简单的正则表达式匹配

Question

我想创建一个正则表达式来生成一个匹配并剥离$和最后两个字符，如果最后一个字符以大写字母加数字结尾。

I'll strip off the $ and then an ending capital letter + number: 

$mytestA1 --> expected output: mytest 
$againD4 --> expected output: again 
$something --> expected output: something 
$name3 --> expected output: name3 // because there was no capital letter before the number digit 
$name2P4 --> expected output: name2 

我会在我的代码中检查'if'检查是否存在$，我甚至会打扰运行正则表达式。

谢谢。

Answer 1

在Java中只使用字符串＃的replaceAll：

String replaced = str.replaceAll("^\\$|[A-Z]\\d$", ""); 

Answer 2

这可能不是最有效的，但它会工作...

\$([^\s]*)(?:[A-Z]\d)|\$([^\s]*) 

它的工作原理，因为第一组发现所有那些国会随后数... ...，第二个发现所有没有后缀的那些...

如果您从捕获组获得您想要的匹配项。

我觉得像这样的工作...

import java.io.Console; 
import java.util.regex.Pattern; 
import java.util.regex.Matcher; 

public class HereYouGo { 
    public static void main (String args[]) { 

     String input = "$mytestA1 --> expected output: mytest\r\n$againD4 --> expected output: again\r\n$something --> expected output: something\r\n$name3 --> expected output: name3 // because there was no capital letter before the number digit\r\n$name2P4 --> expected output: name2\r\n";  

     Pattern myPattern = Pattern.compile("\\$([^ ]*)(?:[A-Z]\\d)|\\$([^ ]*)", Pattern.DOTALL | Pattern.MULTILINE); 

     Matcher myMatcher = myPattern.matcher(input); 

     while(myMatcher.find()) 
     { 
      String group1 = myMatcher.group(1); 
      String group2 = myMatcher.group(2); 

      //note: this should probably be changed in the case neither match is found 
      System.out.println(group1!=null? group1 : group2); 
     } 
    } 
} 

这将输出以下

mytest 
again 
something 
name3 
name2