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Twitter OAuth 2.0错误提示"传递的令牌值无效"
P粉865900994
P粉865900994 2023-12-16 14:06:37
[PHP讨论组]

编辑:查看解决方案。问题在于 Twitter 库的实例化方式。

我正在尝试使用 OAuth2 获取 Twitter 授权。我可以让它在一个简单的测试脚本中工作,但是当我尝试在我的 WordPress 插件中使用它时,它不起作用。

我正在使用 PHP League 的 OAuth2 客户端以及我编写的库来将其连接到 Twitter;测试脚本位于自述文件中。

测试脚本将OAuth2状态存储在$_SESSION中;实际的应用程序将其存储在 WordPress 瞬态中。我已经通过管道确认了数据的完整性:

生成 auth URL 后来自 Twitter 库的数据:

Array
(
    [url] => https://twitter.com/i/oauth2/authorize?redirect_uri=https%3A%2F%2Fsmol.blog%2Fwp-json%2Fsmolblog%2Fv2%2Fconnect%2Fcallback%2Ftwitter&code_challenge=EV7BCVYmkvCnIlVLH6cVzrvjNloQlleAkkYwLLgg41w&code_challenge_method=S256&state=fd5824ef415aa325f1f68d3504bb16b3&scope=tweet.read%20users.read%20offline.access&response_type=code&approval_prompt=auto&client_id=MjVXMnRGVUN5Ym5lcVllcTVKZkk6MTpjaQ
    [state] => fd5824ef415aa325f1f68d3504bb16b3
    [verifier] => u7Zbf1gVEFZLyTgr_2Hk~i5P2pt8VgicyhZgdeO0pAyIZqhSoYqglHaIxsNRjHz0AHpwhlU1~Q
)

回调期间从 WordPress 瞬态提取的数据(存储方式略有不同):

Array
(
    [id] => fd5824ef415aa325f1f68d3504bb16b3
    [userId] => 1
    [info] => Array
        (
                [verifier] => u7Zbf1gVEFZLyTgr_2Hk~i5P2pt8VgicyhZgdeO0pAyIZqhSoYqglHaIxsNRjHz0AHpwhlU1~Q
        )

)

向 Twitter 令牌端点请求对象:

GuzzleHttp\Psr7\Request Object
(
        [method:GuzzleHttp\Psr7\Request:private] => POST
    [requestTarget:GuzzleHttp\Psr7\Request:private] => 
    [uri:GuzzleHttp\Psr7\Request:private] => GuzzleHttp\Psr7\Uri Object
        (
                [scheme:GuzzleHttp\Psr7\Uri:private] => https
            [userInfo:GuzzleHttp\Psr7\Uri:private] => 
            [host:GuzzleHttp\Psr7\Uri:private] => api.twitter.com
            [port:GuzzleHttp\Psr7\Uri:private] => 
            [path:GuzzleHttp\Psr7\Uri:private] => /2/oauth2/token
            [query:GuzzleHttp\Psr7\Uri:private] => 
            [fragment:GuzzleHttp\Psr7\Uri:private] => 
            [composedComponents:GuzzleHttp\Psr7\Uri:private] => 
        )

    [headers:GuzzleHttp\Psr7\Request:private] => Array
        (
                [Host] => Array
                (
                        [0] => api.twitter.com
                )

            [content-type] => Array
                (
                        [0] => application/x-www-form-urlencoded
                )

            [Authorization] => Array
                (
                        [0] => Basic [base64-encoded app id and secret redacted]
                )

        )

    [headerNames:GuzzleHttp\Psr7\Request:private] => Array
        (
                [content-type] => content-type
            [host] => Host
            [authorization] => Authorization
        )

    [protocol:GuzzleHttp\Psr7\Request:private] => 1.1
    [stream:GuzzleHttp\Psr7\Request:private] => GuzzleHttp\Psr7\Stream Object
        (
                [stream:GuzzleHttp\Psr7\Stream:private] => Resource id #101
            [size:GuzzleHttp\Psr7\Stream:private] => 
            [seekable:GuzzleHttp\Psr7\Stream:private] => 1
            [readable:GuzzleHttp\Psr7\Stream:private] => 1
            [writable:GuzzleHttp\Psr7\Stream:private] => 1
            [uri:GuzzleHttp\Psr7\Stream:private] => php://temp
                [customMetadata:GuzzleHttp\Psr7\Stream:private] => Array
                    (

                )

            )

    )

上述请求的正文:

client_id=MjVXMnRGVUN5Ym5lcVllcTVKZkk6MTpjaQ&client_secret=[redacted]&grant_type=authorization_code&code=aTVUMDkybzdsVmExOEQ5MjdrVjVOQVZ3YTVDbUdmTXRDMktZSzBaSGFqVk5LOjE2NjUzNjc1MjIyNjg6MToxOmFjOjE&code_verifier=u7Zbf1gVEFZLyTgr_2Hk~i5P2pt8VgicyhZgdeO0pAyIZqhSoYqglHaIxsNRjHz0AHpwhlU1~Q

错误:

PHP Fatal error:  Uncaught League\OAuth2\Client\Provider\Exception\IdentityProviderException: Value passed for the token was invalid. in /var/www/html/wp-content/plugins/smolblog-wp/vendor/smolblog/oauth2-twitter/src/Twitter.php:169

我知道我在这里错过了一些愚蠢的东西。但我一生都无法弄清楚什么。实际应用程序中的代码比测试脚本中的代码多得多,但我已经在堆栈中的多个点验证了数据,包括在将其发送到 Twitter 之前。我还需要测试什么,或者有什么我忘记了?

P粉865900994
P粉865900994

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