是的,我知道这个问题已经讨论过几次了,但是我没有解决问题。

因此,我使用 org.springframework.web.client.RestTemplate从Http请求中获取了JSONObject:

JSONObject j = RestTemplate.getForObject(url, JSONObject.class);

但是我得到这个错误:
    Exception in thread "main" org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Unrecognized field "uri" (Class org.json.JSONObject), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@6f526c5f; line: 2, column: 12] (through reference chain: org.json.JSONObject["uri"]); nested exception is org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "uri" (Class org.json.JSONObject), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@6f526c5f; line: 2, column: 12] (through reference chain: org.json.JSONObject["uri"])
    at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.readJavaType(MappingJacksonHttpMessageConverter.java:181)
    at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.read(MappingJacksonHttpMessageConverter.java:173)
    at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:94)
    at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:517)
    at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:472)
    at org.springframework.web.client.RestTemplate.getForObject(RestTemplate.java:237)

我想访问Rest-Api,并且Json对象可以具有不同的字段名称。
我已经尝试过@JsonIgnoreProperties(ignoreUnknown=true)。但这行不通...

如何将响应放入JSONObject?

最佳答案

您可以在Jackson 2.0中使用它:

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(
DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

如果您的版本是2.0之前的版本,请使用:
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(
DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);

09-11 19:30