问题描述
我有一个正规的前pression
I have a regular expression
^[a-zA-Z+#-.0-9]{1,5}$
这将验证字包含字母数字字符和一些特殊字符，长度
不应该超过 5 字符。
which validates that the word contains alpha-numeric characters and few special characters and length should not be more than 5 characters.
我如何让这个普通的前pression接受匹配上述正前pression最多五个字。
How do I make this regular expression to accept a maximum of five words matching the above regular expression.
推荐答案
^ [A-ZA-Z +＃\\ - 0-9] {1,5}（\\ S [A-ZA -Z +＃\\ - 0-9] {1,5}）{0,4} $
此外，您还可以使用例如 [] 而不是 \\ S 如果你只是想接受的空间，没有标签和换行符。你可以写 [] + （或 \\ S + ）的任何数量的空格（或空格），而不仅仅是之一。
Also, you could use for example [ ] instead of \s if you just want to accept space, not tab and newline. And you could write [ ]+ (or \s+) for any number of spaces (or whitespaces), not just one.
编辑：删除无效的解决方案，并通过固定提到unicornaddict的bug
Removed the invalid solution and fixed the bug mentioned by unicornaddict.
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