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解释

我需要在lvm.conf文件中找到以 string 开头的行，global_filter并删除方括号之间的所有内容（除了"r/.*/".

只有 1 行以 global_filter 开头。

拆除前：

global_filter = [ "a|^/dev/sda.*$|", "a|^/dev/sdb.*$|", "r/.*/"]

删除后，应该是 -

global_filter = [, "r/.*/"]

答案1

sed -i -e '/^global_filter/{\|"r/\.\*/"|s|.*|global_filter = [, "r/.*/"]|;}' data

解释

sed -i -e '
   # look at only lines that begin with global_filter
   /^global_filter/{
      # and also they should contain the string "r/.*/"
      # change everything to the following expression
      \|"r/\.\*/"| s|.*|global_filter = [, "r/.*/"]|

      ## alternatively could be written:
      #\|"r/\.\*/"|c\
#global_filter = [, "r/.*/"]
   }
' data

答案2

尝试这个：

$ echo 'global_filter = [ "a|^/dev/sda.$|", "a|^/dev/sdb.$|", "r/.*/"]'|\
sed -E 's/(^global_filter = \[)(.*)(, "r.*$)/\1\3/'

输出：

global_filter = [, "r/.*/"]

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