正则表达式包括括号和排除空格

Question

正则表达式，方法名称只包含一个开头和一个右括号，不接受任何空格。我试过了 -

（（（。）））+

但它也接受了空格

Answer 1

请举几个示例，但您的主要问题似乎是要排除\ s和括号。这可能意味着你需要

\([^\s\(\)]*\)

根据regex101.com（https://regex101.com/r/ZtMNbO/1）说明：

\([^\s\(\)]*\) \( matches the character ( literally (case sensitive) Match a single character not present in the list below [^\s\(\)]* * Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy) \s matches any whitespace character (equal to [\r\n\t\f\v ]) \( matches the character ( literally (case sensitive) \) matches the character ) literally (case sensitive) \) matches the character ) literally (case sensitive)

或非常长的字符排除数组

我的建议已经给出了：

string * match

example() * ()
example(a) * (a)
example(a,a) * (a,a)
example(a, a) * no match
example(a(a)) * (a)

Answer 2

^\s - 将省略空格 在＆＃39;（＆＃39;和＆＃39;）＆＃39;前面使用\包含，因为两者都保留，所以你需要＆＃39; \＆＃39;提到它的匹配角色。

\(^\s\)