Parcelable对象中属性的NULL值。为什么?

时间:2012-08-24 14:45:19

标签: java android serialization android-intent parcelable

我有以下Recipe类,它正在实现Parcelable类。但是当我将对象从一个类传递给另一个类时,其属性的值为null。为什么呢?

食谱类:

package mobile.bh.classes;

import java.util.ArrayList;

import mobile.bh.activities.MethodStep;
import android.graphics.Bitmap;
import android.os.Parcel;
import android.os.Parcelable;

//simple class that just has one member property as an example
public class Recipe implements Parcelable {
    public int id;
    public String name;
    public ArrayList<Ingredient> ingredients;
    public ArrayList<MethodStep> method;
    public String comment;
    public String image;
    public Bitmap image2;

    public Recipe(){}
    /* everything below here is for implementing Parcelable */

    // 99.9% of the time you can just ignore this
    public int describeContents() {
        return 0;
    }

    // write your object's data to the passed-in Parcel
    public void writeToParcel(Parcel out, int flags) {
        out.writeInt(id);
        out.writeString(name);
        out.writeList(ingredients);
        out.writeList(method);
        out.writeString(comment);
        out.writeString(image);
    }

    // this is used to regenerate your object. All Parcelables must have a CREATOR that implements these two methods
    public static final Parcelable.Creator<Recipe> CREATOR = new Parcelable.Creator<Recipe>() {
        public Recipe createFromParcel(Parcel in) {
            return new Recipe(in);
        }

        public Recipe[] newArray(int size) {
            return new Recipe[size];
        }
    };

    // example constructor that takes a Parcel and gives you an object populated with it's values
    private Recipe(Parcel in) {
        in.writeInt(id);
        in.writeString(name);
        in.writeList(ingredients);
        in.writeList(method);
        in.writeString(comment);
        in.writeString(image);
        }
}

通过intent

发送对象
    Intent i = new Intent(context,RecipeInfoActivity.class);
    i.putExtra("recipeObj", recipe);

接收另一方的对象

Recipe p = (Recipe) getIntent().getParcelableExtra("recipeObj");

p.name的值为null

3 个答案:

答案 0 :(得分:1)

在Parcelable Constructor中,您需要从包裹中回读。

private Recipe(Parcel in) {
        id = in.readInt();
        name =in.readString();
        ingredients = in.readList();
        method = in.readList();
        comment = in.readString();
        }

答案 1 :(得分:0)

您应该在构造函数中使用readInt而不是writeInt(其他字段使用etc)

答案 2 :(得分:0)

首先,在您的构造函数中,您似乎试图将所有属性写入包裹,但据我所知,它们尚未设置;可能你打算从那里读包裹?现在我不确定这个包究竟是什么,但我在想它的类似属性的类?如果是这样,Java不是通过引用传递的。意思是只修改传递给您方法的值不会修改实际包裹的值,您必须返回修改后的包裹