java - 如何子类化 Java 构建器类？

Question

我有两个构建器 - PayloadA 和 PayloadB。为了使示例更简单，我删除了许多其他字段。

• PayloadA.Builder 构造函数将 processName、genericRecord 作为输入参数，然后从 genericRecord 中提取一些内容。我正在对此进行验证。
• PayloadB.Builder 构造函数还采用 processName、genericRecord 作为输入参数，然后从 genericRecord< 中提取一些不同的内容 与上面相比。我正在这些不同的领域进行验证。

如您所见，这两个 Payload?.Builder 之间的共同点是 processName、genericRecord、提取 oldTimestamp code> 值，然后是 isValid 方法。

下面是我的 PayloadA 类:

public final class PayloadA {
  private final String clientId;
  private final String deviceId;
  private final String processName;
  private final GenericRecord genericRecord;
  private final Long oldTimestamp;

  private PayloadA(Builder builder) {
    this.clientId = builder.clientId;
    this.deviceId = builder.deviceId;
    this.processName = builder.processName;
    this.genericRecord = builder.genericRecord;
    this.oldTimestamp = builder.oldTimestamp;
  }

  public static class Builder {
    private final String processName;
    private final GenericRecord genericRecord;
    private final String clientId;
    private final String deviceId;
    private final Long oldTimestamp;

    public Builder(PayloadA payload) {
      this.processName = payload.processName;
      this.genericRecord = payload.genericRecord;
      this.clientId = payload.clientId;
      this.deviceId = payload.deviceId;
      this.oldTimestamp = payload.oldTimestamp;
    }

    public Builder(String processName, GenericRecord genericRecord) {
      this.processName = processName;
      this.genericRecord = genericRecord;
      this.clientId = (String) DataUtils.parse(genericRecord, "clientId");
      this.deviceId = (String) DataUtils.parse(genericRecord, "deviceId");
      this.oldTimestamp = (Long) DataUtils.parse(genericRecord, "oldTimestamp");
    }

    // calling this method to validate
    public boolean isValid() {
      return isValidClientIdDeviceId();
    }

    private boolean isValidClientIdDeviceId() {
        // validate here
    }

    public PayloadA build() {
      return new PayloadA(this);
    }
  }

  // getter here
}

下面是我的 PayloadB 类:

public final class PayloadB {
  private final GenericRecord genericRecord;
  private final String processName;
  private final String type;
  private final String datumId;
  private final Long oldTimestamp;

  private PayloadB(Builder builder) {
    this.processName = builder.processName;
    this.genericRecord = builder.genericRecord;
    this.type = builder.type;
    this.datumId = builder.datumId;
    this.oldTimestamp = builder.oldTimestamp;
  }

  public static class Builder {
    private final GenericRecord genericRecord;
    private final String processName;
    private final String type;
    private final String datumId;
    private final Long oldTimestamp;

    public Builder(PayloadB payload) {
      this.processName = payload.processName;
      this.genericRecord = payload.genericRecord;
      this.type = payload.type;
      this.datumId = payload.datumId;
      this.oldTimestamp = payload.oldTimestamp;
    }

    public Builder(String processName, GenericRecord genericRecord) {
      this.processName = processName;
      this.genericRecord = genericRecord;
      this.type = (String) DataUtils.parse(genericRecord, "type");
      this.datumId = (String) DataUtils.parse(genericRecord, "datumId");
      this.oldTimestamp = (Long) DataUtils.parse(genericRecord, "oldTimestamp");
    }

    // calling this method to validate
    public boolean isValid() {
      return isValidType() && isValidDatumId();
    }

    private boolean isValidType() {
        // validate here
    }

    private boolean isValidDatumId() {
        // validate here
    }

    public PayloadB build() {
      return new PayloadB(this);
    }
  }

    // getter here

}

现在有什么方法可以在这里使用抽象类的概念吗？我可以创建一个抽象类 Payload 但我的抽象类中应该包含什么内容:

public final class PayloadA extends Payload { ... }
public final class PayloadB extends Payload { ... }

然后，一旦我构建了两个构建器，我会将其传递给其他一些方法，并且我想使用 getter 访问所有字段。假设我已经构建了 PayloadA，因此我将发送到执行方法，如下所示，然后在该方法中，我想提取 PayloadA 的所有字段。同样，如果我发送 PayloadB 到执行方法，那么我想使用 getters 提取 PayloadB 类的所有字段。我怎样才能做到这一点？

private void execute(Payload payload) {

    // How can I access fields of PayloadA or PayloadB 
    // depending on what was passe
}

Answer 1

仅当提到的字段不是巧合时才为有效负载创建一个父类(super class)。您可以在其中移动公共(public)字段和方法(但不能移动构建器)。您甚至可以为构建器创建一个父类(super class)，但它可能会使代码变得过于困惑。

如果您确实需要使用有效负载父类(super class)，那么您可以使用 Visitor Pattern 实现您的 execute 方法。 :

首先，您必须创建一个访问者，您可以在其中访问您的具体类:

public class PayloadVisitor {

    public void visit(PayloadA payloadA) {
        // use payload A here
    }

    public void visit(PayloadB payloadB) {
        // use payload B here
    }
}

然后你必须向你的父类(super class)添加一个方法来接受访问者:

public abstract class Payload {

    // common fields and methods

    public abstract void accept(PayloadVisitor visitor);
}

重写子类中的方法accept:

public final class PayloadA extends Payload {

    // ...

    @Override
    public void accept(PayloadVisitor visitor) {
        visitor.visit(this);
    }
}

public final class PayloadB extends Payload {

    // ...

    @Override
    public void accept(PayloadVisitor visitor) {
        visitor.visit(this);
    }
}

您的方法 execute 只是将调用重定向到相应的 visit 方法:

private void execute(Payload payload) {
    payload.accept(new PayloadVisitor());
}

访客模式可能会令人难以承受。您还可以保持简单并使用 instanceof 来确定具体类。