如何序列android.graphics.Path的对象 [英] how to serialize an object of android.graphics.Path

问题描述

我想保存在内部设备内存Android.graphics.Path的对象。有谁知道如何序列化android.graphics.Path对象？而且，有没有存储路径对象的任何其他方式？谢谢。

I'm trying to store objects of Android.graphics.Path in internal device memory. Does anyone know how to serialize a android.graphics.Path object? And also, is there any other way of storing a Path object? Thanks.

推荐答案

我做的方式是确定，我需要从原来的Path类，然后简单地覆盖这些方法的方法如下：

The way I did it is to identify the methods that I needed from the original Path class and then simply override those methods as follows:

public class CustomPath extends Path implements Serializable {

private static final long serialVersionUID = -5974912367682897467L;

private ArrayList<PathAction> actions = new ArrayList<CustomPath.PathAction>();

private void readObject(ObjectInputStream in) throws IOException, ClassNotFoundException{
    in.defaultReadObject();
    drawThisPath();
}

@Override
public void moveTo(float x, float y) {
    actions.add(new ActionMove(x, y));
    super.moveTo(x, y);
}

@Override
public void lineTo(float x, float y){
    actions.add(new ActionLine(x, y));
    super.lineTo(x, y);
}

private void drawThisPath(){
    for(PathAction p : actions){
        if(p.getType().equals(PathActionType.MOVE_TO)){
            super.moveTo(p.getX(), p.getY());
        } else if(p.getType().equals(PathActionType.LINE_TO)){
            super.lineTo(p.getX(), p.getY());
        }
    }
}

public interface PathAction {
    public enum PathActionType {LINE_TO,MOVE_TO};
    public PathActionType getType();
    public float getX();
    public float getY();
}

public class ActionMove implements PathAction, Serializable{
    private static final long serialVersionUID = -7198142191254133295L;

    private float x,y;

    public ActionMove(float x, float y){
        this.x = x;
        this.y = y;
    }

    @Override
    public PathActionType getType() {
        return PathActionType.MOVE_TO;
    }

    @Override
    public float getX() {
        return x;
    }

    @Override
    public float getY() {
        return y;
    }

}

public class ActionLine implements PathAction, Serializable{
    private static final long serialVersionUID = 8307137961494172589L;

    private float x,y;

    public ActionLine(float x, float y){
        this.x = x;
        this.y = y;
    }

    @Override
    public PathActionType getType() {
        return PathActionType.LINE_TO;
    }

    @Override
    public float getX() {
        return x;
    }

    @Override
    public float getY() {
        return y;
    }

}
}

在我的例子中，我需要的moveTo和的lineTo，所以在这种情况下，我只是拿在列表中的图纸信息。此列表中包含的描绘信息（我称之为行动），以恢复绘图，因为它看起来就像物体序列化之前需要路径。然后，当我反序列化类型CustomPath我的目标我做到让默认的序列化协议称之为drawThisPath ，因此该路径被重新绘制。

In my example I need "moveTo" and "lineTo", so in this case I simply hold the drawing information in a list. This list contains the drawing information (what I call "action") needed by Path in order to restore the drawing as it looked like before the object was serialized. Then when I deserialize my objects of type CustomPath i make sure to let the default serialization protocol call "drawThisPath", so the path is redrawn.

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