小编给大家分享一下怎么解决ObjectMapper.convertValue() 遇到的一些问题，希望大家阅读完这篇文章之后都有所收获，下面让我们一起去探讨吧！
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mapreturnList=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;看完了这篇文章，相信你对“怎么解决ObjectMapper.convertValue() 遇到的一些问题”有了一定的了解，如果想了解更多相关知识，欢迎关注编程网行业资讯频道，感谢各位的阅读！
