Json Web令牌verify()返回错误的jwt [英] Json Web Token verify() return jwt malformed

问题描述

const jwt = require("jsonwebtoken");
const SECRET = "superSuperSecret";

module.exports = function(req, res, next) {
    const token = req.body.token || req.query.token || req.headers[ "x-access-token" ];
    if (token) {
        return jwt.verify(token, SECRET, function(err, decoded) {
            if (err) {
                return res.json({
                    success: false,
                    message: "Failed to authenticate token.",
                });
            }
            req.user = decoded;
            return next();
        });
    }
    return res.unauthorized();
};

我正在使用Postman来测试我的API.我用x-access-token键和值superSuperSecret设置了标题.我收到错误{"name":"JsonWebTokenError","message":"jwt malformed","level":"error"}.我正在使用 https://github.com/FortechRomania/express-mongo-example-project/blob/master/src/middlewares/validateToken.js

I'm using Postman to test my API. I setup the header with a x-access-token key and the value superSuperSecret. I got the error {"name":"JsonWebTokenError","message":"jwt malformed","level":"error"}. I'm using this https://github.com/FortechRomania/express-mongo-example-project/blob/master/src/middlewares/validateToken.js

推荐答案

您不能将任何值作为令牌传递.您需要jwt.sign()来创建令牌.有关更多信息，请参见 JWT 的文档.

You cannot pass any value as token. You need jwt.sign() to create a token. Look at the documentation of JWT for more information.

还

对于请求标头名称，只需使用Authorization而不是x-access-token.将承载置于令牌之前.

For the request Header name just use Authorization not x-access-token. Place Bearer before the Token.

Authorization: Bearer TOKEN_STRING

JWT的每个部分都是base64url编码的值.您可以通过以下方式获取令牌:

Each part of the JWT is a base64url encoded value. You can get your token as:

var token = req.headers.authorization.split(' ')[1];

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