Question

我在亚马逊中使用Cognito来验证我的移动用户，一旦他们完成登录，Cognito会提供一组令牌，我在后端使用id令牌。我已经按照Web API中使用ID令牌和访问令牌部分的步骤进行操作 https://docs.aws.amazon.com/cognito/latest/developerguide/amazon-cognito-user-pools-using-tokens-with-identity-providers.html我被困在第6步。

据我所见，我从String中获取了亚马逊的模数和指数，我必须用这些来构建一个PublicKey，以验证JWT签名。

我不知道如何使用String中的这两个参数构建PublicKey。

Answer 1

我终于找到了一个解决方案，在aws论坛https://forums.aws.amazon.com/message.jspa?messageID=728870中有一个例子，但代码在Kotlin中。我只是将它移植到java并进行一些测试我终于验证了我的JWT签名：

byte[] decodedModulus = Base64.getUrlDecoder().decode(yourModulus);

byte[] decodedExponent = Base64.getUrlDecoder().decode(yourExponent);

BigInteger modulus = new BigInteger(1, decodedModulus);
BigInteger exponent = new BigInteger(1, decodedExponent);

RSAPublicKeySpec publicKeySpec = new RSAPublicKeySpec(modulus, exponent);
KeyFactory keyFactory;

keyFactory = KeyFactory.getInstance("RSA");
PublicKey publicKey = keyFactory.generatePublic(publicKeySpec);

JWSVerifier verifier = new RSASSAVerifier((RSAPublicKey) publicKey);
Boolean verify = parsedToken.verify(verifier);}

希望对遇到同样麻烦的人有所帮助。

Answer 2

我编程的方式是这样，

    // Parse the Cognito Keys and get the key by kid
    // Key is just a class that is used for parsing JSON to POJO
    Key key = this.keyService.getKeyByKeyId(JWT.decode(token).getKeyId());

    // Use Key's N and E
    BigInteger modulus = new BigInteger(1, Base64.decodeBase64(key.getN()));
    BigInteger exponent = new BigInteger(1, Base64.decodeBase64(key.getE()));

    // Create a publick key
    PublicKey publicKey = null;
    try {
        publicKey = KeyFactory.getInstance("RSA").generatePublic(new RSAPublicKeySpec(modulus, exponent));
    } catch (InvalidKeySpecException e) {
        // Throw error
    } catch (NoSuchAlgorithmException e) {
        // Throw error
    }

    // get an algorithm instance
    Algorithm algorithm = Algorithm.RSA256((RSAPublicKey) publicKey, null);

    // I verify ISS field of the token to make sure it's from the Cognito source
    String iss = String.format("https://cognito-idp.%s.amazonaws.com/%s", REGION, POOL_ID);

    JWTVerifier verifier = JWT.require(algorithm)
            .withIssuer(iss)
            .withClaim("token_use", "id") // make sure you're verifying id token
            .build();

    // Verify the token
    DecodedJWT jwt = verifier.verify(token);

    // Parse various fields
    String username = jwt.getClaim("sub").asString();
    String email = jwt.getClaim("email").asString();
    String phone = jwt.getClaim("phone_number").asString();
    String[] groups = jwt.getClaim("cognito:groups").asArray(String.class);

我正在使用此存储库来验证和解析令牌，

    <dependency>
        <groupId>com.auth0</groupId>
        <artifactId>java-jwt</artifactId>
        <version>3.4.1</version>
    </dependency>

确保要导入以下内容，

    import com.auth0.jwt.JWT;
    import com.auth0.jwt.JWTVerifier;
    import com.auth0.jwt.algorithms.Algorithm;
    import com.auth0.jwt.exceptions.JWTDecodeException;
    import com.auth0.jwt.interfaces.DecodedJWT;

    import java.math.BigInteger;
    import java.security.KeyFactory;
    import java.security.NoSuchAlgorithmException;
    import java.security.PublicKey;
    import java.security.interfaces.RSAPublicKey;
    import java.security.spec.InvalidKeySpecException;
    import java.security.spec.RSAPublicKeySpec;
    import org.apache.commons.codec.binary.Base64;

验证Java上的AWS id Token

2 个答案: