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问题

您如何正确组合 ReactFX 中的多个属性更改流以用于 UndoFX(或任何用例)?

细节

这是我要完成的工作的简短说明(完整的示例代码发布在 GitHub 上):

有一个示例模型具有两个属性。为简单起见,它们是双重属性

public class DataModel {
    private DoubleProperty a, b;
    //...
    //with appropriate getters, setters, equals, hashcode
    //...
}

在示例代码中,有用于更改其中一个或两个属性的按钮。如果这就是更改,我想撤消对两者的更改。

根据 UndoFX 示例,每个继承自基类的更改类(此处也缩写为):

public abstract class ChangeBase<T> implements UndoChange {
   protected final T oldValue, newValue;
   protected final DataModel model;

   protected ChangeBase(DataModel model, T oldValue, T newValue) {
      this.model = model;
      this.oldValue = oldValue;
      this.newValue = newValue;
   }

   public abstract ChangeBase<T> invert();
   public abstract void redo();

   public Optional<ChangeBase<?>> mergeWith(ChangeBase<?> other) {
      return Optional.empty();
   }

   @Override
   public int hashCode() {
      return Objects.hash(this.oldValue, this.newValue);
   }      

   @Override
   public boolean equals(Object obj) {
      if (this == obj) {
         return true;
      }
      if (obj == null) {
         return false;
      }
      if (getClass() != obj.getClass()) {
         return false;
      }
      final ChangeBase<?> other = (ChangeBase<?>) obj;
      if (!Objects.equals(this.oldValue, other.oldValue)) {
         return false;
      }
      if (!Objects.equals(this.newValue, other.newValue)) {
         return false;
      }
      if (!Objects.equals(this.model, other.model)) {
         return false;
      }
      return true;
   }
}

public class ChangeA extends ChangeBase<Double> {
   //...
   //constructors and other method implementations
   //..
   @Override
   public void redo() {
      System.out.println("ChangeA redo "+this);
      this.model.setA(this.newValue);
   }   
}

public class ChangeB extends ChangeBase<Double> {
   //...
   //constructors and other method implementations
   //...
   @Override
   public void redo() {
      System.out.println("ChangeA redo "+this);
      this.model.setB(this.newValue);
   }   
}

所有更改都实现一个接口

public interface UndoChange {
   public void redo();
   public UndoChange invert();
   public Optional<UndoChange> mergeWith(UndoChange other);

}

在阅读了文档之后,我首先创建了一个事件流来捕获对每个属性的更改:

EventStream<UndoChange> changeAStream = 
    EventStreams.changesOf(model.aProperty())
              .map(c -> new ChangeA(model, (Change<Number>)c));
EventStream<UndoChange> changeBStream = 
    EventStreams.changesOf(model.bProperty())
              .map(c -> new ChangeB(model, (Change<Number>)c));

我的第一次尝试是像这样合并流

EventStream<UndoChange> bothStream = EventStreams.merge(changeAStream, changeBStream); 

在这种情况下发生的情况是,如果同时更改 A 和 B 属性,则流中将有两个更改,并且每个更改将单独撤消而不是一起撤消。对 setter 的每次调用都会在适当的流中进行更改,然后将其发送到bothStream,然后包含两个单独的事件而不是一个。

经过更多阅读后,我尝试将流和映射组合成一个单独的更改对象:

 EventStream<UndoChange> bothStream = EventStreams.combine(changeAStream, changeBStream).map(ChangeBoth::new); 

其中ChangeBoth定义为:

public class ChangeBoth implements UndoChange {

   private final ChangeA aChange;
   private final ChangeB bChange;

   public ChangeBoth(ChangeA ac, ChangeB bc) {
      this.aChange = ac;
      this.bChange = bc;
   }

   public ChangeBoth(Tuple2<UndoChange, UndoChange> tuple) {
      this.aChange = ((ChangeBoth)tuple.get1()).aChange;
      this.bChange = ((ChangeBoth)tuple.get2()).bChange;
   }

   @Override
   public UndoChange invert() {
      System.out.println("ChangeBoth invert "+this);
      return new ChangeBoth(new ChangeA(this.aChange.model, this.aChange.newValue, this.aChange.oldValue), 
                            new ChangeB(this.bChange.model, this.bChange.newValue, this.bChange.oldValue));
   }

   @Override
   public void redo() {
      System.out.println("ChangeBoth redo "+this);
      DataModel model = this.aChange.model;
      model.setA(this.aChange.newValue);
      model.setB(this.bChange.newValue);
   }

   //...
   // plus appropriate mergeWith, hashcode, equals
   //...
}

这导致IllegalStateException: Unexpected change received被抛出。经过一番挖掘,我确定了为什么会发生这种情况:当ChangeBoth正在撤消(通过invert()redo()调用)时,它将每个属性设置回旧值。但是,当它设置每个属性时,更改会通过流发回,从而在ChangeBoth将两个属性设置回旧值之间将新的内容放入流中。

概括

所以回到我的问题:这样做的正确方法是什么?有没有一种方法可以将两个属性的流更改为一个不会导致此问题的更改对象?

