对于具有属性的枚举,例如:
public enum Thing {
THING_A("a"),
THING_B("b");
private String thing;
private Thing(String thing) {
this.thing = thing;
}
// Getters...
}
Jackson
序列化为值的名称,例如:
mapper.writeValueAsString(Thing.THING_A)); // "THING_A"
如果我们添加注释将序列化视为对象:
@JsonFormat(shape = JsonFormat.Shape.OBJECT)
它将序列化属性:
mapper.writeValueAsString(Thing.THING_A)); // "{"thing":"a"}"
我希望能够在序列化期间决定使用这些方法中的哪一个。因为这涉及大量枚举,所以我不想编辑每个枚举。有什么好的办法吗?
例如:这样的事情会很棒:
mapper.writeValueAsString(Thing.THING_A, JsonFormat.Shape.OBJECT); // "{"thing":"a"}"
mapper.writeValueAsString(Thing.THING_A, JsonFormat.Enum.DEFAULT); // "THING_A"
最佳答案
因为,com.fasterxml.jackson.annotation.JsonFormat
是一个注释,您可以实现自己的com.fasterxml.jackson.databind.AnnotationIntrospector
并返回您想要的值对于你所有的枚举。您可以在下面找到简单的示例:
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.core.Version;
import com.fasterxml.jackson.databind.AnnotationIntrospector;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.introspect.Annotated;
import com.fasterxml.jackson.databind.introspect.JacksonAnnotationIntrospector;
public class JsonPathApp {
public static void main(String[] args) throws Exception {
ObjectMapper mapper = new ObjectMapper();
mapper.setAnnotationIntrospector(AnnotationIntrospector.pair(new DynamicEnumAnnotationIntrospector(), new JacksonAnnotationIntrospector()));
System.out.println(mapper.writeValueAsString(Thing.THING_A));
}
}
class DynamicEnumAnnotationIntrospector extends AnnotationIntrospector {
@Override
public Version version() {
return new Version(1, 0, 0, "Dynamic enum object", "your.package", "jackson.dynamic.enum");
}
@Override
public JsonFormat.Value findFormat(Annotated memberOrClass) {
final Class<?> rawType = memberOrClass.getRawType();
if (rawType.isEnum() && rawType.getPackage().getName().startsWith("your.package")) {
return JsonFormat.Value.forShape(JsonFormat.Shape.OBJECT);
}
return super.findFormat(memberOrClass);
}
}
上面的代码打印:
{"thing":"a"}
现在,您可以创建两个 ObjectMapper
实例,其中一个配置您自己的注释内省(introspection)器,第二个则保留默认值。如果您确实想以动态方式使用它,您可以为每个可用的 Shape
值创建一个 ObjectMapper
并为给定形状选择所需的值:
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.annotation.JsonFormat.Shape;
import com.fasterxml.jackson.core.Version;
import com.fasterxml.jackson.databind.AnnotationIntrospector;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.introspect.Annotated;
import com.fasterxml.jackson.databind.introspect.JacksonAnnotationIntrospector;
import java.util.Arrays;
import java.util.EnumMap;
import java.util.List;
import java.util.Objects;
public class JsonPathApp {
public static void main(String[] args) throws Exception {
JsonFactory factory = new JsonFactory();
for (Shape shape : Shape.values()) {
ObjectMapper mapper = factory.getWithEnumShapeSetTo(shape);
System.out.println(shape + " => " + mapper.writeValueAsString(Thing.THING_A));
}
}
}
class JsonFactory {
private final AnnotationIntrospector defaultIntrospector = new JacksonAnnotationIntrospector();
private final EnumMap<Shape, ObjectMapper> instances = new EnumMap<>(Shape.class);
public JsonFactory() {
final List<Shape> notAllowed = Arrays.asList(Shape.BOOLEAN, Shape.BINARY);
Arrays.stream(Shape.values())
.filter(shape -> !notAllowed.contains(shape))
.forEach(shape -> instances.put(shape, createNewWithEnumShape(shape)));
}
private ObjectMapper createNewWithEnumShape(Shape shape) {
DynamicEnumAnnotationIntrospector enumIntrospector = new DynamicEnumAnnotationIntrospector(shape);
ObjectMapper mapper = new ObjectMapper();
mapper.setAnnotationIntrospector(AnnotationIntrospector.pair(enumIntrospector, defaultIntrospector));
return mapper;
}
public ObjectMapper getWithEnumShapeSetTo(Shape shape) {
Objects.requireNonNull(shape);
final ObjectMapper mapper = instances.get(shape);
if (mapper == null) {
return new ObjectMapper();
}
return mapper;
}
}
class DynamicEnumAnnotationIntrospector extends AnnotationIntrospector {
private final Shape shape;
public DynamicEnumAnnotationIntrospector(Shape shape) {
this.shape = Objects.requireNonNull(shape);
}
@Override
public Version version() {
return new Version(1, 0, 0, "Dynamic enum shape", "your.package", "jackson.dynamic.enum");
}
@Override
public JsonFormat.Value findFormat(Annotated memberOrClass) {
final Class<?> rawType = memberOrClass.getRawType();
if (rawType.isEnum() && rawType.getPackage().getName().startsWith("your.package")) {
return JsonFormat.Value.forShape(shape);
}
return super.findFormat(memberOrClass);
}
}
上面的代码打印:
ANY => "THING_A"
NATURAL => "THING_A"
SCALAR => "THING_A"
ARRAY => 0
OBJECT => {"thing":"a"}
NUMBER => 0
NUMBER_FLOAT => 0
NUMBER_INT => 0
STRING => "THING_A"
BOOLEAN => "THING_A"
BINARY => "THING_A"
上面的代码当然有点矫枉过正,但我想展示我们拥有的可能性。我们只有 3 个不同的输出,因此您可以将具有相同输出的值分组并创建最多 3 个不同的 ObjectMappers
。
关于java - 如何将枚举序列化为对象形状和默认字符串?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61281952/