解决ObjectMapper.convertValue() 遇到的一些问题
作者：STRIVE.ZCR
这篇文章主要介绍了解决ObjectMapper.convertValue() 遇到的一些问题，具有很好的参考价值，希望对大家有所帮助。如有错误或未考虑完全的地方，望不吝赐教
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脚本之家。