我要做的是通过从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<Person> 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<Person> getContacts() {
ArrayList<Person> 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<Person>();
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<Person> list =(ArrayList<Person>) 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)
我猜我的问题是序列化?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?
其他问题。您可以在我的代码中看到我试图侦听不同的InputStreams。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。
答案 0 :(得分:0)
Person
课程必须implement Serializable
:
import java.io.Serializable;
public class Person implements Serializable {
/**
*
*/
private static final long serialVersionUID = 1L;
}
答案 1 :(得分:0)
是:
- &GT;让它实现Serializable; - &GT;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)。