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Curl令牌请求到Spring restemplate转换

resttemplate curl spring java

vicky · 技术社区 · 5 年前

我的卷发请求

curl -d "grant_type=client_credentials&client_id=nu5yzeth9tektzf5egxuntp7&client_secret=uP2Xvr6SCKYgXgxxJsv2QkUG" 
  -H "Content-Type: application/x-www-form-urlencoded" 
  -X POST https://cloudsso.example.com/as/token.oauth2

卷曲响应:

{"access_token":"HVURQ845OPJqs8UpOlef5m2ZCNwR","token_type":"Bearer","expires_in":3599}

RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));

MultiValueMap<String, String> bodyParamMap = new LinkedMultiValueMap<String, String>();
bodyParamMap.add("grant_type", "client_credentials");
bodyParamMap.add("client_id", "nu5yzeth9tektzf5egxuntp7");
bodyParamMap.add("client_secret", "uP2Xvr6SCKYgXgxxJsv2QkUG");

entity = new HttpEntity<>(reqBodyData, bodyParamMap);

restTemplate.postForEntity(url, entity, TokenDTO.class)

Resttemplate呼叫响应:

2019-12-04 11:10:16,483[0;39m [39mDEBUG[0;39m [[34mrestartedMain[0;39m] [33morg.springframework.core.log.CompositeLog[0;39m: Accept=[application/json, application/*+json]
[30m2019-12-04 11:10:16,484[0;39m [39mDEBUG[0;39m [[34mrestartedMain[0;39m] [33morg.springframework.core.log.CompositeLog[0;39m: Writing [{"grant_type":["client_credentials"],"client_id":["nu5yzeth9tektzf5egxuntp7"],"client_secret":["uP2Xvr6SCKYgXgxxJsv2QkUG"]}] with org.springframework.http.converter.StringHttpMessageConverter
[30m2019-12-04 11:10:17,976[0;39m [39mDEBUG[0;39m [[34mrestartedMain[0;39m] [33morg.springframework.core.log.CompositeLog[0;39m: Response 400 BAD_REQUEST
[30m2019-12-04 11:10:17,981[0;39m [34mINFO [0;39m [[34mrestartedMain[0;39m] [33mcom.ibm.ciscoApiIntegration.service.impl.CiscoAPIServiceImpl[0;39m: e::: 400 Bad Request

如何在中发出上述卷曲请求 RestTemplate 工作?

注: https://cloudsso.example.com/as/token.oauth2

2 回复 | 直到 5 年前

M. Deinum 5 年前

如果你看一下文档 HttpEntity

entity = new HttpEntity<>(reqBodyData, bodyParamMap);

bodyParamMap )作为头(因为第二个参数是用于请求的头)。事实上,你甚至没有使用 HttpHeaders 因为你没有把它们传给 HttpEntity公司 . 我不知道这是怎么回事 reqBodyData toString 的 体参数图 .

你应该做的是

entity = new HttpEntity<>(bodyParamMap, headers);

HttpHeaders文件 弹劾 MultiValueMap 您可以直接在构造函数中使用它们。你还想通过考试吗 体参数图 照原样。

注: 我还建议明确设置 Content-Type 上所以你确定要送什么。

headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

因此,总的结果代码应该是

RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

MultiValueMap<String, String> bodyParamMap = new LinkedMultiValueMap<>();
bodyParamMap.add("grant_type", "client_credentials");
bodyParamMap.add("client_id", "nu5yzeth9tektzf5egxuntp7");
bodyParamMap.add("client_secret", "uP2Xvr6SCKYgXgxxJsv2QkUG");

entity = new HttpEntity<>(bodyParamMap, headers);

restTemplate.postForEntity(url, entity, TokenDTO.class)

注: 你不应该构造一个单一的用途 RestTemplate

Shivam Puri 5 年前

https://stackoverflow.com/a/49127760/11226302

您应该将头中的请求内容类型设置为; headers.setContentType文件(MediaType.APPLICATION\表单\URLENCODED);

希望这有帮助!

vicky 5 年前

我刚刚解决了这个问题:

// wrong
//private HttpEntity<String> entity;

我改成这样:

private MultiValueMap<String, String> parametersMap;

改变了它

// not working
entity = new HttpEntity<>(reqBodyData, bodyParamMap);

为了这个

// working
entity = new HttpEntity<>(bodyParamMap, headers);

现在它工作得很好。

答复:

2019-12-04 11:52:46,503 DEBUG [restartedMain] org.springframework.core.log.CompositeLog: HTTP POST https://cloudsso.example.com/as/token.oauth2
2019-12-04 11:52:46,521 DEBUG [restartedMain] org.springframework.core.log.CompositeLog: Accept=[application/json, application/*+json]
2019-12-04 11:52:46,522 DEBUG [restartedMain] org.springframework.core.log.CompositeLog: Writing [{grant_type=[client_credentials], client_id=[nu5yzeth9tektzf5egxuntp7], client_secret=[uP2Xvr6SCKYgXgxxJsv2QkUG]}] with org.springframework.http.converter.support.AllEncompassingFormHttpMessageConverter
2019-12-04 11:52:47,831 DEBUG [restartedMain] org.springframework.core.log.CompositeLog: Response 200 OK

//headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);