Question

我有一个这样的文件：

device1;interface Gi0/1
device1;interface Gi0/2
device1;interface Gi0/3
device1; description Hello
device1;interface Gi0/4
device2;interface Gi0/1
device2; ip vrf forwarding xxx-vrf
device2;interface Gi0/2
device2;interface Gi0/3
device2;interface Gi0/4
device2; ip address 192.168.1.1 255.255.255.254

我希望这会返回每个没有其他配置的接口，这些接口在下一行没有以'interface'开头（见上文） - 即。在这种情况下，我想要：

device1;interface Gi0/1
device1;interface Gi0/2
device1;interface Gi0/4
device1;interface Gi0/2
device2;interface Gi0/3

我发现这很棘手，我不知道如何去做。为我的业余道歉。

第

Answer 1

尝试：

awk '/interface/{if(p)print p; p=$0; next}{p=x} END{if(p)print p}' file

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以下是解释：

awk '
  /interface/ {      # if a line contains "interface"
    if(p"")print p   # then if variable p is non-empty (and if the previous line contained
                     # "interface"), then print its content.
    p=$0             # store the current line in variable "p" (if it contains "interface)
    next             # read the next line
  }
  {                  # if the line does not contain "interface"
    p=x              # empty variable p by assigning an empty variable to it
  }
  END {              # After all lines have been processed
    if(p"")print p   # then if variable p still contains something, print it.
  }
' file

编辑：如果条件强制将其加入字符串上下文，则在p in中添加双引号。

Answer 2

另一个awk

awk 'a=/interface/{if(b)print b;b=$0}!a{b=""}END{print b}' file

制作

device1;interface Gi0/1
device1;interface Gi0/2
device1;interface Gi0/4
device2;interface Gi0/2
device2;interface Gi0/3

返回包含字符串的行

2 个答案:

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