我正在尝试使用 Cognito 对 Java 应用程序进行身份验证。我已经为 python 使用了运行良好的保证库。但我现在想在java中做同样的事情。
我的 Python 函数用于使用授权库进行身份验证
def SRPauthentication(organizationAdmin,
password,
pool_id,
client_id,
client):
aws = AWSSRP(username=organizationAdmin,
password=password,
pool_id=pool_id,
client_id=client_id,
client=client)
tokens = aws.authenticate_user()
authorization_token= tokens['AuthenticationResult']['IdToken']
return authorization_token
有了这个,我可以轻松访问一些安全的 API。现在我想对 Java 做同样的事情,但我遇到了问题。
到目前为止,这是我的解决方案是这种方法:
public static void GetCreds()
{
AWSCognitoIdentityProvider identityProvider = AWSCognitoIdentityProviderClientBuilder.defaultClient();
AdminInitiateAuthRequest adminInitiateAuthRequest = new AdminInitiateAuthRequest().
withAuthFlow(AuthFlowType.USER_SRP_AUTH).
withClientId("234234234234").withUserPoolId("eu-central-1_sdfsdfdsf")
.addAuthParametersEntry("USERNAME", "UserK").
addAuthParametersEntry("PASSWORD","#######);
adminInitiateAuthRequest.getAuthFlow();
AdminInitiateAuthResult adminInitiateAuth = identityProvider.adminInitiateAuth(adminInitiateAuthRequest);
System.out.println(adminInitiateAuth.getAuthenticationResult().getIdToken());
}
当我运行它时,我得到一个异常:
Exception in thread "main" `com.amazonaws.services.cognitoidp.model.AWSCognitoIdentityProviderException: User: arn:aws:iam::XXXXXXXXXXXXXXXXX:user/khan is not authorized to perform: cognito-idp:AdminInitiateAuth on resource: arn:aws:cognito-idp:eu-central-1:XXXXXXXX:userpool/eu-central-1_XXXXXXX with an explicit deny (Service: AWSCognitoIdentityProvider; Status Code: 400; Error Code: AccessDeniedException; Request ID: 21be0b8e-adec-11e8-ad45-234234234)`
它说我无权执行这种指令。所以我想我做的事情通常是错误的。因为它与我的 python 代码和 Java 一起工作,所以它可以从凭据中识别我的用户名。Cognito 调用实际上应该独立于我的 aws 凭据/用户帐户,对吗?
如何使用 Java 向 Cognito 进行身份验证以获取令牌以访问安全的 aws 服务?
编辑:
AWSCognitoIdentityProvider identityProvider = AWSCognitoIdentityProviderClientBuilder.standard()
.build();
InitiateAuthRequest adminInitiateAuthRequest = new InitiateAuthRequest()
.withAuthFlow(AuthFlowType.USER_SRP_AUTH)
.withClientId("XXXXXXXXXXXXXXXXX")
.addAuthParametersEntry("USERNAME", "user").
addAuthParametersEntry("PASSWORD","za$Lwn")
.addAuthParametersEntry("SRP_A",new AuthenticationHelper("eu-central-1XXXXXXXXX").getA().toString(16));
adminInitiateAuthRequest.getAuthFlow();
InitiateAuthResult adminInitiateAuth = identityProvider.initiateAuth(adminInitiateAuthRequest);
System.out.println(adminInitiateAuth);
我将 AdminInitateAuthRequest 更改为 InitateAuthRequest。之后我有错误丢失 SRP_A 参数,我在这里用一个类似的问题以某种方式修复了 现在我收到了这个:
{ChallengeName: PASSWORD_VERIFIER,ChallengeParameters: {SALT=877734234324234ed68300f39bc5b, SECRET_BLOCK=lrkwejrlewrjlewkjrewlrkjwerlewkjrewlrkjewrlkewjrlewkrjZ+Q==, USER_ID_FOR_SRP=user, USERNAME=user, SRP_B=43ecc1lwkerjwelrkjewlrjewrlkewjrpoipweoriwe9r873jr34h9r834hr3455f7d079d71e5012f1623ed54dd10b832792dafa3438cca3f59c0f462cbaee255d5b7c2werwerwerkjweorkjwerwerewrf5020e4f8b5452f3b89caef4a797456743602b80b5259261f90e52374adc06b456521a9026cce9c1cbe8b9ffd6040e8c1589d35546861422110ac7e38c1c93389b802a03e3e2e4a50e75d088275195f836f66e25f1a431dd56bb2},}
我已经用所有键缩短了结果,但下一步该怎么做?