[英]Java security access control exception
我正在嘗試執行此程序, http://java.sun.com/developer/technicalArticles/ThirdParty/WebCrawler/WebCrawler.java在我引用此頁面http://www.velocityreviews.com之后,該程序編譯無誤/forums/t146972-web-crawler.html
但是在使用命令“ appletviewer WebCrawler.html”執行時,出現此異常。
Exception in thread "Thread-4"
java.security.AccessControlException:access denied(java.net.SocketPermission java.sun.com:80 connect,resolve)
at java.security.AccessControlContext.checkPermission(AccessControlContext.java:323)
at java.security.AccessController.checkPermission(AccessController.java:546)
at java.lang.SecurityManager.checkPermission(SecurityManager.java:532)
at java.lang.SecurityManager.checkConnect(SecurityManager.java:1034)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:527)
at sun.net.www.http.HttpClient.<init>(HttpClient.java:233)
at sun.net.www.http.HttpClient.New(HttpClient.java:306)
at sun.net.www.http.HttpClient.New(HttpClient.java:323)
at sun.net.www.protocol.http.HttpURLConnection.getNewHttpClient(HttpURLConnection.java:860)
at sun.net.www.protocol.http.HttpURLConnection.plainConnect(HttpURLConnection.java:801)
at sun.net.www.protocol.http.HttpURLConnection.connect(HttpURLConnection.java:726)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1049)
at java.net.URL.openStream(URL.java:1010)
at WebCrawler.robotSafe(WebCrawler.java:139)
at WebCrawler.run(WebCrawler.java:235)
at java.lang.Thread.run(Thread.java:619)
我如何使其工作。
默認情況下,小程序只能連接到它所在的同一服務器+端口。 您可以執行以下操作之一:
您也可以看看如何為applet配置跨域,請參閱:
http://weblogs.java.net/blog/2008/05/28/java-doodle-crossdomainxml-support
自簽名小程序非常簡單(您可以免費簽名)。 如果不對小程序進行簽名,則只會獲得最低限度的安全性訪問權限,並且不會授予套接字訪問權限。
簡單的3步指南在這里:
http://www.narendranaidu.com/2007/11/3-easy-steps-to-self-sign-applet-jar.html
或更深入的探討: http : //java.sun.com/developer/onlineTraining/Programming/JDCBook/signed.html
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