【发布时间】:2014-02-28 19:37:31
【问题描述】:
我有一个应用程序,我希望它同时接受 XML 和 JSON,我该如何编程返回类型?例如这是我的 POJO
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
// Class to marshall and unmarshall the XML and JSON to POJO
// This is a class for the request JSON and XML
@XmlRootElement
public class KeyProvision {
private String Consumer ;
private String API ;
private String AllowedNames ;
public void setConsumer( String Consumer)
{
this.Consumer= Consumer;
}
public void setAPI( String API){
this.API = API;
}
public void setAllowedNames(String AllowedNames){
this.AllowedNames = AllowedNames;
}
@XmlElement(name="Consumer")
public String getConsumer(){
return Consumer;
}
@XmlElement(name="API")
public String getAPI(){
return API;
}
@XmlElement(name="AllowedNames")
public String getAllowedNames(){
return AllowedNames;
}
}
我的休息界面是
import javax.ws.rs.Consumes;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
@POST
@Path("/request")
@Consumes({MediaType.APPLICATION_XML,MediaType.APPLICATION_JSON})
@Produces({MediaType.APPLICATION_XML,MediaType.APPLICATION_JSON})
public Response getRequest(KeyProvision keyInfo){
/* StringReader reader = new StringReader(keyInfo); // this code just leads to an execution failure for some reason
try{
JAXBContext jaxbContext = JAXBContext.newInstance(KeyProvision.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
KeyProvision api = (KeyProvision) jaxbUnmarshaller.unmarshal(reader);
System.out.println(api);
} catch(JAXBException e){
e.printStackTrace();
}
*/
String result = "Track saved : " + keyInfo;
return Response.status(201).entity(result).build() ;
// return "success" ;
}
我的 XML 是
<?xml version="1.0" encoding="UTF-8"?>
<KeyProvision>
<Consumer> testConsumer </Consumer>
<API>posting</API>
<AllowedNames> google</AllowedNames>
</KeyProvision>
我的 JSON 是
{
"KeyProvision": {
"Consumer": "testConsumer",
"API": "posting",
"AllowedNames": "google",
}
}
我的问题/疑问是
1) 我在使用 JSON 时不断收到 415 错误,为什么这不能正确解组? 2) 返回类型是否由 JAXB 决定?
【问题讨论】:
-
与您的问题无关,但您应该考虑仔细查看您的
setAllowedNames(String)方法。
标签: java json rest jaxb jersey