Question

我正在使用Ruby中的HEX中的动态大小的列读取数据库文件。我可以使用以下脚本成功地将文件拆分为记录：

open_file = IO.binread(path + file_name)
record_delimeters = ['FAFA', 'FEFE', 'FDFD']

# regex with bytes is kinda finicky.. So I went this route to avoid the pitfalls of escape characters and gsub... If anyone knows a better way to do this part, I am up for suggestions as well..
final_reg = '['
record_delimeters.each_with_index do |delim, index|
  standard_string = '\xFA-\xFA'
  standard_string[2,2] = delim[0,2]
  standard_string[7,2] = delim[2,2]
  unless index == 0
    final_reg += '|'
  end
  final_reg += standard_string
end

final_reg += ']+'

reg = Regexp.new final_reg.encode('UTF-8'), Regexp::IGNORECASE | Regexp::MULTILINE, 'n'

records = open_file.split(reg);nil

但是，我想保留我的分隔符作为参考，因为分隔符表示所有＆＃39;类型＆＃39;记录的内容。即：＆＃39; uint，int，word等...＆＃39;。

最终我希望记录看起来像这样：

 ["\xFE\xFE\x00\xF4\x35...", "\xFA\xFA\x03\x4F\x7A...", ...]

或者这个：

 ["\xFE\xFE", "\x00\xF4\x35...", "\xFA\xFA", "\x03\x4F\x7A...", ...]

但绝对不是这个（这就是我所拥有的）：

 ["\x00\xF4\x35...", "\x03\x4F\x7A...", ...]

如何在ruby中拆分字节序列并保留分隔符？

0 个答案: