正則表達式：匹配不重複使用以前匹配的單詞的列表

Question

我想寫一個正則表達式模式，它將匹配任何包含單詞'28'，'獎金'和'日'的字符串。

目前，我想出了這一點：

(bonus|(days|day)|(28th|28)|twenty[ \-\t]*(eighth|eight))[ \ta-z]*(bonus|days|day|(28th|28)|twenty[ \-\t]*(eighth|eight))[ \ta-z]*(bonus|days|day|(28th|28)|twenty[ \-\t]*(eighth|eight)) 

您可以查看結果這裏：https://regex101.com/r/oOcGqk/8

我遇到的問題是，任何詞可以多次使用，並且仍然匹配。例如：'日日紅利'，'紅利紅利獎金'。我怎樣才能排除不止一次使用這些詞（'28'，'獎勵'，'日'）的字符串？

Answer 1

我覺得this正則表達式expresion是解決方案：

(?=.*bonus)(?=.*day)(?=.*28|twenty\s*-?\s*eight).* 

Answer 2

一個像樣的正則表達式引擎，你可以利用一個漂亮的絕招：

^  # Start of string 
(?=(?:(?!bonus).)*bonus()(?:(?!bonus).)*$) 
# Explanation: This lookahead assertion makes sure that "bonus" occurs exactly once 
# in the string. It doesn't actually match any text, it just "looks ahead" to see if 
# that condition is met. However, it contains an empty capturing group "()" that only 
# participates in the match if the lookahead assertion succeeds. We can check this later. 
(?=(?:(?!days?).)*days?()(?:(?!days?).)*$) 
(?=(?:(?!28(?:th)?|twenty-eighth?).)*(?:28(?:th)?|twenty-eighth?)()(?:(?!28(?:th)?|twenty-eighth?).)*$) 
[\w\s]* # Match a string that only contains alnum character or whitespace 
\1\2\3 # Assert that all three words participated in the match 
$  # End of string. 

您可以測試這個here

在JavaScript中，你必須闡明所有可能的排列。不幸的是，JS甚至不允許冗長的正則表達式，所以它會變得怪異。

只是作爲一個起點：下面的正則表達式將匹配包含bonus，days和28正好一次字符串，但只允許他們在秩序「bonus，days和28」或「days，bonus和28」。你需要添加其他四個排列組合，以獲得完整的正則表達式（和一個完整的混亂）。以編程方式執行此操作，而不是使用正則表達式。

^(?:(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*bonus(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*days?(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*(?:28(?:th)?|twenty-eighth?)(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*|(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*days?(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*bonus(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*(?:28(?:th)?|twenty-eighth?)(?:(?!bonus|days?|28(?:th)?|twenty-eighth?).)*)$ 

測試它here。你被警告了。