6.4.7.2. Grouping API を使用する方法

Configuration c = new ConfigurationBuilder()
   .clustering().hash().groups().enabled()
   .build();

Configuration c = new ConfigurationBuilder()
   .clustering().hash().groups().enabled()
   .build();

<distributed-cache>
   <groups enabled="true"/>
</distributed-cache>

<distributed-cache>
   <groups enabled="true"/>
</distributed-cache>

class User {
   ...
   String office;
   ...

   public int hashCode() {
      // Defines the hash for the key, normally used to determine location
      ...
   }

   // Override the location by specifying a group
   // All keys in the same group end up with the same owners
   @Group
   public String getOffice() {
      return office;
   }
   }
}

class User {
   ...
   String office;
   ...

   public int hashCode() {
      // Defines the hash for the key, normally used to determine location
      ...
   }

   // Override the location by specifying a group
   // All keys in the same group end up with the same owners
   @Group
   public String getOffice() {
      return office;
   }
   }
}

public interface Grouper<T> {
    String computeGroup(T key, String group);

    Class<T> getKeyType();
}

public interface Grouper<T> {
    String computeGroup(T key, String group);

    Class<T> getKeyType();
}

public class KXGrouper implements Grouper<String> {

   // The pattern requires a String key, of length 2, where the first character is
   // "k" and the second character is a digit. We take that digit, and perform
   // modular arithmetic on it to assign it to group "0" or group "1".
   private static Pattern kPattern = Pattern.compile("(^k)(<a>\\d</a>)$");

   public String computeGroup(String key, String group) {
      Matcher matcher = kPattern.matcher(key);
      if (matcher.matches()) {
         String g = Integer.parseInt(matcher.group(2)) % 2 + "";
         return g;
      } else {
         return null;
      }
   }

   public Class<String> getKeyType() {
      return String.class;
   }
}

public class KXGrouper implements Grouper<String> {

   // The pattern requires a String key, of length 2, where the first character is
   // "k" and the second character is a digit. We take that digit, and perform
   // modular arithmetic on it to assign it to group "0" or group "1".
   private static Pattern kPattern = Pattern.compile("(^k)(<a>\\d</a>)$");

   public String computeGroup(String key, String group) {
      Matcher matcher = kPattern.matcher(key);
      if (matcher.matches()) {
         String g = Integer.parseInt(matcher.group(2)) % 2 + "";
         return g;
      } else {
         return null;
      }
   }

   public Class<String> getKeyType() {
      return String.class;
   }
}

Configuration c = new ConfigurationBuilder()
   .clustering().hash().groups().enabled().addGrouper(new KXGrouper())
   .build();

Configuration c = new ConfigurationBuilder()
   .clustering().hash().groups().enabled().addGrouper(new KXGrouper())
   .build();

<distributed-cache>
   <groups enabled="true">
      <grouper class="com.acme.KXGrouper" />
   </groups>
</distributed-cache>

<distributed-cache>
   <groups enabled="true">
      <grouper class="com.acme.KXGrouper" />
   </groups>
</distributed-cache>