[英]Using GSON for generic Parcelable -- how to retrieve class type
There may be an easier / smarter way to do this.可能有一种更简单/更智能的方法来做到这一点。 I'm not a good Android developer.
我不是一个好的 Android 开发者。 Please advise
请指教
I have a library with about 100 different models.我有一个包含大约 100 个不同模型的库。 I now have to make all of these models parcelable.
我现在必须使所有这些模型都可以打包。
Rather than implement writeToParcel
for every single one , I thought it would be nice if I could just create a super class that used generic serialization logic. writeToParcel
为每一个都实现writeToParcel
,我认为如果我能创建一个使用通用序列化逻辑的超类就好了。 Something like the following:类似于以下内容:
public abstract class Model implements Parcelable
{
public final static Parcelable.Creator<Model> CREATOR = new Parcelable.Creator<Model>() {
@Override
public Model createFromParcel ( Parcel parcel )
{
Gson gson = new Gson();
String type = parcel.readString();
String serialized = parcel.readString();
gson.fromJson(serialized, /* HELP ME */)
}
@Override
public Model[] newArray ( int i )
{
// TODO: I haven't even read the docs on this method yet
}
};
@Override
public int describeContents ()
{
return 0;
}
@Override
public void writeToParcel ( Parcel parcel, int i )
{
Gson gson = new Gson();
String type = this.getClass().getName();
String serialized = gson.toJson(this);
parcel.writeString(type);
parcel.writeString(serialized);
}
}
Then in each of my models I just say:然后在我的每个模型中我只是说:
public class Whatever extends Model {
The problem is that Gson requires me to provide Type
in order to de-serialize.问题是 Gson 要求我提供
Type
以进行反序列化。 How can I get this Type
value?我怎样才能得到这个
Type
值? As you can see, I tried using this.getClass().getName()
to get the name of the class, but I have no idea how to reverse this and go from a string back to a type.如您所见,我尝试使用
this.getClass().getName()
来获取类的名称,但我不知道如何将其反转并将其从字符串返回到类型。 I also don't fully understand the difference between Class<T>
and Type
.我也不完全理解
Class<T>
和Type
之间的区别。
I solved this by using:我通过使用解决了这个问题:
String type = this.getClass().getCanonicalName();
to get the type, and then:获取类型,然后:
return (Model) gson.fromJson(serialized, Class.forName(type));
in my createFromParcel
method在我的
createFromParcel
方法中
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