编辑 - 尝试 1

根据 Tomas 的回答,我添加/更改了以下代码(注意:repo 中的代码已更新):

changeAStreamchangeBstream保持不变。

我没有合并流,而是按照 Tomas 的建议进行操作,并创建了一个二元运算符以将两个更改减少为一个:

BinaryOperator<UndoChange> abOp = (c1, c2) -> {
   ChangeA ca = null;
   if(c1 instanceof ChangeA) {
      ca = (ChangeA)c1;
   }
   ChangeB cb = null;
   if(c2 instanceof ChangeB) {
      cb = (ChangeB)c2;
   }
   return new ChangeBoth(ca, cb);
};

并将事件流更改为

SuspendableEventStream<UndoChange> bothStream = EventStreams.merge(changeAStream, changeBStream).reducible(abOp);

现在按钮操作没有实现,setonAction而是使用事件流处理

  EventStreams.eventsOf(bothButton, ActionEvent.ACTION)
          .suspenderOf(bothStream)
          .subscribe((ActionEvent event) ->{
             System.out.println("stream action");
              model.setA(Math.random()*10.0);
              model.setB(Math.random()*10.0);                 
          });

这对于适当地组合事件非常有用,但是对于 A+B 更改,撤消仍然会中断。它适用于单独的 A 和 B 更改。这是两个 A+B 更改然后撤消的示例

A+B Button Action in event stream
Change in A stream
Change in B stream
A+B Button Action in event stream
Change in A stream
Change in B stream
ChangeBoth attempting merge with combinedeventstreamtest.ChangeBoth@775ec8e8... merged
undo 6.897901340713284  2.853416510829745
ChangeBoth invert combinedeventstreamtest.ChangeBoth@aae83334
ChangeA invert combinedeventstreamtest.ChangeA@32ee049a
ChangeB invert combinedeventstreamtest.ChangeB@4919dd13
ChangeBoth redo combinedeventstreamtest.ChangeBoth@b2155b1e
Change in A stream
Exception in thread "JavaFX Application Thread" java.lang.IllegalArgumentException: Unexpected change received.
Expected:
combinedeventstreamtest.ChangeBoth@b2155b1e
Received:
combinedeventstreamtest.ChangeA@2ad21e08
Change in B stream

编辑 - 尝试 2 - 成功!!

Tomas 很好地指出了解决方案(我应该意识到这一点)。只是在做的时候暂停redo()

  UndoManager<UndoChange> um = UndoManagerFactory.unlimitedHistoryUndoManager(
          bothStream, 
          c -> c.invert(), 
          c -> bothStream.suspendWhile(c::redo),
          (c1, c2) -> c1.mergeWith(c2)
  );
4

1 回答 1

0

bothStream因此,任务是将处理按钮单击期间发出的更改合并为一个。

您将需要一个将两个UndoChanges 减为一个的函数:

BinaryOperator<UndoChange> reduction = ???; // implement as appropriate

bothStream在“暂停”时将reduce 更改为一个:

SuspendableEventStream<UndoChange> bothStream =
        EventStreams.merge(changeAStream, changeBStream).reducible(reduction);

现在您只需要bothStream在处理按钮单击时暂停。这可以这样做:

EventStreams.eventsOf(bothButton, ActionEvent.ACTION) // Observe actions of bothButton ...
        .suspenderOf(bothStream)                      // but suspend bothStream ...
        .subscribe((ActionEvent event) -> {           // before handling the action.
            model.setA(Math.random()*10.0);
            model.setB(Math.random()*10.0);
        })

从撤消管理器中撤消/重做更改时也暂停bothStream,以便在撤消组合更改时发出(组合)更改的精确逆bothStream(这是使UndoManager快乐所必需的)。这可以通过将apply参数包装到UndoManager构造函数中来完成bothStream.suspendWhile(),例如:

UndoManagerFactory.unlimitedHistoryUndoManager(
        bothStream, 
        c -> c.invert(), 
        c -> bothStream.suspendWhile(c::redo)  // suspend while applying the change
)
于 2017-12-14T17:21:32.243 回答