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Ta的回复 :大神问下 获取assacctoken出错 String authHost = "https://aip.baidubce.com/oauth/2.0/token?"; String getAccessTokenUrl = authHost // 1. grant_type为固定参数 + "grant_type=client_credentials" // 2. 官网获取的 API Key + "&client_id=" + "SUGktFzGCA2wwcbfO5xWDySL" // 3. 官网获取的 Secret Key + "&client_secret=" + "u5g7LPFltdXXyHusC7RbrDy6I6PPV9YS"; 为什么我用官网上的代码基本没改只填充了key 运行会出现java.net.MalformedURLException: Illegal character in URL at sun.net.www.protocol.https.HttpsURLConnectionImpl.checkURL(HttpsURLConnectionImpl.java:86) at sun.net.www.protocol.https.HttpsURLConnectionImpl.<init>(HttpsURLConnectionImpl.java:94) 包是官网的最新的 Postman 提交https://aip.baidubce.com/oauth/2.0/token?grant_type=client_credentials&client_id=SUGktFzGCA2wwcbfO5xWDySL&client_secret=u5g7LPFltdXXyHusC7RbrDy6I6PPV9YS 却成功返回assecc token