Question

为此，我使用了如下所示的awk命令

awk '/H.*/{x="F"++i;next}{print NR-1 "," $0 > x;}' words.txt

当任何“页眉”模式匹配时都会拆分为多个文件。

words.txt
Header
LLLL
AAAA
Header
SSSS
DDDD

现在拆分后通过上述命令获取输出

File1.txt
1. LLLL
2. AAAA
File2.txt
3. SSSS
4. DDDD

期望的是每个文件中从1开始的行号，如下所示

File1.txt
1. LLLL
2. AAAA
File2.txt
1. SSSS
2. DDDD

Answer 1

如果需要打印台虎钳（与您的条件匹配），请使用以下内容。

awk '/H.*/{count=1;close(x);x="F"++i;next}{print count++ "," $0 > x;}' words.txt

还要添加close以避免错误，这有时会给我们“打开的文件过多”

说明： 现在也添加了上述代码的说明。

awk '                             ##Starting awk program here.
/H.*/{                            ##Checking condition from H.* to till everything it covers in line.
  count=1                         ##Setting variable named count value to 1 here.
  close(x)                        ##Closing the file(in case it is opened) whose value is variable x value. To avoid too many opened files error.
  x="F"++i                        ##Creating variable x whose value is character F with increasing value of variable F each time with 1.
  next                            ##next will skip all further statements.
}
{                                 ##Following statements will be executed when above condition is NOT TRUE.
  print count++ "," $0 > x        ##Printing variable count value with comma and current line value into file named x here.
}
' words.txt                       ##Mentioning Input_file name here.

Answer 2

这里

awk '/^Header$/{close(f); f="File"++n".txt"; l=0; next}{print ++l". "$0 > f}' words.txt

结果

$ cat File1.txt 
1. LLLL
2. AAAA
$ 
$ cat File2.txt 
1. SSSS
2. DDDD

如何根据条件匹配将一个文件从UNIX中所有分割文件的开头开始按行号分割为多个文件？

2 个答案: