我有以下几点 Recipe
类,它正在实现 Parcelable
班级。但是当我将对象从一个类传递到另一个类时,它的属性值是 null
. 为什么?
配方类别:
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);
}
}
发送对象 intent
```
Intent i = new Intent(context,RecipeInfoActivity.class);
i.putExtra("recipeObj", recipe);
在另一边接收对象
Recipe p = (Recipe) getIntent().getParcelableExtra("recipeObj");
但它的价值 `p.name` 是 `null`
3条答案
按热度按时间6l7fqoea1#
在parcelable构造函数中,您需要从包中读回数据。
b5buobof2#
首先,在您的构造函数中,似乎您正在尝试将所有属性写入包,但据我所知,它们尚未设置;你可能是想看包裹里的东西?现在我不确定这个包裹到底是什么,但我在想它的一些属性,比如类?如果是这样,java就不是通过引用传递的。这意味着仅仅修改传递给您的方法的值不会修改真正的包的值,您必须返回修改后的包
c0vxltue3#
您应该在构造函数中使用readint而不是writeint(对于其他字段是etc)