正则表达式匹配具有唯一(非重复)字符的单词

Question

我正在寻找一个正则表达式，它只在所有字符都是唯一的情况下才匹配一个单词，意思是，单词中的每个字符只出现一次。

示例 :abcdefg -> 将返回 匹配 abcdefgbh -> 将返回 没有匹配项 (因为字母 b 重复不止一次)

Answer 1

试试这个，可能有用

^(?:([A-Za-z])(?!.*\1))*$

解释

Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match the regular expression below «(?:([A-Z])(?!.*\1))*»
   Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Match the regular expression below and capture its match into backreference number 1 «([A-Z])»
      Match a single character in the range between “A” and “Z” «[A-Z]»
   Assert that it is impossible to match the regex below starting at this position (negative lookahead) «(?!.*\1)»
      Match any single character that is not a line break character «.*»
         Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
      Match the same text as most recently matched by capturing group number 1 «\1»
Assert position at the end of a line (at the end of the string or before a line break character) «$»