Foo1
public class Foo1{
private Long id;
private String code;
private String name;
private Boolean rState;
private String comments; // Contain json data
private String attachments; // Contain json data
}
Foo2
public class Foo2{
private Long id;
private String code;
private String name;
private Boolean rState;
private List comments;
private List attachments;
}
转换值
new ObjectMapper().convertValue(foo1, Foo2.class);
是否有可能,当我调用转换值时,json字符串自动转换为列表吗?
答案 0 :(得分:0)
选项1:仅注释Foo1.class的getter方法
@JsonProperty("comments")
public List getComments() {
List list = new ArrayList();
list.add(comments);
return list;
}
@JsonProperty("attachments")
public List getAttachments() {
List list = new ArrayList();
list.add(attachments);
return list;
}
Foo1 foo1 = new Foo1(Long.valueOf(1),"a","aaa",true,"abc","def");
System.out.println(new ObjectMapper().writeValueAsString(foo1));
Foo2 foo2 = new ObjectMapper().convertValue(foo1, Foo2.class);
System.out.println(new ObjectMapper().writeValueAsString(foo2));
选项2:使用jackson-custom-serialization
ObjectMapper mapper = new ObjectMapper();
SimpleModule simpleModule = new SimpleModule();
simpleModule.addSerializer(Foo1.class,new ListSerializer());
mapper.registerModule(simpleModule);
public class ListSerializer extends StdSerializer<Foo1> {
public ListSerializer() {
this(null);
}
protected ListSerializer(Class<Blah.Foo1> t) {
super(t);
}
public void serialize(Blah.Foo1 foo1, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException {
jsonGenerator.writeStartObject();
List list = new ArrayList();
list.add(foo1.getComments());
List list1 = new ArrayList();
list1.add(foo1.getAttachments());
jsonGenerator.writeObjectField("id",foo1.getId());
jsonGenerator.writeObjectField("code",foo1.getCode());
jsonGenerator.writeObjectField("name",foo1.getName());
jsonGenerator.writeObjectField("rState",foo1.getrState());
jsonGenerator.writeObjectField("comments",list);
jsonGenerator.writeObjectField("attachments",list1);
jsonGenerator.writeEndObject();
}
}
Foo1 foo1 = new Foo1(Long.valueOf(1),"a","aaa",true,"abc","def");
System.out.println(mapper.writeValueAsString(foo1));
Foo2 foo2 = mapper.convertValue(foo1, Foo2.class);
System.out.println(new ObjectMapper().writeValueAsString(foo2));
答案 1 :(得分:0)
我建议您使用Deserializer
,因为Foo1
可能实际上已在其他地方进行序列化,而我们可能只有序列化的数据要转换。这就是我们在这里的目标。
创建一个反序列化SimpleModule
类型的List
public class ListDeserializer extends StdDeserializer<List> {
public ListDeserializer() {
super(List.class);
}
@Override
public List deserialize(JsonParser p, DeserializationContext ctxt)
throws IOException {
if (p.getCurrentName().equals("result") || p.getCurrentName().equals("attachments")) {
JsonNode node = p.getCodec().readTree(p);
return new ObjectMapper().readValue(node.asText(), List.class);
}
return null;
}
}
上述反序列化器仅识别result
和attachments
,而忽略其余部分。因此,这是List
类中Foo2
的非常特定的反序列化器。
我的测试Foo1
和Foo2
如下
public class Foo1 {
String result = "[\"abcd\", \"xyz\"]";
public String getResult() {
return result;
}
public void setResult(String result) {
this.result = result;
}
}
public class Foo2 {
List result;
public List getResult() {
return result;
}
public void setResult(List result) {
this.result = result;
}
}
我测试了以下代码
// Module to help deserialize List objects
SimpleModule listModule = new SimpleModule();
listModule.addDeserializer(List.class, new ListDeserializer());
ObjectMapper objectMapper = new ObjectMapper().registerModules(listModule);
Foo2 foo2 = objectMapper.convertValue(new Foo1(), Foo2.class);
System.out.println(foo2.getResult());
输出为
[abcd, xyz]