基于 Java 的正则表达式允许字母数字字符和 ', 和
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Java-based regular expression to allow alphanumeric chars and ', and
提问by Mr Albany Caxton
I'm new to regular expressions in Java and I need to validate if a string has alphanumeric chars, commas, apostrophes and full stops (periods) only. Anything else should equate to false.
我是 Java 正则表达式的新手，我需要验证字符串是否仅包含字母数字字符、逗号、撇号和句号（句点）。其他任何东西都应该等同于错误。
Can anyone give any pointers?
任何人都可以提供任何指示吗？
I have this at the moment which I believe does alphanumerics for each char in the string:
我现在有这个，我相信对字符串中的每个字符都有字母数字：
Pattern p = Pattern.compile("^[a-zA-Z0-9_\s]{1," + s.length() + "}");
Thanks
谢谢
Mr Albany Caxton
奥尔巴尼·卡克斯顿先生
回答by aioobe
I'm new to regular expressions in Java and I need to validate if a string has alphanumeric chars, commas, apostrophesand full stops(periods) only.
我是新来的Java正则表达式，我需要验证一个字符串有字母数字字符，逗号，撇号和句号只（期）。
I suggest you use the \p{Alnum}class to match alpha-numeric characters:
我建议您使用\p{Alnum}该类来匹配字母数字字符：
Pattern p = Pattern.compile("[\p{Alnum},.']*");
(I noticed that you included \sin your current pattern. If you want to allow white-space too, just add \sin the character class.)
（我注意到你包含\s在你当前的模式中。如果你也想允许空格，只需添加\s字符类。）
From documentation of Pattern:
[...]
\p{Alnum}An alphanumeric character:[\p{Alpha}\p{Digit}][...]
[...]
\p{Alnum}一个字母数字字符：[\p{Alpha}\p{Digit}][...]
You don't need to include ^and {1, ...}. Just use methods like Matcher.matchesor String.matchesto match the full pattern.
您不需要包含^和{1, ...}。只需使用像Matcher.matches或String.matches这样的方法来匹配完整的模式。
Also, note that you don't need to escape .within a character class ([...]).
另请注意，您不需要.在字符类 ( [... ]) 中转义。
回答by Herring
Pattern p = Pattern.compile("^[a-zA-Z0-9_\s\.,]{1," + s.length() + "}$");
回答by Bohemian
Keep it simple:
把事情简单化：
String x = "some string";
boolean matches = x.matches("^[\w.,']*$");
