解决ObjectMapper.convertValue() 遇到的一些问题

源代码：

public <T> T convertValue(Object fromValue, TypeReference<?> toValueTypeRef) throws IllegalArgumentException { return (T) _convert(fromValue, _typeFactory.constructType(toValueTypeRef)); }

该方法用于用jackson将bean转换为map

例子：

List<SObject> sObjects = new ObjectMapper().convertValue(map.get("list"), new TypeReference<List<SObject>>() { });

微服务中从其他服务获取过来的对象，如果从Object强转为自定义的类型会报错，利用ObjectMapper转换。

ObjectMapper mapper = new ObjectMapper();
DefaultResponse defaultResponse = proxy.getData();
List<Resource> resources = (<Resource>) defaultResponse.getData();  //这里的场景是：data是一个Object类型的，但是它其实是一个List<Resouce>，想把List中的每个对象分别转成可用的对象
for (int i = 0; i < serviceDateResources.size(); i++) {
    Resource resource = mapper.convertValue(resources.get(i), Resource.class);   //经过这步处理，resource就是可用的类型了，如果不转化会报错
}

在转换过程中有些属性被设置为空，这样就不需要转化

处理方法：

在需要转化的实体类商添加如下注解

@JsonInclude(Include.NON_NULL) 
@JsonInclude(Include.Include.ALWAYS) 默认 
@JsonInclude(Include.NON_DEFAULT) 属性为默认值不序列化 
@JsonInclude(Include.NON_EMPTY) 属性为 空（“”） 或者为 NULL 都不序列化 
@JsonInclude(Include.NON_NULL) 属性为NULL 不序列化

jackson objectMapper json字符串、对象bean、map、数组list互相转换常用的方法列举：

ObjectMapper mapper = new ObjectMapper();

1.对象转json字符串

User user=new User();
String userJson=mapper.writeValueAsString(user);

2.Map转json字符串

Map map=new HashMap();  
String json=mapper.writeValueAsString(map);

3.数组list转json字符串

Student[] stuArr = {student1, student3};  
String jsonfromArr =  mapper.writeValueAsString(stuArr);

4.json字符串转对象

String expected = "{\"name\":\"Test\"}";
User user = mapper.readValue(expected, User.class);

5.json字符串转Map

String expected = "{\"name\":\"Test\"}";
Map userMap = mapper.readValue(expected, Map.class);

6.json字符串转对象数组List

String expected="[{\"a\":12},{\"b\":23},{\"name\":\"Ryan\"}]";
CollectionType listType = mapper.getTypeFactory().constructCollectionType(ArrayList.class, User.class);
List<User> userList = mapper.readValue(expected, listType);

7.json字符串转Map数组List<Map<String,Object>>

String expected="[{\"a\":12},{\"b\":23},{\"name\":\"Ryan\"}]";
CollectionType listType = mapper.getTypeFactory().constructCollectionType(ArrayList.class, Map.class);
List<Map<String,Object>> userMapList = mapper.readValue(expected, listType);

8.jackson默认将对象转换为LinkedHashMap：

String expected = "[{\"name\":\"Ryan\"},{\"name\":\"Test\"},{\"name\":\"Leslie\"}]";
ArrayList arrayList = mapper.readValue(expected, ArrayList.class);

9.json字符串与list或map互转的方法

ObjectMapper objectMapper = new ObjectMapper();
 //遇到date按照这种格式转换
 SimpleDateFormat fmt = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
 objectMapper.setDateFormat(fmt);
 
  String preference = "{name:'侯勇'}";
        //json字符串转map
  Map<String, String> preferenceMap = new HashMap<String, String>();
  preferenceMap = objectMapper.readValue(preference, preferenceMap.getClass());
  
  //map转json字符串
  String result=objectMapper.writeValueAsString(preferenceMap);

10.bean转换为map

List<Map<String,String>> returnList=new ArrayList<Map<String,String>>();
List<Menu> menuList=menuDAOImpl.findByParentId(parentId);
ObjectMapper mapper = new ObjectMapper();
//用jackson将bean转换为map
returnList=mapper.convertValue(menuList,new TypeReference<List<Map<String, String>>>(){});

objectMapper.convertValue() 报错

报错信息如下：

com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of java.time.LocalDateTime (no Creators, like default constructor, exist): cannot deserialize from Object value (no delegate- or property-based Creator) at [Source: UNKNOWN; line: -1, column: -1] (through reference chain: net.too1.tplus.user.user.entity.User[“createTime”])

根据以上报错得知, 是java.time.LocalDateTime类型的原因. ObjectMapper 不能对LocalDateTime 序列化. 加上以下注解即可解决

@JsonDeserialize(using = LocalDateTimeDeserializer.class)
@JsonSerialize(using = LocalDateTimeSerializer.class)

@ApiModelProperty(value = "创建时间")
@JsonDeserialize(using = LocalDateTimeDeserializer.class)
@JsonSerialize(using = LocalDateTimeSerializer.class)
private LocalDateTime createTime;

以上为个人经验，希望能给大家一个参考，也希望大家多多支持三水点靠木。