当在块中匹配模式时，删除两个正则表达式标记之间的块

Question

假设以下结构：

  -   key1: value11
      key2:
      - value21
      - value22
      - value23
      key3: value31
      key4:
      - value41
      - value42
      key5: value51
  -   key1: value12
      key2:
      - value24
      - value25
      key3: value32
      key5: value52
  -   key1: value13
      key2:
      - value26
      key3: value33
      key4:
      - value43
      - value44
      - value45
      key5: value53

是否有可能删除开始标记和结束标记正则表达式之间（并包括）之间的所有块：

 - begin marker: '^[[:blank:]]{2}-[[:blank:]]{3}key1:[[:blank:]].+$'
 - end marker:   '^[[:blank:]]{6}key5:[[:blank:]].+$'

当以下正则表达式在块内匹配时：

matching pattern: '^[[:blank:]]{6}key3:[[:blank:]]value32$'?

目标是获得：

  -   key1: value11
      key2:
      - value21
      - value22
      - value23
      key3: value31
      key4:
      - value41
      - value42
      key5: value51
  -   key1: value13
      key2:
      - value26
      key3: value33
      key4:
      - value43
      - value44
      - value45
      key5: value53

开始标记还可以用作结束标记，因为在删除块期间未删除第二个标记。

我未成功使用sed / awk尝试多种方法，例如在post处从4.21段落启发而来的这种方法：

sed ':t
/^[[:blank:]]{2}-[[:blank:]]{3}key1:[[:blank:]].+$/,/^[[:blank:]]{6}key5:[[:blank:]].+$/ {      # For each line between these block markers
        /^[[:blank:]]{6}key5:[[:blank:]].+$/!{                                                  # If we are not at the /end/ marker
                $!{                                                                             # nor the last line of the file
                        N;                                                                      # add the Next line to the pattern space
                        bt
                }                                                                               # and branch (loop back) to the :t label
        }                                                                                       # This line matches the /end/ marker
        /^[[:blank:]]{6}key3:[[:blank:]]value32$/d;                                             # If /regex/ matches, delete the block
}' file

Answer 1

sed是对单个字符串执行s/old/new/的正确工具，仅此而已。对于更有趣的事情，您应该使用awk以获得清晰度，可移植性，鲁棒性，效率等。

您实际上不需要给定您发布的示例输入/输出所指定的第一个条件，例如使用GNU awk进行多字符RS和RT：

awk -v RS='[[:blank:]]{6}key5:[[:blank:]][^\n]+\n' -v ORS= '
    !/\n[[:blank:]]{6}key3:[[:blank:]]value32\n/{ print $0 RT }
' file
  -   key1: value11
      key2:
      - value21
      - value22
      - value23
      key3: value31
      key4:
      - value41
      - value42
      key5: value51
  -   key1: value13
      key2:
      - value26
      key3: value33
      key4:
      - value43
      - value44
      - value45
      key5: value53

或任何awk：

awk '
{ rec = rec $0 ORS }
/^[[:blank:]]{6}key5:[[:blank:]].+$/ {
    if ( rec !~ /\n[[:blank:]]{6}key3:[[:blank:]]value32\n/ ) {
        printf "%s", rec
    }
    rec=""
}
' file
  -   key1: value11
      key2:
      - value21
      - value22
      - value23
      key3: value31
      key4:
      - value41
      - value42
      key5: value51
  -   key1: value13
      key2:
      - value26
      key3: value33
      key4:
      - value43
      - value44
      - value45
      key5: value53

但是您可以根据需要使用它：

awk '
/^[[:blank:]]{2}-[[:blank:]]{3}key1:[[:blank:]].+$/ { inBlock=1 }
inBlock { rec = rec $0 ORS }
/^[[:blank:]]{6}key5:[[:blank:]].+$/ {
    if ( rec !~ /\n[[:blank:]]{6}key3:[[:blank:]]value32\n/ ) {
        printf "%s", rec
    }
    rec=""
    inBlock=0
}
' file
  -   key1: value11
      key2:
      - value21
      - value22
      - value23
      key3: value31
      key4:
      - value41
      - value42
      key5: value51
  -   key1: value13
      key2:
      - value26
      key3: value33
      key4:
      - value43
      - value44
      - value45
      key5: value53