Mik*_*key 4 python decode jwt amazon-cognito
我想在 Django 应用程序中使用 Cognito 身份提供商 (Facebook) 对用户进行身份验证。当用户登录时,他会被重定向到带有 和 的access_token主页id_token。这些是 JWT 令牌。我需要对它们进行解码以获取有关用户的信息。如何实现?
我尝试使用jwt图书馆。许多资源说我需要将 PUBLIC_KEY 转换为 PEM 格式来解码这些令牌。我访问了:
https://cognito-idp.{aws-region}.amazonaws.com/{user-pool-id}/.well-known/jwks.json
并得到了字典:
{
"keys": [
{
"alg": "RS256",
"e": "AQAB",
"kid": "cKgEILaVv5nXrJNBtfdaEVfA9Sn+GNdMpXbO58Li+eLyiA=",
"kty": "RSA",
"n": "kpE2A8GfBoDiLcnHme9WLSIMezPAXg8ibZ8hIa4GvmJAT1LmR9CDN0Yt-JlKE_gH73HbldpgqCbflygbcZtdn1RMdfafdafaQq3sOY0RJqC8-jdZTWb0Cbw2E-fmmOtgHqz11ZlbqaZYpcRYvyNwuWQs05kikzaUudwp05TTd4N0VznbMMm9peu_Ghw7nqlgkvU8WWO6tD_LzF8VDTroKdGAnrh3kwRVY8le5JwzoGMDVbNtOQFwla8yyxNfRRFQVkcnOxb14BhrO3N8ZsI-E9eB9ZbUNt27DxAMjxsNzERSUx3gm7zq-d2SYyZ31P_FNxOTS0RlKPN69Jlscack7ghM_nZdnw",
"use": "sig"
},
{
"alg": "RS256",
"e": "AQAB",
"kid": "wMYp8SsLhXBtE2wbYK1hQfdafadfadeiYkDYXctg1GkdmTTFs+I=",
"kty": "RSA",
"n": "i_W1pexAB1FhY3oJ0jEGeks1b8jCFZOWU7PAvzi3kdeP7JR2IP91W8FV9ou6gT1dn6F8ZbDvJEF9PNJkmQxwvHkoeYzgaOhNFXBU5Jfv20rQfdafa56PufchgtGndaI22TnnPDg_L5UElaljJvmzTcDTk0xHqJRxhw2LAqembfij5TV_8mSHpqIbWvppOTqT7s3zI7DdfDLDOX7CHFRht3uesX5drVo-S9IsJCJ2l3rCChTCy8e32YY68iHH40t5vVPkuqr1ffMfKYkyXC3urliCawRKnrk62ngAskjRiAihmIl8E79_Ddfn5O5RYqLY71Q4T6Epep4C1ygjlBrToFd8Haw",
"use": "sig"
}
]
}
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我不知道从这里能抓到什么。
小智 6
使用该PyJWT库,您可以通过以下方式解码 JWT 令牌:
import jwt\n\nencoded = token # replace this with your encoded token\njwt.decode(encoded, algorithms=["RS256"], options={"verify_signature": False})\nRun Code Online (Sandbox Code Playgroud)\n该options配置将告诉 PyJWT 库忽略验证过程的公钥方面,并无论如何解码 Base64 密钥。
注意 \xe2\x80\x93 根据 AWS Cognito 标准,该算法被硬编码为 RS256,但如果您愿意,您可以选择忽略该参数(PyJWT 会弄清楚)。
\n要解码和验证JWT,您可以使用borisrozumnuk/cognitojwt库:
\nimport jwt\n\nencoded = token # replace this with your encoded token\njwt.decode(encoded, algorithms=["RS256"], options={"verify_signature": False})\nRun Code Online (Sandbox Code Playgroud)\n如果您想避免使用第 3 方库,请手动安装 AWSLabs aws-support-tools \'decode-verify-jwt\'模块:
\nfrom cognitojwt import jwt_sync\n\njwt_sync.decode(token)\nRun Code Online (Sandbox Code Playgroud)\n
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