字段，或者a = / 1/2/3/4/5

echo ${a##*/}

选择最后一个字段。但是，如何返回原始字符串减去最后一个字段呢？即

/1/2/3/4

Answer 1

您可以使用%或%%运算符代替##。来自Bash手册页：

 ${parameter%word}
 ${parameter%%word}
          The word is expanded to produce a pattern just as  in  pathname  expan-
          sion.   If the pattern matches a trailing portion of the expanded value
          of parameter, then the result of the expansion is the expanded value of
          parameter  with  the  shortest matching pattern (the ``%'' case) or the
          longest matching pattern (the ``%%'' case) deleted.  If parameter is  @
          or  *,  the  pattern  removal  operation  is applied to each positional
          parameter in turn, and the expansion is the resultant list.  If parame-
          ter  is  an array variable subscripted with @ or *, the pattern removal
          operation is applied to each member of  the  array  in  turn,  and  the
          expansion is the resultant list.

简而言之，${a%/*}可以解决问题。

Answer 2

您可以剪切一系列字段

cut -f 1-4 -d '/'

从一个字符串中切出一个部分并返回相反的部分

2 个答案: