Regex查找一个数字是否在一个范围内，例1,3,10-15,17[重复]

Question

任何解决这个问题的帮助都是感激的。

谢谢

====根据收到的答案修改。====

create or replace and compile java source named "RegExp" as
import java.util.HashSet;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegExp
{
  private static  Pattern P = Pattern.compile("(\\d+)-(\\d+)");

  public static int inRange (int p1, java.lang.String p2)
  {
    int iReturn = -1;
         int input;

         String[] sa;
         sa = p2.split(",");
         Set<Integer> is = new HashSet<Integer>();

         input = p1;

        for ( String s : sa)
        {
             Matcher m = P.matcher(s);
            if (m.find())
            {
                 int start = Integer.parseInt(m.group(1));
                 int end = Integer.parseInt(m.group(2));
                for (int i = start; i <= end; i++)
                {
                    is.add(i);
                }
            }
            else
            {
                is.add(Integer.parseInt(s));
            }
        }

        if (is.contains(input))
        {
           // System.out.format("Found %d in the set %s", input, is);
            iReturn = 1;
        }
        else
        {
           // System.out.format("Didn't find %d in set %s", input, is);
            iReturn = 0;
        }
        return iReturn;

  }

};

CREATE OR REPLACE FUNCTION InRange (p1  IN  number, p2  IN  VARCHAR2) RETURN NUMBER
AS LANGUAGE JAVA 
NAME 'RegExp.inRange (int, java.lang.String) return int';

SELECT InRange(13,'1,3,10-15,17') FROM   DUAL;

SELECT InRange(6,'1,3,10-15,17') FROM   DUAL;

public static int inRangeString (java.lang.String p1, java.lang.String p2)
  {
      int iReturn = -1;


      String[] sa;
      sa = p2.split(",");

      Set<String> is = new HashSet<String>();


      for ( String s : sa)
        {
                 is.add(s);
        }

       if (is.contains(p1)) {
          iReturn = 1;  
              // value is valid  
          } else {  
                    iReturn = 0;
              // value is invalid  
          }  

       return iReturn;
  }

Answer 1

但他们绝不是唯一的部分。正则表达式用于模式匹配，而不是用于执行您需要的任意逻辑。

import javax.annotation.Nonnull;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

import com.google.common.collect.Range;
import com.google.common.collect.RangeSet;
import com.google.common.collect.TreeRangeSet;

public class Q25004732
{
    private static final Pattern RANGE = Pattern.compile("(?i)^(?=[a-z]+-[a-z]+$|\\d+-\\d+$)([a-z\\d]+)-([a-z\\d]+)$");

    public static void main(final String[] args)
    {
        final String[] sa = "1,3,10-15,17,A,b,XX-ZZ,z".split(",");
        final RangeSet<String> rs = TreeRangeSet.create();
        for (final String s : sa)
        {
            final Matcher m = RANGE.matcher(s);
            if (m.find())
            {
                rs.add(Range.closed(m.group(1), m.group(2)));
            }
            else
            {
                rs.add(Range.closed(s, s));
            }
        }
        report("13", rs);
        report("A", rs);
        report("XY", rs);
        report("c", rs);
        report("42", rs);
    }

    private static void report(@Nonnull final String input, @Nonnull final RangeSet<String> rs)
    {
        if (rs.contains(input))
        {
            System.out.format("Found %s in the set %s", input, rs);
        }
        else
        {
            System.out.format("Didn't find %s in set %s", input, rs);
        }
        System.out.println();
    }
}

Found 13 in the set [[1‥1], [10‥15], [17‥17], [3‥3], [A‥A], [XX‥ZZ], [b‥b], [z‥z]]
Found A in the set [[1‥1], [10‥15], [17‥17], [3‥3], [A‥A], [XX‥ZZ], [b‥b], [z‥z]]
Found XY in the set [[1‥1], [10‥15], [17‥17], [3‥3], [A‥A], [XX‥ZZ], [b‥b], [z‥z]]
Didn't find c in set [[1‥1], [10‥15], [17‥17], [3‥3], [A‥A], [XX‥ZZ], [b‥b], [z‥z]]
Didn't find 42 in set [[1‥1], [10‥15], [17‥17], [3‥3], [A‥A], [XX‥ZZ], [b‥b], [z‥z]]

谷歌番石榴

<dependency>
    <groupId>com.google.guava</groupId>
    <artifactId>guava</artifactId>
    <version>[17.0,)</version>
</dependency>

芬德布格斯

<dependency>
    <groupId>com.google.code.findbugs</groupId>
    <artifactId>findbugs</artifactId>
    <version>[3.0.0,)</version>
</dependency>