根据模式将一个文件拆分为多个文件 [英] Split one file into multiple files based on pattern

问题描述

我有一个二进制文件，我使用 hexdump 和几个 awk 和 sed 命令将其转换为常规文件.输出文件看起来像这样 -

I have a binary file which I convert into a regular file using hexdump and few awk and sed commands. The output file looks something like this -

$cat temp
3d3d01f87347545002f1d5b2be4ee4d700010100018000cc57e5820000000000000000000
000000087d3f513000000000000000000000000000000000001001001010f000000000026 
58783100b354c52658783100b43d3d0000ad6413400103231665f301010b9130194899f2f
fffffffffff02007c00dc015800a040402802f1d5b2b8ca5674504f433031000000000004
6363070000000000000000000000000065450000b4fb6b4000393d3d1116cdcc57e58287d
3f55285a1084b

临时文件很少有吸引眼球的东西 (3d3d)，它们不会经常重复.它们有点表示新二进制记录的开始.我需要根据那些引人注目的东西拆分文件.

The temp file has few eye catchers (3d3d) which don't repeat that often. They kinda denote a start of new binary record. I need to split the file based on those eye catchers.

我想要的输出是有多个文件(基于我的临时文件中引人注目的数量).

My desired output is to have multiple files (based on the number of eyecatchers in my temp file).

所以我的输出看起来像这样 -

So my output would look something like this -

$cat temp1
3d3d01f87347545002f1d5b2be4ee4d700010100018000cc57e582000000000000000
0000000000087d3f513000000000000000000000000000000000001001001010f00000000
002658783100b354c52658783100b4

$cat temp2
3d3d0000ad6413400103231665f301010b9130194899f2ffffffffffff02007c00dc0
15800a040402802f1d5b2b8ca5674504f4330310000000000046363070000000000000000
000000000065450000b4fb6b400039

$cat temp3
3d3d1116cdcc57e58287d3f55285a1084b

推荐答案

#!/usr/bin/perl

undef $/;
$_ = <>;
$n = 0;

for $match (split(/(?=3d3d)/)) {
      open(O, '>temp' . ++$n);
      print O $match;
      close(O);
}

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