【发布时间】:2017-05-31 05:01:32
【问题描述】:
我试图通过如下所示的 spring rest 模板获取 getAccessToken。我在调用 oAuthRestTemplate.getAccessToken()
时收到以下错误Exception in thread "main" org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Current token (VALUE_STRING) not numeric, can not use numeric value accessors
代码:
ResourceOwnerPasswordResourceDetails resource = new ResourceOwnerPasswordResourceDetails();
resource.setAccessTokenUri("https://../oauth2/token");
resource.setUsername("u");
resource.setPassword("p");
resource.setClientSecret("...s");
resource.setClientId("...i");
resource.setGrantType("password");
List<String> scopes = new ArrayList<String>(2);
scopes.add("write");
scopes.add("read");
resource.setScope(scopes);
DefaultOAuth2ClientContext clientContext = new DefaultOAuth2ClientContext();
List<HttpMessageConverter<?>> messageConverters = new ArrayList<HttpMessageConverter<?>>();
messageConverters.add(new MappingJackson2HttpMessageConverter());
messageConverters.add(new FormHttpMessageConverter());
OAuth2RestTemplate oAuthRestTemplate = new OAuth2RestTemplate(resource,clientContext);
oAuthRestTemplate.setMessageConverters(messageConverters);
try {
SSLUtil.turnOffSslChecking();
System.out.println(oAuthRestTemplate.getAccessToken());
String result = oAuthRestTemplate.getForObject("url" + "/api/version", String.class);
} catch (RestClientException e) {
e.printStackTrace();
}
完整的错误跟踪:
Exception in thread "main" org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Current token (VALUE_STRING) not numeric, can not use numeric value accessors
at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@44908881; line: 1, column: 159]; nested exception is com.fasterxml.jackson.core.JsonParseException: Current token (VALUE_STRING) not numeric, can not use numeric value accessors
at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@44908881; line: 1, column: 159]
at org.springframework.http.converter.json.MappingJackson2HttpMessageConverter.readJavaType(MappingJackson2HttpMessageConverter.java:171)
at org.springframework.http.converter.json.MappingJackson2HttpMessageConverter.read(MappingJackson2HttpMessageConverter.java:163)
at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:94)
at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:491)
at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:454)
at org.springframework.security.oauth2.client.token.OAuth2AccessTokenSupport.retrieveToken(OAuth2AccessTokenSupport.java:113)
at org.springframework.security.oauth2.client.token.grant.password.ResourceOwnerPasswordAccessTokenProvider.obtainAccessToken(ResourceOwnerPasswordAccessTokenProvider.java:47)
at org.springframework.security.oauth2.client.token.AccessTokenProviderChain.obtainNewAccessTokenInternal(AccessTokenProviderChain.java:142)
at org.springframework.security.oauth2.client.token.AccessTokenProviderChain.obtainAccessToken(AccessTokenProviderChain.java:118)
at org.springframework.security.oauth2.client.OAuth2RestTemplate.acquireAccessToken(OAuth2RestTemplate.java:216)
at org.springframework.security.oauth2.client.OAuth2RestTemplate.getAccessToken(OAuth2RestTemplate.java:168)
at com.ClientTemplate.main(ClientTemplate.java:69)
Caused by: com.fasterxml.jackson.core.JsonParseException: Current token (VALUE_STRING) not numeric, can not use numeric value accessors
at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@44908881; line: 1, column: 159]
at com.fasterxml.jackson.core.JsonParser._constructError(JsonParser.java:1419)
at com.fasterxml.jackson.core.base.ParserMinimalBase._reportError(ParserMinimalBase.java:508)
at com.fasterxml.jackson.core.base.ParserBase._parseNumericValue(ParserBase.java:748)
at com.fasterxml.jackson.core.base.ParserBase.getLongValue(ParserBase.java:619)
at org.springframework.security.oauth2.common.OAuth2AccessTokenJackson2Deserializer.deserialize(OAuth2AccessTokenJackson2Deserializer.java:75)
at org.springframework.security.oauth2.common.OAuth2AccessTokenJackson2Deserializer.deserialize(OAuth2AccessTokenJackson2Deserializer.java:44)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:3051)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2206)
at org.springframework.http.converter.json.MappingJackson2HttpMessageConverter.readJavaType(MappingJackson2HttpMessageConverter.java:168)
... 11 more
在没有 messageConverters 的情况下尝试过。仍然是同样的错误。
【问题讨论】:
-
你能发布回复吗?
-
尝试得到响应
String result = oAuthRestTemplate.getForObject("url" + "/api/version", String.class);.. 但同样的错误。 -
您是否尝试使用任何工具(浏览器、wget 等)调用 url 并检查纯响应?
-
我在浏览器中得到了结果。但在程序 oAuthRestTemplate.getAccessToken() 本身失败
-
请发表回复。
标签: java spring oauth-2.0 jackson resttemplate