用新浪sdk开发java时 获取access token时出现如下问题
Exceptioninthread"main"java.lang.NumberFormatException:Forinputstring:"2345582042"atj...
Exception in thread "main" java.lang.NumberFormatException: For input string: "2345582042"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:459)
at java.lang.Integer.parseInt(Integer.java:497)
at weibo4j.http.AccessToken.<init>(AccessToken.java:50)
at weibo4j.http.AccessToken.<init>(AccessToken.java:42)
at weibo4j.http.HttpClient.getOAuthAccessToken(HttpClient.java:221)
at weibo4j.http.RequestToken.getAccessToken(RequestToken.java:60)
at weibo4j.examples.OAuthUpdate.main(OAuthUpdate.java:77) 展开
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:459)
at java.lang.Integer.parseInt(Integer.java:497)
at weibo4j.http.AccessToken.<init>(AccessToken.java:50)
at weibo4j.http.AccessToken.<init>(AccessToken.java:42)
at weibo4j.http.HttpClient.getOAuthAccessToken(HttpClient.java:221)
at weibo4j.http.RequestToken.getAccessToken(RequestToken.java:60)
at weibo4j.examples.OAuthUpdate.main(OAuthUpdate.java:77) 展开
3个回答
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询