小编Suh*_*pta的帖子

我想让我的电脑和服务器兼容.这是我的代码

import java.net.*;
class tester {
static int pos=0; 
 static byte buffer[]=new byte[100];
   static void Client() throws Exception {
    InetAddress address=InetAddress.getLocalHost();
  DatagramSocket ds=new DatagramSocket(3000,address);
   while(true) {
    int c=System.in.read();
    buffer[pos++]=(byte)c;
    if((char)c=='\n')
      break;
   }
   ds.send(new DatagramPacket(buffer,pos,address,3000));
  Server();
}                   

static void Server() throws Exception {
 InetAddress address=InetAddress.getLocalHost();
 DatagramSocket ds=new DatagramSocket(3001,address);
 DatagramPacket dp=new DatagramPacket(buffer,buffer.length);
 ds.receive(dp);
 System.out.print(new String(dp.getData(),0,dp.getLength()));
}
  public static void main(String args[])throws Exception {

 if(args.length==1) {
  Client();
   } 
 }

}

在这里,我试图使我的计算机既服务器又客户端. 我在cmd上运行这个程序, java tester hello但程序一直在等待.我应该怎么做以接收键入的消息.

*如果代码中有任何修改,请提出建议.请注意,目的是使我的电脑和服务器兼容.

这是片段:

protected void paintComponent(final Graphics g) {

 Runnable r=new Runnable() {

 @Override

  public void run() {
   while(true) {
     super.paintComponent(g);  // <----- line of error
     g.setColor(Color.red);
     g.drawOval(x,y,width,height);
     g.fillOval(x,y,width,height);
     x++;
     y++;
     width++;
     height++;
       if(width==20)
          break;
     try {
        Thread.sleep(100);
     } catch(Exception exc) {
         System.out.println(exc);
       }
   }
  }
};
 Thread moveIt=new Thread(r);
 moveIt.start();
}

编译完整代码时会产生以下错误:

d:\UnderTest>javac mainClass.java
mainClass.java:18: cannot find symbol
     super.paintComponent(g);
          ^
symbol:   method paintComponent(Graphics)
location: class Object
1 error

为什么我会收到此错误？

如果这是我的完整代码:

import java.awt.*;
import javax.swing.*;
import java.lang.Thread;

class movingObjects extends JPanel { …

如何从目录中删除文件c++？

我知道这个函数int remove ( const char * filename )删除了在参数中指定了文件名的文件.但它只接受char*.c ++中还有其他函数可以接受string它的参数吗？

这是获取电子邮件sender和subject电子邮件的代码.使用此代码,我看到正确的主题显示,但我看到不同格式的发件人的地址.

Properties props = new Properties();
    props.put("mail.imap.host" , "imap.gmail.com" );
    props.put("mail.imap.user" , "username");
    // User SSL
    props.put("mail.imap.socketFactory" , 993);
    props.put("mail.imap.socketFactory.class" , "javax.net.ssl.SSLSocketFactory" );
    props.put("mail.imap.port" , 993 );
    Session session = Session.getDefaultInstance(props , new Authenticator() {
        @Override
                protected PasswordAuthentication getPasswordAuthentication() {
                        return new PasswordAuthentication("username" , "password");
        }
    });

    try {
      Store store = session.getStore("imap");
      store.connect("imap.gmail.com" , "username" , "password");
      Folder fldr = store.getFolder("Inbox");
      fldr.open(Folder.READ_ONLY);
      Message msgs[] = fldr.getMessages();
        for(int i = 0 ; i < msgs.length ; …

