嗯,我知道这个问题已经被讨论了几次,但我还是没能解决我的问题。
所以我正在使用org.springframework.web.client.RestTemplate从http请求中获取一个JSONObject:
JSONObject j = RestTemplate.getForObject(url, JSONObject.class);
但是我遇到了以下错误:
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)
我想访问一个Rest-Api,Json对象可能有不同的字段名。
我已经尝试过@JsonIgnoreProperties(ignoreUnknown=true)
,但这并不起作用...
如何将响应转换为JSONObject?
Exception in thread "main" org.codehaus.jackson.JsonParseException: Unexpected character ('h' (code 104)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
。 - Maik