Java Jackson 仅序列化接口方法
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Hymanson serialize only interface methods
提问by Jordi P.S.
I have one object Awith some methods ma, mb, mcand this object implements an interface Bwith only maand mb.
我有一个对象A有一些方法ma，mb，mc并且这个对象实现了一个只有ma和mb的接口B。
When I serialize BI expect only maand mbas a json response but I get as well mc.
当我序列化B 时，我只希望ma和mb作为json响应，但我也得到了mc。
I'd like to automatize this behaviour so that all the classes I serialize get serialized based on the interface and not on the implementation.
我想自动化这种行为，以便我序列化的所有类都基于接口而不是实现进行序列化。
How should I do that?
我该怎么做？
Example:
例子：
public interface Interf {
public boolean isNo();
public int getCountI();
public long getLonGuis();
}
Implementation:
执行：
public class Impl implements Interf {
private final String patata = "Patata";
private final Integer count = 231321;
private final Boolean yes = true;
private final boolean no = false;
private final int countI = 23;
private final long lonGuis = 4324523423423423432L;
public String getPatata() {
return patata;
}
public Integer getCount() {
return count;
}
public Boolean getYes() {
return yes;
}
public boolean isNo() {
return no;
}
public int getCountI() {
return countI;
}
public long getLonGuis() {
return lonGuis;
}
}
Serialization:
序列化：
ObjectMapper mapper = new ObjectMapper();
Interf interf = new Impl();
String str = mapper.writeValueAsString(interf);
System.out.println(str);
Response:
回复：
{
"patata": "Patata",
"count": 231321,
"yes": true,
"no": false,
"countI": 23,
"lonGuis": 4324523423423423500
}
Expected response:
预期回应：
{
"no": false,
"countI": 23,
"lonGuis": 4324523423423423500
}
采纳答案by broc.seib
Just annotate your interface such that Hymanson constructs data fields according to the interface's class and not the underlying object's class.
只需注释您的接口，以便 Hymanson 根据接口的类而不是底层对象的类来构造数据字段。
@JsonSerialize(as=Interf.class)
public interface Interf {
public boolean isNo();
public int getCountI();
public long getLonGuis();
}
回答by Lu55
You have two options:
您有两个选择：
1) put @JsonSerializeannotation on your interface (see @broc.seib answer)
1）@JsonSerialize在您的界面上添加注释（请参阅@broc.seib 答案）
2) or use specific writer for the serialization (as of Hymanson 2.9.6):
2) 或使用特定的 writer 进行序列化（从 Hymanson 2.9.6 开始）：
ObjectMapper mapper = new ObjectMapper();
String str = mapper.writerFor(Interf.class).writeValueAsString(interf);
