Java ( applet )

Question

我有以下HttpGet()函数：

    HttpGet request = new HttpGet(url);

    request.setHeader("User-Agent", userAgent);
    request.setHeader("Accept", "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8");
    request.setHeader("Accept-Language", "en-US,en;q=0.5");
    if (!cookies.equals(""))
        request.setHeader("Cookie", this.cookies);

    for (Map.Entry<String, String> header : tempHeaders.entrySet())
    {
        request.setHeader(header.getKey(), header.getValue());
    }
    tempHeaders.clear();

    HttpResponse response = client.execute(request);

此函数是包装Apache的HttpClient的Browser()类的一部分。当我在applet查看器中运行Applet时，一切都正常。但是，当我将applet嵌入到页面中时，会得到以下异常：

java.security.AccessControlException: access denied ("java.net.SocketPermission" "127.0.0.1:80" "connect,resolve")
    at java.security.AccessControlContext.checkPermission(Unknown Source)
    at java.security.AccessController.checkPermission(Unknown Source)
    at java.lang.SecurityManager.checkPermission(Unknown Source)
    at java.lang.SecurityManager.checkConnect(Unknown Source)
    at sun.plugin2.applet.SecurityManagerHelper.checkConnectHelper(Unknown Source)
    at sun.plugin2.applet.AWTAppletSecurityManager.checkConnect(Unknown Source)
    at java.net.Socket.connect(Unknown Source)
    at org.apache.http.conn.scheme.PlainSocketFactory.connectSocket(PlainSocketFactory.java:117)
    at org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:178)
    at org.apache.http.impl.conn.ManagedClientConnectionImpl.open(ManagedClientConnectionImpl.java:304)
    at org.apache.http.impl.client.DefaultRequestDirector.tryConnect(DefaultRequestDirector.java:610)
    at org.apache.http.impl.client.DefaultRequestDirector.execute(DefaultRequestDirector.java:445)
    at org.apache.http.impl.client.AbstractHttpClient.doExecute(AbstractHttpClient.java:863)
    at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:82)
    at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:106)
    **at vidinstant.HttpBrowser.Get(HttpBrowser.java:60)**
    at vidinstant.ServerThread$1.run(ServerThread.java:201)
    at vidinstant.ServerThread$1.run(ServerThread.java:197)
    at java.security.AccessController.doPrivileged(Native Method)
    at vidinstant.ServerThread.GetLink(ServerThread.java:196)
    at vidinstant.ServerThread.run(ServerThread.java:95)

您可以看到，异常的“源”位于浏览器类的第60行，更准确地说，这是代码的上述摘录中的一行：

  HttpResponse response = client.execute(request);

我已经对applet .jar进行了自签名，并且清单文件中有一行“权限:所有权限”。

为什么我会得到这个异常，为什么访问被拒绝？在浏览器中运行applet的用户单击“允许”和“不阻止”，但仍然无法工作。Apache的库.jars也需要签名吗？他们需要在他们的名单中有“所有的惩罚”吗？如何获得运行这些函数的权限，而不需要用户篡改他们的Java策略文件？

Answer 1

Apache的库.jars也需要签名吗？

是。没有90%的安全许可。部署的代码要么被认为是安全的，而..or则不安全。