代码

import javax.sound.sampled.*;
import java.io.*;

public class Tester {
static Thread th;


public static void main(String[] args) {
    startNewThread();   
   while( th.isAlive() == true) {
       System.out.println("sound thread is working");
   }
}

public static void startNewThread() {
   Runnable r = new Runnable() {
       public void run() {
           startPlaying();
       }
   };
   th =new Thread(r);
   th.start();
} 
public static void startPlaying() {
   try {            
        AudioInputStream ais = AudioSystem.getAudioInputStream(new File("d:/UnderTest/wavtester.wav"));
        Clip clip = AudioSystem.getClip();
        clip.open(ais);
        clip.loop(-1); // keep playing the sound                  
   } catch(Exception exc) { …

这是产生错误的片段:

')' expected
';' expected
not a statement
cannot find symbol
symbol : variable ActionEvent

片段:

private void jMenuItem5ActionPerformed(java.awt.event.ActionEvent evt) {
    JFileChooser chooseToAdd = new JFileChooser();
    int option = chooseToAdd.showOpenDialog(this);
    if( option == JFileChooser.APPROVE_OPTION ) {
        nameOfAudioFile = chooseToAdd.getSelectedFile().getAbsolutePath();
        //clonejTree1ValueChanged( TreeSelectionEvent evt );
        tester(java.awt.event.ActionEvent evt);
    }
}

private void tester(java.awt.event.ActionEvent evt) {
    System.out.println("tester");
}

有语法错误吗？

我有一个JDialog名为的班级Preferences.这个类创建一个构造函数,如:

class Preferences extends javax.swing.JDialog {
          Preferences(java.awt.Frame parent,modal)  {
                      super(parent,modal);
                      //......
          }
}

在我的程序中,我希望在单击JFrame表单中的按钮时打开此首选项对话框.在按钮上注册动作监听器后,我将代码写入:

Frame fr = new Frame();
Preferences p = new Preferences(fr,false);
fr.add(p);
fr.setVisible(true);

当我运行此代码时,我得到以下异常(当我单击按钮时):

Exception in thread "AWT-EventQueue-0" 
    java.lang.IllegalArgumentException: adding a window to a container

这是什么意思,我该如何解决？

到现在为止,每当我查询数据库时,我都会打开一个与数据库的新连接.我如何实现该属性,一旦我打开连接,我可以重用它？

完成后,请告诉我是否可以泄漏资源.

单击netbeans项目时,Deploy to Google App Engine这些是Google App Deployment控制台上显示的消息:

May 28, 2012 4:22:06 PM com.google.apphosting.utils.config.AppEngineWebXmlReader readAppEngineWebXml
INFO: Successfully processed /home/non-admin/NetBeansProjects/Guestbook/build/web/WEB-INF/appengine-web.xml
May 28, 2012 4:22:06 PM com.google.apphosting.utils.config.AbstractConfigXmlReader readConfigXml
INFO: Successfully processed /home/non-admin/NetBeansProjects/Guestbook/build/web/WEB-INF/web.xml
Beginning server interaction for ...
Password for suhailgupta03@gmail.com: 
com.google.appengine.tools.admin.HttpIoException: Error posting to URL:  https://appengine.google.com/api/appversion/getresourcelimits?app_id=&version=1&
400 Bad Request
app_id GET query string parameter must be supplied.

Unable to update app: Error posting to URL: https://appengine.google.com/api/appversion/getresourcelimits?app_id=&version=1&
400 Bad Request
app_id GET query string parameter must be supplied.

Please see the logs …

以下片段尝试将某些目录中存在的目录和文件的名称写入文本文件.每个名称应写入单独的行.而不是在同一行上打印每个名称.为什么会这样？

        try {
        File listFile = new File("E:" + System.getProperty("file.separator") + "Shiv Kumar Sharma Torrent"+ System.getProperty("file.separator") +"list.txt");
        FileWriter writer = new FileWriter(listFile,true);
        Iterator iterator = directoryList.iterator();
        while(iterator.hasNext()) {
            writer.write((String)iterator.next());
            writer.write("\n"); // Did this so each name is on a new line
        }
        writer.close();
    }catch(Exception exc) {
        exc.printStackTrace();
    }

输出:

我在哪里弄错了？