Question

对从Location扩展的AddressLocation和AirportLocation进行分类。当json中的“类型”字段显示为“地址”或“机场”时，杰克逊分别将其正确反序列化为AddressLocation和AirportLocation类。缺少“类型”时，杰克逊不知道如何反序列化对象。有没有一种方法可以配置杰克逊，如果“类型”不存在或为空，则使用默认类型作为“地址”？

///从以下内容中删除了getter，setter和构造函数

@JsonInclude(value= JsonInclude.Include.NON_EMPTY)
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type", visible=false)
@JsonSubTypes({ 
        @JsonSubTypes.Type(value = AirportLocation.class, name = "AIRPORT"),
        @JsonSubTypes.Type(value = AddressLocation.class, name = "ADDRESS")
})

public abstract class Location {
    private List<Float> geoCoordinates;
    private String city;
    private String countryCode;  
}

public class AddressLocation extends Location {
    private String province;
    private String postalCode;
    private String streetAddress;

    public AddressLocation(final List<Float> geoCoordinates, final String city, final String countryCode,
                           final String province,
                           final String postalCode, final String streetAddress) {
        super(geoCoordinates, city, countryCode);
        this.province = province;
        this.postalCode = postalCode;
        this.streetAddress = streetAddress;
    }
}

public class AirportLocation extends Location {
 String airportCode;
}

我得到的错误是-

Missing type id when trying to resolve subtype of [simple type, class ***]: missing type id property 'type'
 at [Source: (String)"{"geo_coordinates":[1.17549435E-38],"city":"Whateverville","country_code":"WW"}"; line: 1, column: 79]

Answer 1

当JSON由客户端发送并由诸如控制器之类的框架反序列化时，我遇到了类似的问题。如果是这种情况，那是因为框架管理的Jackson不受您的代码管理。您可以通过注入自定义反序列化器，也可以通过传输对象简单地控制代码中的所有实例初始化

@POST
public String createAddress(AddressLocationRequest requestObj) {

   AddressLocation address = new AddressLocation();
   BeanUtils.copyProperties(address, requestObj);
   ...
}

杰克逊反序列化为默认子类型

1 个答案: