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java - OAuth 401 未经授权

转载 作者:行者123 更新时间:2023-11-30 06:24:38 26 4
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仍在学习,但我从下面的代码中收到未经授权的错误 401。我知道 OAuth header 的工作方式与 postman 中的工作方式相同,因此我假设 POST 请求/Auth header 存在问题?有什么想法吗?

//set timestamp
Long timestamp = System.currentTimeMillis()/1000;
//set nonce ***** call from main system*************************************************************
String aString = randomAlphaNumeric(11);
// other stuff
RestTemplate restTemplate = new RestTemplate();
restTemplate.getMessageConverters().add(new StringHttpMessageConverter());
HttpHeaders headers = new HttpHeaders();
String url = "aURL";
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
// String auth = Base64.getEncoder().encodeToString(credentials.getBytes());
List<NameValuePair> oauthHeaders = new ArrayList<>(9);
oauthHeaders.add(new BasicNameValuePair("oauth_consumer_key", "aKey"));
oauthHeaders.add(new BasicNameValuePair("oauth_nonce", aString));
oauthHeaders.add(new BasicNameValuePair("oauth_timestamp", String.valueOf(timestamp)));
oauthHeaders.add(new BasicNameValuePair("oauth_signature_method", "HMAC-SHA1"));
oauthHeaders.add(new BasicNameValuePair("oauth_version", "1.0"));
//generate signature
//encode
String encodedURL = encode(oauthHeaders.toString());
System.out.println("encoded URL:" +encodedURL);
//form base string
String baseString = "POST&"+encode(url).toString()+encodedURL;
System.out.println("Base String: "+baseString);
//form signature
byte[] byteHMAC = null;
try {

Mac mac = Mac.getInstance("HmacSHA1");
SecretKeySpec spec;
if (null == secretKey) {
String signingKey = encode(secretKey) + '&';
spec = new SecretKeySpec(signingKey.getBytes(), "HmacSHA1");
} else {
String signingKey = encode(secretKey) + '&' + encode(secretKey);
spec = new SecretKeySpec(signingKey.getBytes(), "HmacSHA1");
}
mac.init(spec);
byteHMAC = mac.doFinal(baseString.getBytes());
} catch (Exception e) {
e.printStackTrace();
}
String signature = new BASE64Encoder().encode(byteHMAC);
System.out.println("oauth signature: "+signature);


//set signature to params
oauthHeaders.add(new BasicNameValuePair("oauth_signature", signature));
String test = "OAuth "+oauthHeaders.toString();
headers.set("Authorization", test);
MultiValueMap<String, String> map = new LinkedMultiValueMap<String, String>();
map.add("Name",name.toString());
map.add("Region",region.toString());


HttpEntity<MultiValueMap<String, String>> requestEntity= new HttpEntity<MultiValueMap<String, String>>(headers, map);
System.out.println(requestEntity);
ResponseEntity<String> response= restTemplate.exchange(url ,HttpMethod.POST, requestEntity, String.class);
System.out.println(response.toString());
HttpStatus status = response.getStatusCode();
status.toString();
if(status.equals("200")){
Notification.show("Employer" + name +" added successfully");
}
else{
Notification.show("Unsuccessful, error: "+status);
}


}

出于显而易见的原因删除了 URL 和消费者 key /签名。

以下系统输出打印也可能有帮助:

编码参数: %5Boauth_consumer_key%3aKey%2C%20oauth_nonce%3DWZU8H1B5JA6%2C%20oauth_timestamp%3D1511621759%2C%20oauth_signature_method%3DHMAC-SHA1%2C%20oauth_version%3D1.0%5D

基本字符串:POST&https%3A%2F%2Fapi.test.payrun.io%2FEmployer%5Boauth_consumer_key%3aKey%2C%20oauth_nonce%3DWZU8H1B5JA6%2C%20oauth_timestamp%3D1511621759%2C%20oauth_signature_method%3DHMAC-SHA1 %2C%20oauth_version% 3D1.0%5D

oauth 签名:DlRJGSzgRIItzz+LzMbgnIfbOqU=

最佳答案

oauth_signature 的值错误。您使用 asignature 作为 oauth_signature 的值,但您必须计算请求的正确值并将其设置为 oauth_signature。如果oauth_signature的值错误,服务器将拒绝您的请求。请参阅 3.4. Signature 中的"RFC 5849 " (OAuth 1.0 协议(protocol))了解详细信息。

关于java - OAuth 401 未经授权,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47460242/

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