Question

我正在使用shell变量来调用sed调用。该变量包含文件名为的路径名：

sed "file name is '$variable'"
...
variable=/path/path/file.txt

问题在于我不需要/path/path/部分。我只需输出file.txt部分。我的路径也是动态的所以我猜我需要在某个字符串中搜索（不知何故）从结尾处获得第一个斜杠。我该怎么做？

Answer 1

您可以使用basename来执行此操作：

basename /tmp/a.jpg
a.jpg

Answer 2

您可以使用shell的变量替换功能删除与glob模式匹配的部分：

$ variable=/path/path/file.txt
$ echo ${variable##*/}               # Remove longest left part matching "*/"
file.txt

来自bash手册：

${parameter#word}
${parameter##word}
   The word is expanded to produce a pattern just as in pathname expansion.  If
   the pattern matches the beginning of the value of parameter, then the result
   of the expansion is the expanded value of parameter with the shortest match-
   ing  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 parameter 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.

${parameter%word}
${parameter%%word}
   The word is expanded to produce a pattern just as in pathname expansion.  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 parameter 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.

从BASH调用的SED脚本：如何缩短输出中的变量路径？

2 个答案: