我正在使用Congnito用户池执行API网关授权,效果很好。现在,我试图将Instagram添加为登录提供程序之一,为此我在联合身份中使用以下代码创建了自定义身份验证提供程序:
var cognitoidentity = new AWS.CognitoIdentity({ apiVersion: '2014-06-30' });
var params = {
IdentityPoolId: 'us-east-1:7d99e750-.....',
Logins: {
'login.instagram': 'Access-Token-Returned-By-Instagram',
},
TokenDuration: 60
};
cognitoidentity.getOpenIdTokenForDeveloperIdentity(params, function (err, data) {
if (err) {
console.log(err, err.stack);
} else {
console.log(data);
var idParams = {
IdentityId: data['IdentityId'],
Logins: {
'cognito-identity.amazonaws.com': data['Token']
}
};
cognitoidentity.getCredentialsForIdentity(idParams, function (err2, data2) {
if (err2) console.log(err2, err2.stack); // an error occurred
else console.log(data2); // successful response
});
}
});
我能够获得 accessToken 和 sessionToken ,但是,我仍然找不到找到 idToken 和的方法。 API网关用来授权传入请求的accessToken 。
我尝试研究SDK和AWS论坛,但仍然找不到使用自定义联合身份提供者来授权使用认知用户池的API Gateway的方法。
答案 0 :(得分:0)
我要对此一想。...
您有一个认知用户吗?
import { CognitoUser, CognitoUserPool, AuthenticationDetails } from "amazon-cognito-identity-js";
let cognitoUser;
const userPool = new CognitoUserPool({
UserPoolId: config.USER_POOL.pool_Id, //your userpool id
ClientId: config.appClientId, //your appClient
});
const userData = {
Username: 'user name',
Pool: userPool
};
cognitoUser = new CognitoUser(userData);
/* Should now have a cognitoUser */
const authenticationData = {
Username : payload.userName,
Password : payload.password,
};
const authenticationDetails = new AuthenticationDetails(authenticationData);
cognitoUser.authenticateUser(authenticationDetails, {
onSuccess: function (result) {
const accessToken = result.getAccessToken().getJwtToken();
/* Use the idToken for Logins Map when Federating User Pools with identity pools or when passing through an Authorization Header to an API Gateway Authorizer*/
const idToken = result.idToken.jwtToken;
/*
Do something with
idToken
accessToken
I would write them to a file or encrypt them and store them on the localStorage OR encrypt and save in REDUX store.
*/
},
onFailure: function(err) {
//handle error
},
});