我正在使用RestTemplate并解决对象的反序列化问题.这就是我在做的事情. JSON响应看起来像,
{
"response": {
"Time": "Wed 2013.01.23 at 03:35:25 PM UTC","Total_Input_Records": 5,},-
"message": "Succeeded","code": "200"
}
使用jsonschema2pojo将此Json有效负载转换为POJO
public class MyClass {
@JsonProperty("response")
private Response response;
@JsonProperty("message")
private Object message;
@JsonProperty("code")
private Object code;
private Map这是我获得异常的请求处理,
String url = "http://my.site.com/someparams";
RestTemplate template = new RestTemplate(
new HttpComponentsClientHttpRequestFactory());
FormHttpMessageConverter converter = new FormHttpMessageConverter();
List这是令人沮丧的部分,例外(故意修剪,不完整).我错过了什么？
org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Unrecognized field "Time" (Class com.temp.pointtests.Response),not marked as ignorable
at [Source: org.apache.http.conn.EofSensorInputStream@340ae1cf; line: 1,column: 22 (through reference chain: com.temp.pointtests.MyClass["response"]->com.temp.pointtests.Response["Time"]);]
更新已解决
我看到Spring添加了使用Jackson 2的MappingJackson2HttpMessageConverter.因为我上面的代码中的MappingJacksonHttpMessageConverter使用的是Jackson Pre2.0版本,但它不起作用.但它适用于Jackson 2.0.现在可以使用MappingJackson2HttpMessageConverter,我现在可以将它添加到我的RestTemplate中,一切正常:-).以下是具有相同问题的人的代码,
String url = "http://mysite.com/someparams";
RestTemplate template = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
HttpEntity request = new HttpEntity(headers);
List最佳答案
原文链接：https://www.f2er.com/spring/432097.html