【发布时间】:2015-05-01 03:34:27
【问题描述】:
我有以 xml 形式返回响应的 restlet 网络服务。我正在使用杰克逊作为粘合剂。 下面是我要返回的课程。
import java.io.Serializable;
import java.util.ArrayList;
import java.util.List;
public class ApiResponse<T> implements Serializable {
/**
*
*/
private static final long serialVersionUID = -2736991050157565598L;
private int responseCode;
private String reponseMessage;
private List<T> body = new ArrayList<T>();
public int getResponseCode() {
return responseCode;
}
public void setResponseCode(int responseCode) {
this.responseCode = responseCode;
}
public String getReponseMessage() {
return reponseMessage;
}
public void setReponseMessage(String reponseMessage) {
this.reponseMessage = reponseMessage;
}
public List<T> getBody() {
return body;
}
public void setBody(List<T> body) {
this.body = body;
}
}
以下是服务的响应。除了将嵌套对象的属性名称与父对象相同之外,一切几乎都很好。它显示嵌套标签名称的主体,但我希望它是 T 模板。有什么想法吗?
<ApiResponse>
<responseCode>1</responseCode>
<reponseMessage />
<body>
<body>
<reportId>1</reportId>
<reportName>name1</reportName>
</body>
<body>
<reportId>2</reportId>
<reportName>name2</reportName>
</body>
</body>
</ApiResponse>
【问题讨论】:
标签: java xml serialization deserialization restlet-2.0