使用正则表达式匹配非单词字符但不要笑脸 [英] Using regex to match non-word characters BUT NOT smiley faces

问题描述

我有一个Java程序，它应该从字符串中删除所有非字母字符，除非它们是笑脸，例如=）或=]或：P

I have a Java program which is supposed to remove all non-letter characters from a string, except when they are a smiley face such as =) or =] or :P

很容易匹配 [a-zA-Z] | = \）| = \] |：P ，但我无法弄清楚如何否定这种表达方式。由于我使用的是String.replaceAll（）函数，因此它必须是否定形式。

It's very easy to match the opposite with [a-zA-Z ]|=\)|=\]|:P but I cannot figure out how to negate this expression. Since I am using the String.replaceAll() function it must be in the negated form.

我认为部分问题可能来自微笑一般为2的事实字符长，我一次只匹配1个字符？

I believe part of the issue may come from the fact that smiles are generally 2 characters long, and I am only matching 1 character at a time?

有趣的是， replaceAll（（？！[Tt]）[Oo] ，）删除字母O的每次出现，即使在单词to中也是如此。这是否意味着我的replaceAll函数不能理解正则表达式的前瞻性？它没有抛出任何错误...

Interestingly, replaceAll("(?![Tt])[Oo]","") removes every occurrence of the letter O, even in the word "to." Does this mean my replaceAll function does not understand regex lookahead? It doesn't throw any errors...

我最终使用

replaceAll("(?<![=:;])[\\]\\[\\(\\)\\/]","")
.replaceAll("[=:;](?![\\]\\[\\(\\)o0OpPxX\\/])","")
.replaceAll("[^a-zA-Z=:;\\(\\)\\[\\]\\/ ]","")

这非常凌乱，但效果很好。 ......快！ （棕色）狐狸跳过[]懒狗。 ：] = O; X 变成在懒惰的狗身上快速布朗的狐狸笑声：] = O; X

which is extremely messy but works perfectly. The... quick! (brown) fox jump's over the[] lazy dog. :] =O ;X becomes THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG :] =O ;X

编辑：忽略该修复，请参阅下面接受的答案。

Ignore that fix, see the accepted answer below.

推荐答案

使用负向前瞻应该很容易。基本上，匹配将在（？！...）组内的正则表达式匹配的任何位置失败。您应该使用单个通配符（。）跟随负向前瞻，以便在前瞻不匹配时使用字符（意味着下一个字符是非字母字符，即不是笑脸的一部分。）

It should be pretty easy to due this using a negative lookahead. Basically the match will fail at any position where the regex inside of the (?!...) group matches. You should follow the negative lookahead with a single wildcard (.) to consume a character if the lookahead did not match (meaning that the next character is a non-letter character that is not part of a smiley face).

编辑显然我没有彻底测试我的原始正则表达式，你还需要一个负面的背后在。之后，确保你消费的角色不是笑脸中的第二个角色：

edit: Clearly I hadn't tested my original regex very thoroughly, you also need a negative lookbehind following the . to make sure that the character you consumed was not the second character in a smiley:

(?![a-zA-Z ]|=\)|=\]|:P).(?<!=\)|=\]|:P)

请注意，您可以通过使用眼睛和嘴巴的字符类来缩短正则表达式，例如：

Note that you might be able to shorten the regex by using character classes for the eyes and the mouth, for example:

[:=][\(\)\[\]]
  ^    ^-----mouth
  |--eyes

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