通过套接字发送ArrayList 。 Java的

我想要做的是通过从Android客户端到Java服务器的套接字发送和ArrayList。 以下是发送ArrayList的客户端代码:

private void sendContacts(){ AppHelper helperClass = new AppHelper(getApplicationContext()); final ArrayList list = helperClass.getContacts(); System.out.println("Lenght of an contacts array : " +list.size()); // for (Person person : list) { // System.out.println("Name "+person.getName()+"\nNumber "+ person.getNr()); // } handler.post(new Runnable() { @Override public void run() { // TODO Auto-generated method stub try { //**151 line** os.writeObject(list); os.flush(); } catch (IOException e) { // TODO Auto-generated catch block Log.e(TAG, "Sending Contact list has failed"); e.printStackTrace(); } } }); } public ArrayList getContacts() { ArrayList alContacts = null; ContentResolver cr = mContext.getContentResolver(); //Activity/Application android.content.Context Cursor cursor = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, null); if(cursor.moveToFirst()) { alContacts = new ArrayList(); do { String id = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts._ID)); if(Integer.parseInt(cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) { Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,null,ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = ?",new String[]{ id }, null); while (pCur.moveToNext()) { String contactNumber = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER)); String contactName = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME)); Person temp = new Person(contactName, contactNumber); alContacts.add(temp); break; } pCur.close(); } } while (cursor.moveToNext()) ; } return alContacts; } 

这是服务器代码:

 private void whileChatting() throws IOException { ableToType(true); String message = "You are now connected "; sendMessage(message); do {// have conversation try { message = (String) input.readObject(); // message = (String) input.readLine(); showMessage("\n" + message); } catch (ClassNotFoundException e) { showMessage("It is not a String\n"); // TODO: handle exception } try{ ArrayList list =(ArrayList) input.readObject(); showMessage("GOT A LIST OF PERSON WITH SIZE :" + list.size()); }catch(ClassNotFoundException e){ showMessage("It is not a List of Person\n"); } } while (!message.equals("client - end")); } 

错误代码 :

 Caused by: java.io.NotSerializableException: com.lauris.client.Person at java.io.ObjectOutputStream.writeNewObject(ObjectOutputStream.java:1344) at java.io.ObjectOutputStream.writeObjectInternal(ObjectOutputStream.java:1651) at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:1497) at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:1461) at java.util.ArrayList.writeObject(ArrayList.java:648) at java.lang.reflect.Method.invoke(Native Method) at java.lang.reflect.Method.invoke(Method.java:372) at java.io.ObjectOutputStream.writeHierarchy(ObjectOutputStream.java:1033) at java.io.ObjectOutputStream.writeNewObject(ObjectOutputStream.java:1384) at java.io.ObjectOutputStream.writeObjectInternal(ObjectOutputStream.java:1651) at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:1497) at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:1461) **at com.lauris.client.MainActivity$ClientThread$4.run(MainActivity.java:151)** at android.os.Handler.handleCallback(Handler.java:739) at android.os.Handler.dispatchMessage(Handler.java:95) at android.os.Looper.loop(Looper.java:135) at android.app.ActivityThread.main(ActivityThread.java:5221) at java.lang.reflect.Method.invoke(Native Method) at java.lang.reflect.Method.invoke(Method.java:372) at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:899) at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:694) 

我想我的问题是序列化? 我真的不明白这个意思吗? 有人可以解释一下它是什么,为什么我需要它? 我怎么能修复我的代码?

补充问题。 您可以在我的代码中看到我试图侦听不同的InputStream。 我想我做错了。 更高级的人可以解释一下如何做到正确吗?

感谢你的帮助,我真的坚持这个。

Person类必须implement Serializable

 import java.io.Serializable; public class Person implements Serializable { /** * */ private static final long serialVersionUID = 1L; } 

是:

– >让它实现Serializable; – >分配随机SerialVersionUID是明智的。

除了序列化之外,您还可以考虑XML序列化(检查Java API)。 XML更具可读性且更易于移植(例如,在VM之间)。