使 Gson 使用枚举字符串值而不是常量名称
Make Gson to use Enum string value instead of constant name
有没有办法告诉 Gson 使用字符串值本身,而不是它的 Java 常量名称?
理想情况下在 Gson 配置中是全局的,因此它会为所有枚举执行此操作。
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
public class Main {
public static class Dress {
public Color color;
}
public static enum Color {
RED("red"),
BLUE("blue");
private final String type;
Color(final String type) { this.type = type; }
public String toString() { return type; }
}
public static void main(String[] args) throws InterruptedException {
Dress dress = new Dress();
dress.color = Color.RED;
GsonBuilder builder = new GsonBuilder();
builder.setPrettyPrinting();
Gson gson = builder.create();
System.out.println(gson.toJson(dress));
// ==> { "color": "RED" }
}
}
它打印 { "color": "RED" }
而不是 { "color": "red" }
。
使用带有序列化值的@SerializedName
import com.google.gson.annotations.SerializedName;
public static enum Color {
@SerializedName("red")
RED("red"),
@SerializedName("blue")
BLUE("blue");
...
}
另一种方法是使用自定义序列化程序
class ColorSerializer implements JsonSerializer {
@Override
public JsonElement serialize(Color src, Type typeOfSrc, JsonSerializationContext context) {
return new JsonPrimitive(src.toString());
}
}
并在构建器中注册自定义序列化程序
builder.registerTypeAdapter(Color.class, new ColorSerializer());
创建自定义序列化程序,例如:
public class EnumSerializer<T extends Enum<T>>
implements JsonSerializer<T> {
@Override
public JsonElement serialize(T src, Type typeOfSrc,
JsonSerializationContext context) {
return new JsonPrimitive(src.toString());
}
}
像这样注册:
builder.registerTypeAdapter(Color.class, new EnumSerializer<>());
// there is a reason for generics: that way you can easily register it to
// other Enum types also, for example:
// builder.registerTypeAdapter(SomeEnum.class, new EnumSerializer<>());
如果您需要将它应用于一个以上或所有 Enum,并且您不希望将它单独注册到每个 Enum,您可以使用 TypeAdapterFactory(有点复杂)。首先你需要TypeAdapter:
public class EnumTypeAdapter<T extends Enum<T>> extends TypeAdapter<T> {
@Override
public void write(JsonWriter out, T value) throws IOException {
out.jsonValue(value.toString());
}
@Override
public T read(JsonReader in) throws IOException {
return null;
}
}
然后工厂:
public static class EnumTypeAdapterFactory implements TypeAdapterFactory {
@Override
// Actually it might be easier to just make EnumTypeAdapter non-generic
// but on the other hand it might be better if used in some other contexts
// and in some other ways. Thus these suppressions
@SuppressWarnings({"rawtypes", "unchecked"})
public <T> TypeAdapter<T> create(Gson gson, TypeToken<T> type) {
// This if applies to all Enums. Change if not wanted.
if(Enum.class.isAssignableFrom(type.getRawType())) {
return new EnumTypeAdapter();
} else {
return null;
}
}
}
并注册:
builder.registerTypeAdapterFactory(new EnumTypeAdapterFactory());
有没有办法告诉 Gson 使用字符串值本身,而不是它的 Java 常量名称? 理想情况下在 Gson 配置中是全局的,因此它会为所有枚举执行此操作。
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
public class Main {
public static class Dress {
public Color color;
}
public static enum Color {
RED("red"),
BLUE("blue");
private final String type;
Color(final String type) { this.type = type; }
public String toString() { return type; }
}
public static void main(String[] args) throws InterruptedException {
Dress dress = new Dress();
dress.color = Color.RED;
GsonBuilder builder = new GsonBuilder();
builder.setPrettyPrinting();
Gson gson = builder.create();
System.out.println(gson.toJson(dress));
// ==> { "color": "RED" }
}
}
它打印 { "color": "RED" }
而不是 { "color": "red" }
。
使用带有序列化值的@SerializedName
import com.google.gson.annotations.SerializedName;
public static enum Color {
@SerializedName("red")
RED("red"),
@SerializedName("blue")
BLUE("blue");
...
}
另一种方法是使用自定义序列化程序
class ColorSerializer implements JsonSerializer {
@Override
public JsonElement serialize(Color src, Type typeOfSrc, JsonSerializationContext context) {
return new JsonPrimitive(src.toString());
}
}
并在构建器中注册自定义序列化程序
builder.registerTypeAdapter(Color.class, new ColorSerializer());
创建自定义序列化程序,例如:
public class EnumSerializer<T extends Enum<T>>
implements JsonSerializer<T> {
@Override
public JsonElement serialize(T src, Type typeOfSrc,
JsonSerializationContext context) {
return new JsonPrimitive(src.toString());
}
}
像这样注册:
builder.registerTypeAdapter(Color.class, new EnumSerializer<>());
// there is a reason for generics: that way you can easily register it to
// other Enum types also, for example:
// builder.registerTypeAdapter(SomeEnum.class, new EnumSerializer<>());
如果您需要将它应用于一个以上或所有 Enum,并且您不希望将它单独注册到每个 Enum,您可以使用 TypeAdapterFactory(有点复杂)。首先你需要TypeAdapter:
public class EnumTypeAdapter<T extends Enum<T>> extends TypeAdapter<T> {
@Override
public void write(JsonWriter out, T value) throws IOException {
out.jsonValue(value.toString());
}
@Override
public T read(JsonReader in) throws IOException {
return null;
}
}
然后工厂:
public static class EnumTypeAdapterFactory implements TypeAdapterFactory {
@Override
// Actually it might be easier to just make EnumTypeAdapter non-generic
// but on the other hand it might be better if used in some other contexts
// and in some other ways. Thus these suppressions
@SuppressWarnings({"rawtypes", "unchecked"})
public <T> TypeAdapter<T> create(Gson gson, TypeToken<T> type) {
// This if applies to all Enums. Change if not wanted.
if(Enum.class.isAssignableFrom(type.getRawType())) {
return new EnumTypeAdapter();
} else {
return null;
}
}
}
并注册:
builder.registerTypeAdapterFactory(new EnumTypeAdapterFactory());