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TA贡献1772条经验 获得超5个赞
您应该将JSON反序列化与应用程序的其他部分分开。您不能为所有响应实现一种方法,但您的响应数量可能有限,您可以为每个类声明一些简单的方法。通常,您只能使用一种声明如下的方法:
public <T> T deserialise(String payload, Class<T> expectedClass) {
Objects.requireNonNull(payload);
Objects.requireNonNull(expectedClass);
try {
return mapper.readValue(payload, expectedClass);
} catch (IOException e) {
throw new IllegalStateException("JSON is not valid!", e);
}
}
现在,您可以反序列化您想要的所有有效负载。您需要提供要接收的JSON负载和类。POJO
显示该概念的简单工作解决方案:
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import java.io.IOException;
import java.util.Objects;
public class JsonMapper {
private final ObjectMapper mapper = new ObjectMapper();
public JsonMapper() {
// configure mapper instance if required
mapper.enable(SerializationFeature.INDENT_OUTPUT);
mapper.enable(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT);
// etc...
}
public String serialise(Object value) {
try {
return mapper.writeValueAsString(value);
} catch (JsonProcessingException e) {
throw new IllegalStateException("Could not generate JSON!", e);
}
}
public <T> T deserialise(String payload, Class<T> expectedClass) {
Objects.requireNonNull(payload);
Objects.requireNonNull(expectedClass);
try {
return mapper.readValue(payload, expectedClass);
} catch (IOException e) {
throw new IllegalStateException("JSON is not valid!", e);
}
}
public Foo parseResponseFoo(String payload) {
return deserialise(payload, Foo.class);
}
public Bar parseResponseBar(String payload) {
return deserialise(payload, Bar.class);
}
public static void main(String[] args) {
JsonMapper jsonMapper = new JsonMapper();
String bar = "{\"bar\" : 2}";
System.out.println(jsonMapper.parseResponseBar(bar));
String foo = "{\"foo\" : 1}";
System.out.println(jsonMapper.parseResponseFoo(foo));
System.out.println("General method:");
System.out.println(jsonMapper.deserialise(foo, Foo.class));
System.out.println(jsonMapper.deserialise(bar, Bar.class));
}
}
class Foo {
public int foo;
@Override
public String toString() {
return "Foo{" +
"foo=" + foo +
'}';
}
}
class Bar {
public int bar;
@Override
public String toString() {
return "Bar{" +
"bar=" + bar +
'}';
}
}
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