java 没有任何空格的字母数字、连字符和下划线的正则表达式
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Regular Expression for AlphaNumeric, Hyphen and Underscore without any space
提问by Gautam R
I would like to have a regular expression that checks if the string contains Alphanumerics, Hyphen and Underscore. There should not be any spaces or other special characters other these three. My string will be under either of these 2 patterns.
我想要一个正则表达式来检查字符串是否包含字母数字、连字符和下划线。除了这三个之外，不应有任何空格或其他特殊字符。我的字符串将在这两种模式中的任何一种之下。
XYZ0123_123456ABCdefGHI-727
XYZ0123_123456ABCdefGHI-727
I have already tried this expression. But it didnt workout. [[a-zA-Z0-9_-]*]
我已经尝试过这种表达方式。但它没有锻炼。[[a-zA-Z0-9_-]*]
回答by abubaker
Working
Here is the quick solution.
在这里工作是快速解决方案。
String regex = "^[a-zA-Z0-9_-]*$";
sampleString.matches(regex);
(^) anchor means start of the string
($) anchor means end of the string
A-Z , a-z represent range of characters
0-9 represent range of digit
( _ ) represent underscore
( - ) represent hyphen
(^) 锚表示字符串的开始
($) 锚表示字符串的结尾
AZ , az 代表字符范围
0-9 代表数字范围
( _ ) 代表下划线
( - ) 代表连字符
回答by Thomas Ayoub
You can use ^[\w-]*$where \wmeans:
您可以使用^[\w-]*$where\w表示：
- any letter in the range
a-z,A-Z, - any digit from
0 to 9, underscore
- 范围内的任何字母
a-z,A-Z, - 来自 的任何数字
0 to 9， underscore
The error you made was to encapsulate you correct regex ([a-zA-Z0-9_-]*== [\w-]*) within a character class, loosing the quantifier (*) meaning
您犯的错误是将正确的正则表达式 ( [a-zA-Z0-9_-]*== [\w-]*)封装在字符类中，从而丢失了量词 ( *) 的含义
