正则表达式:通过排除匹配，没有前瞻 - 有可能吗? [英] Regex: Matching by exclusion, without look-ahead - is it possible?

问题描述

在某些正则表达式风格中，不支持 [否定] 零宽度断言(前瞻/后视).

In some regex flavors, [negative] zero-width assertions (look-ahead/look-behind) are not supported.

这使得声明排除非常困难(不可能?).例如没有的每一行都有foo"，就像这样:

This makes it extremely difficult (impossible?) to state an exclusion. For example "every line that does not have "foo" on it", like this:

^((?!foo).)*$

完全不使用环视是否可以实现同样的事情(暂时搁置复杂性和性能问题)?

Can the same thing be achieved without using look-around at all (complexity and performance concerns set aside for the moment)?

推荐答案

UPDATE: 它失败了在 oo 之前有两个 ff"作为 @Ciantic 在评论中指出.

UPDATE: It fails "with two ff before oo" as @Ciantic pointed out in the comments.

^(f(o[^o]|[^o])|[^f])*$

注意:在客户端否定匹配比使用上面的正则表达式要容易得多.

NOTE: It is much much easier just to negate a match on the client side instead of using the above regex.

正则表达式假设每一行都以换行符结尾，如果不是，请查看 C++ 和 grep 的正则表达式.

The regex assumes that each line ends with a newline char if it is not then see C++'s and grep's regexs.

Perl、Python、C++ 和 grep 中的示例程序都给出相同的输出.

Sample programs in Perl, Python, C++, and grep all give the same output.

• perl

#!/usr/bin/perl -wn
print if /^(f(o[^o]|[^o])|[^f])*$/;

蟒蛇

#!/usr/bin/env python
import fileinput, re, sys
from itertools import ifilter

re_not_foo = re.compile(r"^(f(o[^o]|[^o])|[^f])*$")
for line in ifilter(re_not_foo.match, fileinput.input()):
    sys.stdout.write(line)

c++

c++

#include <iostream>
#include <string>
#include <boost/regex.hpp>

int main()
{
  boost::regex re("^(f(o([^o]|$)|([^o]|$))|[^f])*$");
  //NOTE: "|$"s are there due to `getline()` strips newline char

  std::string line;
  while (std::getline(std::cin, line)) 
    if (boost::regex_match(line, re))
      std::cout << line << std::endl;
}

grep

$ grep "^\(f\(o\([^o]\|$\)\|\([^o]\|$\)\)\|[^f]\)*$" in.txt

示例文件:

foo
'foo'
abdfoode
abdfode
abdfde
abcde
f

fo
foo
fooo
ofooa
ofo
ofoo

输出:

abdfode
abdfde
abcde
f

fo
ofo

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