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) «$»

您可以检查字符串中是否有2个字符示例：

^.*(.).*\1.*$

（我只是简单地抓取其中一个字符，然后检查它是否在其他地方有一个带有反向引用的副本。其余的 .* 是不关心）。
如果上面的正则表达式匹配，则字符串具有重复字符。如果上面的正则表达式不匹配，则所有字符都是唯一的。
上面正则表达式的优点是正则表达式引擎不支持环顾四周。
显然，john woo的解决方案是一种直接检查唯一性的漂亮方法。它在每个字符处Assert前面的字符串不包含当前字符。

这一个也将提供与任何长度的单词完全匹配的非重复字母：

^(?!.*(.).*\1)[a-z]+$

不久前，我对@bohemian提供的另一个问题的答案进行了轻微修改，以得到这个答案。
上面的问题也提出了一段时间了，但我认为在这里也有这个正则表达式模式会很好。