Google OAuth2中签名中的JWT"invalid_grant"

Question

我正在编写一些代码,试图在OAuth2中获取Google使用的令牌.这是针对服务帐户的,因此说明如下:

https://developers.google.com/identity/protocols/OAuth2ServiceAccount

当我将JWT发布到Google时,我一直收到此错误:

{"error":"invalid_grant","error_description":"无效的JWT签名." }

这是代码:

try{        
    var nowInSeconds : Number = (Date.now() / 1000);
    nowInSeconds = Math.round(nowInSeconds);
    var fiftyNineMinutesFromNowInSeconds : Number = nowInSeconds + (59 * 60);


    var claimSet : Object = {};
    claimSet.iss   = "{{RemovedForPrivacy}}";        
    claimSet.scope = "https://www.googleapis.com/auth/plus.business.manage";
    claimSet.aud   = "https://www.googleapis.com/oauth2/v4/token";
    claimSet.iat   = nowInSeconds; 
    claimSet.exp   = fiftyNineMinutesFromNowInSeconds;

    var header : Object = {};
    header.alg = "RS256";
    header.typ = "JWT";

    /* Stringify These */
    var claimSetString = JSON.stringify(claimSet);
    var headerString = JSON.stringify(header);

    /* Base64 Encode These */
    var claimSetBaseSixtyFour = StringUtils.encodeBase64(claimSetString);
    var headerBaseSixtyFour = StringUtils.encodeBase64(headerString);

    var privateKey = "{{RemovedForPrivacy}}";

    /* Create the signature */
    var signature : Signature = Signature();
    signature =  signature.sign(headerBaseSixtyFour + "." + claimSetBaseSixtyFour, privateKey , "SHA256withRSA");

    /* Concatenate the whole JWT */
    var JWT = headerBaseSixtyFour + "." + claimSetBaseSixtyFour + "." + signature;

    /* Set Grant Type */
    var grantType = "urn:ietf:params:oauth:grant-type:jwt-bearer"

    /* Create and encode the body of the token post request */
    var assertions : String = "grant_type=" + dw.crypto.Encoding.toURI(grantType) + "&assertion=" + dw.crypto.Encoding.toURI(JWT);

    /* Connect to Google And Ask for Token */
    /* TODO Upload Certs? */
    var httpClient : HTTPClient = new HTTPClient();
    httpClient.setRequestHeader("content-type", "application/x-www-form-urlencoded; charset=utf-8");
    httpClient.timeout = 30000;
    httpClient.open('POST', "https://www.googleapis.com/oauth2/v4/token");
    httpClient.send(assertions);

    if (httpClient.statusCode == 200) {
       //nothing
    } else {
       pdict.errorMessage = httpClient.errorText;
    }  

}
catch(e){
    Logger.error("The error with the OAuth Token Generator is --> " + e);
}

有谁知道为什么JWT会失败？

非常感谢!布拉德

Answer 1

该问题可能与您的StringUtils.encodeBase64()方法可能执行标准base64编码这一事实有关。

但是，根据JWT规范，不是需要使用标准的base64编码，而是使用URL和文件名安全的Base64编码，并=省略了填充字符。

如果没有方便的base64URL编码实用程序方法，则可以通过

• 全部替换+为-;
• 全部替换/为_;
• 全部删除 =

在您的base64编码的字符串中。

另外，您的签名是否也以base64编码？必须遵循与上述相同的规则。