如何使用Spring的REST模板传递自定义对象

Question

我需要使用RESTTemplate将自定义对象传递给我的REST服务.

RestTemplate restTemplate = new RestTemplate();
MultiValueMap<String, Object> requestMap = new LinkedMultiValueMap<String, Object>();
...

requestMap.add("file1", new FileSystemResource(..);
requestMap.add("Content-Type","text/html");
requestMap.add("accept", "text/html");
requestMap.add("myobject",new CustomObject()); // This is not working
System.out.println("Before Posting Request........");
restTemplate.postForLocation(url, requestMap);//Posting the data.
System.out.println("Request has been executed........");

我无法将自定义对象添加到MultiValueMap.请求生成失败.

有人可以帮我找到一个方法吗？我可以简单地传递一个字符串对象而没有问题.用户定义的对象会产生问题.

感谢任何帮助!

Answer 1

你可以用杰克逊相当简单地做到这一点.

这是我为一篇简单的POJO撰写的文章.

@XmlRootElement(name="newobject")
@JsonSerialize(include=JsonSerialize.Inclusion.NON_NULL)
public class NewObject{
    private String stuff;

    public String getStuff(){
        return this.stuff;
    }

    public void setStuff(String stuff){
        this.stuff = stuff;
    }
}

....
//make the object
NewObject obj = new NewObject();
obj.setStuff("stuff");

//set your headers
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);

//set your entity to send
HttpEntity entity = new HttpEntity(obj,headers);

// send it!
ResponseEntity<String> out = restTemplate.exchange("url", HttpMethod.POST, entity
    , String.class);

上面的链接应告诉您如何在需要时进行设置.这是一个非常好的教程.