解決ObjectMapper.convertValue() 遇到的一些問題

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源代碼：

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

該方法用於用jackson將bean轉換為map

例子：

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

微服務中從其他服務獲取過來的對象，如果從Object強轉為自定義的類型會報錯，利用ObjectMapper轉換。

ObjectMapper mapper = new ObjectMapper();
DefaultResponse defaultResponse = proxy.getData();
List<Resource> resources = (<Resource>) defaultResponse.getData();  //這裡的場景是：data是一個Object類型的，但是它其實是一個List<Resouce>，想把List中的每個對象分別轉成可用的對象
for (int i = 0; i < serviceDateResources.size(); i++) {
    Resource resource = mapper.convertValue(resources.get(i), Resource.class);   //經過這步處理，resource就是可用的類型瞭，如果不轉化會報錯
}

在轉換過程中有些屬性被設置為空，這樣就不需要轉化

處理方法：

在需要轉化的實體類商添加如下註解

@JsonInclude(Include.NON_NULL) 
@JsonInclude(Include.Include.ALWAYS) 默認 
@JsonInclude(Include.NON_DEFAULT) 屬性為默認值不序列化 
@JsonInclude(Include.NON_EMPTY) 屬性為 空（“”） 或者為 NULL 都不序列化 
@JsonInclude(Include.NON_NULL) 屬性為NULL 不序列化

jackson objectMapper json字符串、對象bean、map、數組list互相轉換常用的方法列舉：

ObjectMapper mapper = new ObjectMapper();

1.對象轉json字符串

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

2.Map轉json字符串

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

3.數組list轉json字符串

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

4.json字符串轉對象

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

5.json字符串轉Map

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

6.json字符串轉對象數組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字符串轉Map數組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默認將對象轉換為LinkedHashMap：

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

9.json字符串與list或map互轉的方法

ObjectMapper objectMapper = new ObjectMapper();
 //遇到date按照這種格式轉換
 SimpleDateFormat fmt = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
 objectMapper.setDateFormat(fmt);
 
  String preference = "{name:'侯勇'}";
        //json字符串轉map
  Map<String, String> preferenceMap = new HashMap<String, String>();
  preferenceMap = objectMapper.readValue(preference, preferenceMap.getClass());
  
  //map轉json字符串
  String result=objectMapper.writeValueAsString(preferenceMap);

10.bean轉換為map

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

objectMapper.convertValue() 報錯

報錯信息如下：

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”])

根據以上報錯得知, 是java.time.LocalDateTime類型的原因. ObjectMapper 不能對LocalDateTime 序列化. 加上以下註解即可解決

@JsonDeserialize(using = LocalDateTimeDeserializer.class)
@JsonSerialize(using = LocalDateTimeSerializer.class)
@ApiModelProperty(value = "創建時間")
@JsonDeserialize(using = LocalDateTimeDeserializer.class)
@JsonSerialize(using = LocalDateTimeSerializer.class)
private LocalDateTime createTime;

以上為個人經驗，希望能給大傢一個參考，也希望大傢多多支持WalkonNet。

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