Question

我有以下XML-RPC实现工作，我从apache website复制并稍加修改。

public class DemoServer { 
  public static void main (String [] args) {
    try {
        WebServer webServer = new WebServer(8080);
        XmlRpcServer xmlRpcServer = webServer.getXmlRpcServer();

        PropertyHandlerMapping phm = new PropertyHandlerMapping();
        phm.addHandler("sample", RequestHandler.class);
        xmlRpcServer.setHandlerMapping(phm);

        XmlRpcServerConfigImpl serverConfig =
                (XmlRpcServerConfigImpl) xmlRpcServer.getConfig();
        serverConfig.setEnabledForExtensions(true);
        serverConfig.setContentLengthOptional(false);

        webServer.start();
    } catch (Exception e) {
        e.printStackTrace();
    }
  }
}

与客户：

public class DemoClient {
  public static void main (String[] args) {
    try {
        XmlRpcClientConfigImpl config = new XmlRpcClientConfigImpl();
        config.setServerURL(new URL("http://127.0.0.1:8080/xmlrpc"));
        config.setEnabledForExtensions(true);  
        config.setConnectionTimeout(60 * 1000);
        config.setReplyTimeout(60 * 1000);

        XmlRpcClient client = new XmlRpcClient();

        // set configuration
        client.setConfig(config);

        // make the a regular call
        Object[] params = new Object[] { new Integer(2), new Integer(3) };

                    //!CRITICAL LINE!
        Integer result = (Integer) client.execute("sample.sum", params);

        System.out.println("2 + 3 = " + result);
    } catch (Exception e) {
        e.printStackTrace();
    }
  }
}

我先运行DemoServer，然后运行DemoClient，然后输出“2 + 3 = 5”。但是，如果我改变

Integer result = (Integer) client.execute("sample.sum", params);

到

client.executeAsync("sample.sum", params, new ClientCallback());

然后我得到以下内容：

In error
java.lang.ExceptionInInitializerError
    at java.lang.Runtime.addShutdownHook(Runtime.java:192)
at java.util.logging.LogManager.<init>(LogManager.java:237)
at java.util.logging.LogManager$1.run(LogManager.java:177)
at java.security.AccessController.doPrivileged(Native Method)
at java.util.logging.LogManager.<clinit>(LogManager.java:158)
at java.util.logging.Logger.getLogger(Logger.java:273)
at sun.net.www.protocol.http.HttpURLConnection.<clinit>(HttpURLConnection.java:62)
at sun.net.www.protocol.http.Handler.openConnection(Handler.java:44)
at sun.net.www.protocol.http.Handler.openConnection(Handler.java:39)
at java.net.URL.openConnection(URL.java:945)
at org.apache.xmlrpc.client.XmlRpcSun15HttpTransport.newURLConnection(XmlRpcSun15HttpTransport.java:62)
at org.apache.xmlrpc.client.XmlRpcSunHttpTransport.sendRequest(XmlRpcSunHttpTransport.java:62)
at org.apache.xmlrpc.client.XmlRpcClientWorker$1.run(XmlRpcClientWorker.java:80)
at java.lang.Thread.run(Thread.java:680)
Caused by: java.lang.IllegalStateException: Shutdown in progress
at java.lang.Shutdown.add(Shutdown.java:62)
at java.lang.ApplicationShutdownHooks.<clinit>(ApplicationShutdownHooks.java:21)
... 14 more

我的ClientCallback类：

public class ClientCallback implements AsyncCallback {

@Override
public void handleError(XmlRpcRequest request, Throwable t) {
    System.out.println("In error");
    t.printStackTrace();

}

@Override
public void handleResult(XmlRpcRequest request, Object result) {
    System.out.println("In result");
    System.out.println(request.getMethodName() + ": " + result);
}
}

这里出了什么问题？我正在使用Apache XML-RPC版本3.1.2，遗憾的是我发现的示例代码是2.x版，不再适用。我也从类的开头省略了import语句（肯定没有语法错误）。任何帮助将不胜感激。

Answer 1

您的主程序正在运行，因为executeAsync会立即返回，而不会等待请求发送或回复。

你想通过使用executeAsync来实现什么？

XML-RPC异常从execute切换到executeAsync

1 个答案: