例如：public void builCarManager(Car car) { Manager<car.getClass()> carManager = new Manager<>();}解决这个问题的一些建议？编辑我的主要问题是：public class CustomSerializer<T> extends StdSerializer<T> { @Override public void serialize(T car, JsonGenerator gen, SerializerProvider sp) { // I get car.getClass() fine but I need add it Manager<T> carManager = new Manager<>(); // Manager<car.getClass()> carManager = new Manager<>(); }}这里我不能通过CustomSerializer<Lamborgini>.class。@JsonSerialize(using = CustomSerializer.class)private Lamborgini car;
3 回答
慕尼黑5688855
TA贡献1848条经验 获得超2个赞
假设您有以下POJO模型：
class Car {
}
class Lamborghini extends Car {
}
class Ferrari extends Car {
}
您的Manager课程如下所示：
class Manager<T extends Car> {
private Class<T> clazz;
public Manager(Class<T> clazz) {
this.clazz = clazz;
}
public String generateCustomCarName() {
if (Lamborghini.class == clazz) {
return "Lambooooo!";
} else if (Ferrari.class == clazz) {
return "Wild horse Ferrari!";
}
return "Not for everyone!";
}
@Override
public String toString() {
return String.format("Manager[clazz=%s]", clazz.getName());
}
}
现在，让我们创建一个Store有 oneFerrari和 one的Lamborghini：
class Store {
private Car ferrari = new Ferrari();
private Car lamborghini = new Lamborghini();
public Car getFerrari() {
return ferrari;
}
public void setFerrari(Car ferrari) {
this.ferrari = ferrari;
}
public Car getLamborghini() {
return lamborghini;
}
public void setLamborghini(Car lamborghini) {
this.lamborghini = lamborghini;
}
}
现在，我们需要class在运行时获取信息。由于Java's Type Erasure我们需要提供class：
class CustomCarSerializer<T extends Car> extends StdSerializer<T> {
public CustomCarSerializer(Class<T> clazz) {
super(clazz);
}
@Override
public void serialize(T value, JsonGenerator jgen, SerializerProvider provider)
throws IOException {
Manager<T> manager = new Manager<>(_handledType);
jgen.writeString(manager.generateCustomCarName());
}
}
请注意，它StdSerializer也有需要的构造函数，class我们可以稍后使用_handledType. 现在我们需要使用序列化器配置我们的实现SimpleModule：
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.databind.ser.std.StdSerializer;
import java.io.IOException;
public class Cars {
public static void main(String[] args) throws Exception {
SimpleModule carsModule = new SimpleModule("CarsModule");
carsModule.addSerializer(Lamborghini.class, new CustomCarSerializer<>(Lamborghini.class));
carsModule.addSerializer(Ferrari.class, new CustomCarSerializer<>(Ferrari.class));
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(carsModule);
mapper.enable(SerializationFeature.INDENT_OUTPUT);
System.out.println(mapper.writeValueAsString(new Store()));
}
}
上面的代码打印：
{
"ferrari" : "Wild horse Ferrari!",
"lamborghini" : "Lambooooo!"
}
人到中年有点甜
TA贡献1895条经验 获得超7个赞
处理这个问题的正确方法是让两个类（例如Car和Desk）实现一个公共接口（例如Item）并将该接口与Manager该类相关联。
public class Instance {
private static interface Item {
String getName();
}
private static class Car implements Item {
@Override
public String getName() {
return "Car";
}
}
private static class Desk implements Item {
@Override
public String getName() {
return "Desk";
}
}
private static class Manager<T extends Item> {
private T item;
public Manager(T item) {
this.item = item;
}
@Override
public String toString() {
return String.format("Manager[item=%s]", item.getName());
}
}
public static <E extends Item> Manager<E> buildItemManager(E item) {
return new Manager<E>(item);
}
public static void main(String[] args) {
Item car = new Car(), desk = new Desk();
Manager<Item> manager1 = buildItemManager(car);
System.out.println(manager1);
Manager<Item> manager2 = buildItemManager(desk);
System.out.println(manager2);
}
}
输出
Manager[item=Car]
Manager[item=Desk]
另一种方法...Manager通过传入一个代表 a 的类来实例化 a Item。
public class Instance {
public static interface Item {
String getName();
}
public static class Car implements Item {
@Override
public String getName() {
return "Car";
}
}
public static class Desk implements Item {
@Override
public String getName() {
return "Desk";
}
}
public static class Manager<T extends Item> {
private T item;
public Manager(T item) {
this.item = item;
}
@Override
public String toString() {
return String.format("Manager[item=%s]", item.getName());
}
}
public static <E extends Item> Manager<E> buildItemManager(Class<E> item) throws InstantiationException, IllegalAccessException {
return new Manager<E>(item.newInstance());
}
public static void main(String[] args) {
try {
/* Cannot initialize array with generic types... */
@SuppressWarnings("unchecked")
Manager<? extends Item>[] managers = new Manager[2];
managers[0] = buildItemManager(Car.class);
managers[1] = buildItemManager(Desk.class);
for (Manager<?> manager : managers) {
System.out.println(manager);
}
} catch (InstantiationException e) {
e.printStackTrace();
} catch (IllegalAccessException e) {
e.printStackTrace();
}
}
}
qq_笑_17
TA贡献1818条经验 获得超7个赞
是的，您可以使用泛型。
public <T> void builCarManager(T car) {
Manager<T> carManager = new Manager<>();
}
现在您可以传递任何类型的对象。创建的管理器将属于同一类型。
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