【发布时间】:2014-10-17 09:28:05
【问题描述】:
是的,我知道这个问题已经讨论过几次了,但我没能解决我的问题。
所以我使用 org.springframework.web.client.RestTemplate 从 http-request 获取 JSONObject:
JSONObject j = RestTemplate.getForObject(url, JSONObject.class);
但我收到此错误:
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 并且 Json 对象可以有不同的字段名称。
我已经尝试过@JsonIgnoreProperties(ignoreUnknown=true)。但这行不通……
如何将响应放入 JSONObject?
【问题讨论】: