Question

我试图使用弹簧休息模板获取getAccessToken，如下所示。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();
}

Full Errortrace：

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

尝试没有messageConverters.Still同样的错误。

Answer 1

通过你的堆栈跟踪，我可以建议你使用spring oauth版本1.0.1，它在第75行OAuth2AccessTokenJackson2Deserializer内有这个代码：

...
else if (OAuth2AccessToken.EXPIRES_IN.equals(name)) {
    expiresIn = jp.getLongValue();
}
...

它希望字段expires_in为数字。而您的例外表明，在您的回复中，它实际上是string。

要解决此问题，您可以升级到最新版本（目前为2.1.1），他们修复此部分以接受字符串值：

....
else if (OAuth2AccessToken.EXPIRES_IN.equals(name)) {
    try {
        expiresIn = jp.getLongValue();
    } catch (JsonParseException e) {
        expiresIn = Long.valueOf(jp.getText());
    }
}
....

无法读取JSON：当前令牌（VALUE_STRING）不是数字，不能使用数值访问器

1 个答案: