Question

这里是反应式编程的新手。

我正在尝试使用ReactFX在JavaFX中实现一个“懒惰”的实时搜索文本区域。懒惰在这里我的意思是它一旦用户停止键入一秒钟就执行搜索。代码非常简单：

EventStream<Change<String>> textEvents = EventStreams.changesOf(textArea.textProperty())
.successionEnds(Duration.ofSeconds(1));

然后订阅该事件流并发表声明。

但是如果用户按Enter键，我也希望它立即执行搜索。我不确定如何以“反应”的方式做到这一点。简单地在Enter键事件上执行搜索会导致搜索触发两次（一次用于键事件，一次用于文本更改），因此这是我当前的解决方案：

BooleanProperty hasSearched = new SimpleBooleanProperty(false);
EventStream<KeyEvent> enterKeyPressedEvents = EventStreams.eventsOf(textArea, KeyEvent.KEY_PRESSED)
        .filter(k -> k.getCode() == KeyCode.ENTER);
AwaitingEventStream<Change<String>> textEvents = EventStreams.changesOf(textArea.textProperty())
        .successionEnds(Duration.ofSeconds(1));

subs = Subscription.multi(
        //Text changed
        textEvents.subscribe(e -> {
            if (hasSearched.get()) {
                hasSearched.set(false);
                System.out.println("ignored text event");
            } else {
                performSearch(textArea.getText());
            }
        }),

        //Enter key pressed
        enterKeyPressedEvents.subscribe(e -> {
            e.consume();
            if (e.isShiftDown()) {
                textArea.insertText(textArea.getCaretPosition(), "\n");
            } else {
                hasSearched.set(true);
                System.out.println("enter pressed");
                performSearch(textArea.getText());
                if (!textEvents.isPending()) {
                    hasSearched.set(false);
                }
            }
        })
);

我已经尝试使用SuspendableEventStream.suspend()认为它会“删除”所有待处理事件，但它没有按预期工作，仍会发出待处理事件：

EventStream<KeyEvent> enterKeyPressedEvents = EventStreams.eventsOf(textArea, KeyEvent.KEY_PRESSED)
        .filter(k -> k.getCode() == KeyCode.ENTER);
SuspendableEventStream<Change<String>> textEvents = EventStreams.changesOf(textArea.textProperty())
        .successionEnds(Duration.ofSeconds(1)).suppressible();

subs = Subscription.multi(
        //Text changed
        textEvents.subscribe(e -> {
                performSearch(textArea.getText());
        }),

        //Enter key pressed
        enterKeyPressedEvents.subscribe(e -> {
            e.consume();
            if (e.isShiftDown()) {
                textArea.insertText(textArea.getCaretPosition(), "\n");
            } else {
                Guard guard = textEvents.suspend();
                System.out.println("enter pressed");
                performSearch(textArea.getText());
                guard.close();
            }
        })
);

我怎样才能想到一个更好（更具反应性？）的解决方案？

Answer 1

这是 a 解决方案。此解决方案的关键部分是观察flatMap内的文本更改，其效果是“重置”文本更改流。

import java.time.Duration;
import java.util.function.Function;

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.control.TextArea;
import javafx.scene.input.KeyCode;
import javafx.scene.input.KeyEvent;
import javafx.stage.Stage;

import org.reactfx.EventStream;
import org.reactfx.EventStreams;

public class AutoSearch extends Application {

    public static void main(String[] args) {
        launch(args);
    }

    @Override
    public void start(Stage stage) throws Exception {
        TextArea area = new TextArea();

        EventStream<KeyEvent> enterPresses = EventStreams.eventsOf(area, KeyEvent.KEY_PRESSED)
                .filter(k -> k.getCode() == KeyCode.ENTER)
                .hook(KeyEvent::consume);

        EventStream<?> searchImpulse = enterPresses.withDefaultEvent(null) // emit an event even before Enter is pressed
                .flatMap(x -> {
                    EventStream<?> edits = EventStreams.changesOf(area.textProperty())
                                                       .successionEnds(Duration.ofSeconds(1));
                    return ((x == null) ? edits : edits.withDefaultEvent(null))
                            .map(Function.identity()); // just to get the proper type of the result
                });

        searchImpulse.subscribe(x -> System.out.println("Search now!"));

        stage.setScene(new Scene(area));
        stage.show();
    }

}

Answer 2

这是另一种解决方案。如果输入计数在此期间没有改变，则只计算Enter按钮，并且只允许编辑事件触发搜索。

import java.time.Duration;

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.control.TextArea;
import javafx.scene.input.KeyCode;
import javafx.scene.input.KeyEvent;
import javafx.stage.Stage;

import org.reactfx.EventStream;
import org.reactfx.EventStreams;

public class AutoSearch2 extends Application {

    public static void main(String[] args) {
        launch(args);
    }

    @Override
    public void start(Stage stage) throws Exception {
        TextArea area = new TextArea();

        EventStream<KeyEvent> enterPressed = EventStreams.eventsOf(area, KeyEvent.KEY_PRESSED)
                .filter(k -> k.getCode() == KeyCode.ENTER)
                .hook(KeyEvent::consume);

        EventStream<Long> enterCount = enterPressed.accumulate(0L, (n, k) -> n + 1)
                                                   .withDefaultEvent(0L);

        EventStream<Long> delayedEdits = enterCount.emitOnEach(EventStreams.changesOf(area.textProperty()))
                                                   .successionEnds(Duration.ofSeconds(1));

        // check that the delayed edit event still has the current value of the Enter counter
        EventStream<?> validEdits = enterCount.emitBothOnEach(delayedEdits)
                                              .filter(cd -> cd.test((current, delayed) -> delayed == current));

        EventStream<?> searchImpulse = EventStreams.merge(enterPressed, validEdits);

        searchImpulse.subscribe(x -> System.out.println("Search now!"));

        stage.setScene(new Scene(area));
        stage.show();
    }

}

Answer 3

未经测试，但如何：

EventStream<KeyEvent> enterKeyPressedEvents = EventStreams.eventsOf(textArea, KeyEvent.KEY_PRESSED)
        .filter(k -> k.getCode() == KeyCode.ENTER);
EventStream<Change<String>> textEvents = EventStreams.changesOf(textArea.textProperty())
        .successionEnds(Duration.ofSeconds(1))
        .filter(c -> ! isAdditionOfNewline(c, textArea.getCaratPosition()));

EventStreams.merge(enterKeyPressedEvents, textEvents)
        .subscribe(o -> performSearch(textArea.getText()));

private boolean isAdditionOfNewline(Change<String> change, int caratPos) {
    // TODO make sure this works as required
    String oldText = change.getOldValue();
    String newText = change.getNewValue();
    if (oldText.length()+1 != newText.length() || caratPos == 0) {
        return false ;
    }

    return oldText.equals(newText.substring(0, caratPos-1) + newText.substring(caratPos));
}

ReactFX - “懒惰”实时搜索文本区域

3 个答案: