为每 N 次出现添加前缀，但来自特定的数字前缀

Question

我的源文件：

0.txt：

[dog$#@.txt ;`ffxx00`]
[dog$#@.txt ;`ffxx00`]
[dog$#@.txt ;`ffxx00`]
[dog$#@.txt ;`ffxx00`]
[dog$#@.txt ;`ffxx00`]
[dog$#@.txt ;`ffxx00`]

我有以下命令使用每 2 个 .txt 计数为数字添加前缀：

$ awk -v n=2 '{ newline=""; while ( x=index($0,".txt")) { if(cnt%n==0) pfx++; newline=newline substr($0,1,x-1) pfx substr($0,x,3) ; $0=substr($0,x+3); cnt++} print newline $0}' 0.txt 

[dog$#@1.txt ;`ffxx00`]
[dog$#@1.txt ;`ffxx00`]
[dog$#@2.txt ;`ffxx00`]
[dog$#@3.txt ;`ffxx00`]
[dog$#@3.txt ;`ffxx00`]
[dog$#@3.txt ;`ffxx00`]

但我想从特定的数字前缀开始计数，例如 3 这样我的结果是这样的：

[dog$#@3.txt ;`ffxx00`]
[dog$#@3.txt ;`ffxx00`]
[dog$#@4.txt ;`ffxx00`]
[dog$#@4.txt ;`ffxx00`]
[dog$#@5.txt ;`ffxx00`]
[dog$#@5.txt ;`ffxx00`]

Answer 1

给定：

printf ".txt%.0s\n" {1..6} >0.txt # makes your example...

你可以试试：

awk -v n=2 -v start=3 '
/^\.txt$/{  print start $O
            if (++cnt%n==0) start++
}
' 0.txt

打印：

3.txt
3.txt
4.txt
4.txt
5.txt
5.txt

---

通过更新（grrr），这将输出所需的输出和输入：

awk -v n=2 -v start=3 '
/\.txt /{
replacement=start ".txt "
sub(/\.txt /, replacement, $0)
if(++cnt%n==0) start++
} 1
' 0.txt

打印：

[dog$#@3.txt ;`ffxx00`]
[dog$#@3.txt ;`ffxx00`]
[dog$#@4.txt ;`ffxx00`]
[dog$#@4.txt ;`ffxx00`]
[dog$#@5.txt ;`ffxx00`]
[dog$#@5.txt ;`ffxx00`]