我有一些类A,B,C,它们都继承自BaseClass类。
我有一个String json,其中包含A,B,C或BaseClass的json表示形式。
我想要某种方法将此字符串反序列化为BaseClass(多态反序列化)。像这样
BaseClass base = ObjectMapper.readValue(jsonString, BaseClass.class);
jsonString 可以是A,B,C或BaseClass中任何一个的Json String表示形式。
jsonString
目前尚不清楚原始海报有什么问题。我猜这是两件事之一:
未绑定JSON元素的反序列化问题,因为JSON包含Java中没有要绑定的元素;要么
要实现多态反序列化。
这是第一个问题的解决方案。
import static org.codehaus.jackson.map.DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES; import org.codehaus.jackson.map.ObjectMapper; public class Foo { public static void main(String[] args) throws Exception { BaseClass base = new BaseClass(); A a = new A(); B b = new B(); C c = new C(); ObjectMapper mapper = new ObjectMapper(); String baseJson = mapper.writeValueAsString(base); System.out.println(baseJson); // {"baseName":"base name"} String aJson = mapper.writeValueAsString(a); System.out.println(aJson); // {"baseName":"base name","aName":"a name"} String bJson = mapper.writeValueAsString(b); System.out.println(bJson); // {"baseName":"base name","bName":"b name"} String cJson = mapper.writeValueAsString(c); System.out.println(cJson); // {"baseName":"base name","cName":"c name"} BaseClass baseCopy = mapper.readValue(baseJson, BaseClass.class); System.out.println(baseCopy); // baseName: base name // BaseClass aCopy = mapper.readValue(aJson, BaseClass.class); // throws UnrecognizedPropertyException: // Unrecognized field "aName", not marked as ignorable // because the JSON contains elements for which no Java field // to bind to was provided. // Need to let Jackson know that not all JSON elements must be bound. // To resolve this, the class can be annotated with // @JsonIgnoreProperties(ignoreUnknown=true) or the ObjectMapper can be // directly configured to not FAIL_ON_UNKNOWN_PROPERTIES mapper = new ObjectMapper(); mapper.configure(FAIL_ON_UNKNOWN_PROPERTIES, false); BaseClass aCopy = mapper.readValue(aJson, BaseClass.class); System.out.println(aCopy); // baseName: base name BaseClass bCopy = mapper.readValue(bJson, BaseClass.class); System.out.println(bCopy); // baseName: base name BaseClass cCopy = mapper.readValue(cJson, BaseClass.class); System.out.println(cCopy); // baseName: base name } } class BaseClass { public String baseName = "base name"; @Override public String toString() {return "baseName: " + baseName;} } class A extends BaseClass { public String aName = "a name"; @Override public String toString() {return super.toString() + ", aName: " + aName;} } class B extends BaseClass { public String bName = "b name"; @Override public String toString() {return super.toString() + ", bName: " + bName;} } class C extends BaseClass { public String cName = "c name"; @Override public String toString() {return super.toString() + ", cName: " + cName;} }
这是第二个问题的解决方案。
import org.codehaus.jackson.annotate.JsonSubTypes; import org.codehaus.jackson.annotate.JsonSubTypes.Type; import org.codehaus.jackson.annotate.JsonTypeInfo; import org.codehaus.jackson.map.ObjectMapper; public class Foo { public static void main(String[] args) throws Exception { BaseClass base = new BaseClass(); A a = new A(); B b = new B(); C c = new C(); ObjectMapper mapper = new ObjectMapper(); String baseJson = mapper.writeValueAsString(base); System.out.println(baseJson); // {"type":"BaseClass","baseName":"base name"} String aJson = mapper.writeValueAsString(a); System.out.println(aJson); // {"type":"a","baseName":"base name","aName":"a name"} String bJson = mapper.writeValueAsString(b); System.out.println(bJson); // {"type":"b","baseName":"base name","bName":"b name"} String cJson = mapper.writeValueAsString(c); System.out.println(cJson); // {"type":"c","baseName":"base name","cName":"c name"} BaseClass baseCopy = mapper.readValue(baseJson, BaseClass.class); System.out.println(baseCopy); // baseName: base name BaseClass aCopy = mapper.readValue(aJson, BaseClass.class); System.out.println(aCopy); // baseName: base name, aName: a name BaseClass bCopy = mapper.readValue(bJson, BaseClass.class); System.out.println(bCopy); // baseName: base name, bName: b name BaseClass cCopy = mapper.readValue(cJson, BaseClass.class); System.out.println(cCopy); // baseName: base name, cName: c name } } @JsonTypeInfo( use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type") @JsonSubTypes({ @Type(value = A.class, name = "a"), @Type(value = B.class, name = "b"), @Type(value = C.class, name = "c") }) class BaseClass { public String baseName = "base name"; @Override public String toString() {return "baseName: " + baseName;} } class A extends BaseClass { public String aName = "a name"; @Override public String toString() {return super.toString() + ", aName: " + aName;} } class B extends BaseClass { public String bName = "b name"; @Override public String toString() {return super.toString() + ", bName: " + bName;} } class C extends BaseClass { public String cName = "c name"; @Override public String toString() {return super.toString() + ", cName: " + cName;} }
如果相反,目标是在没有专门用于指示子类类型的JSON元素的情况下反序列化为子类类型,那么这也是可能的,只要可以使用JSON中的内容来确定应该使用什么子类类型。 。我在http://programmerbruce.blogspot.com/2011/05/deserialize-json-with-jackson- into.html上发布了这种方法的示例。