awk 在特定字段中搜索模式并替换其内容

Question

我需要找到空的密码字段，带空格或制表符，并将其替换为 x(在/etc/passwd 文件上)

我在 awk 中发现了这个语法，它向用户显示第二个字段(使用 : 作为分隔符)是空的，还是里面有空格或制表符:

awk -F":" '($2 == "" || $2 == " " || $2 == "\t") {print $0}' $file

结果如下:

user1::53556:100::/home/user1:/bin/bash
user2: :53557:100::/home/user2:/bin/bash
user3:  :53558:100::/home/user3:/bin/bash

我怎么能告诉 awk 用另一个字符替换这个第二个字段(空的或空格或制表符)？ (例如 x)

Answer 1

能否请您尝试以下。

awk 'BEGIN{FS=OFS=":"} {$2=$2=="" || $2~/^[[:space:]]+$/?"X":$2} 1' Input_file

说明:添加以上代码的说明。

awk '                                          ##Starting awk program here.
BEGIN{                                         ##Starting BEGIN section here which will be executed before Input_file is being read.
  FS=OFS=":"                                   ##Setting FS and OFS as colon here for all lines of Input_file.
}                                              ##Closing BEGIN section block here.
{
  $2=$2=="" || $2~/^[[:space:]]+$/?"X":$2      ##Checking condition if $2(2nd field) of current line is either NULL or having complete space in it then put its vaklue as X or keep $2 value as same as it is.
}
1                                              ##mentioning 1 will print edited/non-edited current line.
' Input_file                                   ##Mentioning Input_file name here.

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编辑:根据 OP，OP 无需触及 Input_file 的最后一行，因此现在添加以下解决方案。

tac Input_file | awk 'BEGIN{FS=OFS=":"} FNR==1{print;next} {$2=$2=="" || $2~/^[[:space:]]+$/?"X":$2} 1' | tac

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EDIT2:如果您想在单个 awk 本身中执行此操作，请尝试执行以下操作。

awk '
BEGIN{
  FS=OFS=":"
}
prev{
  num=split(prev,array,":")
  array[2]=array[2]=="" || array[2]~/^[[:space:]]+$/?"X":array[2]
  for(i=1;i<=num;i++){
    val=(val?val OFS array[i]:array[i])
  }
  print val
  val=""
}
{
  prev=$0
}
END{
  if(prev){
    print prev
  }
}'  Input_file

如果您想更改 Input_file 本身，请在上面的代码中附加 > temp_file && mv temp_file Input_file。