Android 如何在 onSaveInstanceState 中保存自定义类的实例？
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How to save an instance of a custom class in onSaveInstanceState?
提问by jul
I created an instance of a custom class RestaurantList to hold my data (a list of restaurant data received from a web service as json data).
我创建了一个自定义类 RestaurantList 的实例来保存我的数据（从 Web 服务作为 json 数据接收的餐厅数据列表）。
How can I save it in onSaveInstanceState?
我怎样才能保存它onSaveInstanceState？
回答by Maaalte
Custom objects can be saved inside a Bundle when they implement the interface Parcelable.
Then they can be saved via:
自定义对象在实现接口时可以保存在 Bundle 中Parcelable。然后可以通过以下方式保存它们：
@Override
public void onSaveInstanceState(Bundle outState) {
super.onSaveInstanceState(outState);
outState.putParcelable("key", myObject);
}
Basically the following methods must be implemented in the class file:
基本上必须在类文件中实现以下方法：
public class MyParcelable implements Parcelable {
private int mData;
public int describeContents() {
return 0;
}
/** save object in parcel */
public void writeToParcel(Parcel out, int flags) {
out.writeInt(mData);
}
public static final Parcelable.Creator<MyParcelable> CREATOR
= new Parcelable.Creator<MyParcelable>() {
public MyParcelable createFromParcel(Parcel in) {
return new MyParcelable(in);
}
public MyParcelable[] newArray(int size) {
return new MyParcelable[size];
}
};
/** recreate object from parcel */
private MyParcelable(Parcel in) {
mData = in.readInt();
}
}
回答by dsewtz
I know "that this case is cold", but because i found this thread first, when I was searching for exactly the same thing (and found an answer by now):
我知道“这个案例很冷”，但是因为我首先找到了这个线程，当我搜索完全相同的东西时（并且现在找到了答案）：
Imagine Bundle as an XML file.
If you create a new <BUNDLE name="InstanceName" type="ClassName">you can freely add elements and attributes in a fresh and empty namespace.
把 Bundle 想象成一个 XML 文件。如果你创建一个新的，<BUNDLE name="InstanceName" type="ClassName">你可以在一个新的空命名空间中自由添加元素和属性。
When onSaveInstance(Bundle outState)of your MainActivity is called (you can also force this in onPause), you can create a new: Bundle b = new Bundle();
当onSaveInstance(Bundle outState)您的 MainActivity 被调用时（您也可以在 中强制调用onPause），您可以创建一个新的：Bundle b = new Bundle();
Then call your (probably not inherited and not overriden) custom Method onSaveInstance(Bundle b)in your own class with your newly created Bundle b.
Then (in onSaveInstance(Bundle outState)) of your MainActivity, call outState.putBundle("StringClassAndInstanceName", b);
然后onSaveInstance(Bundle b)在您自己的类中使用新创建的 Bundle b调用您的（可能不是继承的和未覆盖的）自定义方法。然后（在onSaveInstance(Bundle outState)）你的 MainActivity，调用outState.putBundle("StringClassAndInstanceName", b);
When you find this string in onCreate, you can either use a switch/case to recreate this object or (better) have a factory function in your custom class to work with Bundle and "StringClassAndInstanceName".
当您在 onCreate 中找到此字符串时，您可以使用 switch/case 来重新创建此对象，或者（更好）在您的自定义类中有一个工厂函数来处理 Bundle 和“StringClassAndInstanceName”。
回答by Andrew Lee
Custom class objects can be converted to JSON and stored in the bundle as a string. The following example is for Fragments.
自定义类对象可以转换为 JSON 并作为字符串存储在包中。以下示例适用于片段。
@Override
public void onSaveInstanceState(Bundle outState) {
super.onSaveInstanceState(outState);
Gson gson = new Gson();
String json= gson.toJson(customClass);
outState.putString("CUSTOM_CLASS", json);
}
public void onActivityCreated(@Nullable Bundle savedInstanceState) {
super.onActivityCreated(savedInstanceState);
if(savedInstanceState != null) {
String json= savedInstanceState.getString("CUSTOM_CLASS");
if(!json.isEmpty()) {
Gson gson = new Gson();
CustomClass customClass = gson.fromJson(json, CustomClass.class);
}
}
}
For Activities, override onRestoreInstanceStatemethod instead.
对于活动，请改写onRestoreInstanceState方法。
