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 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 sObjects = new ObjectMapper().convertValue(map.get("list"), new TypeReference>() { });
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 resources = () 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，想把List中的每个对象分别转成可用的对象 for (int i = 0; i
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：
@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 userList = mapper.readValue(expected, listType);
7.json字符串转Map数组List>
String expected="[{"a":12},{"b":23},{"name":"Ryan"}]"; CollectionType listType = mapper.getTypeFactory().constructCollectionType(ArrayList.class, Map.class); List> 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 preferenceMap = new HashMap(); preferenceMap = objectMapper.readValue(preference, preferenceMap.getClass()); //map转json字符串 String result=objectMapper.writeValueAsstring(preferenceMap);10.bean转换为map
List> returnList=new ArrayList>(); List menuList=menuDAOImpl.findByParentId(parentId); ObjectMapper mapper = new ObjectMapper(); //用jackson将bean转换为map returnList=mapper.convertValue(menuList,new TypeReference>>(){});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;版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
