以下是我的代码。编辑器:Eclipse,平台:Windows。
它是一个聊天应用程序,其中 2 个 android 模拟器通过 tcp 套接字连接。
UI 由发送按钮、 TextView 和文本框组成。
问题:只要我输入文本并点击发送,应用程序就会崩溃。
服务器端口是8000。 所以我的重定向是 redir add tcp:8081:8000 和 redir add tcp:8082:8000。
我不知道我的代码有什么问题。请建议我需要更改的内容。
public class HelloandroidActivity extends Activity
{
/** Called when the activity is first created. */
public int serverport=8000;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
final EditText nameField = (EditText) findViewById(R.id.editText1);
final Button button2 = (Button) findViewById(R.id.button1);
Integer severport=8000;
new Server().executeOnExecutor(AsyncTask.THREAD_POOL_EXECUTOR,severport);
button2.setOnClickListener(new OnClickListener() {
public void onClick(View v) {
final String name = nameField.getText().toString();
final TextView tv = (TextView) findViewById(R.id.textView1);
//tv.setText(name);
String s=null;
new Client().executeOnExecutor(AsyncTask.THREAD_POOL_EXECUTOR,s);
}
});// end onclicklis
}//end oncreate
class Server extends AsyncTask <Integer, String, String>
{
public InetAddress byIpAsName ;
int r=0;
@Override
protected String doInBackground(Integer... serverport) {
//i[0]=serverport;
Integer[] sp=serverport;
BufferedReader in=null;
ServerSocket s=null;
r=sp[0];
String cIn="";
try {
//byIpAsName = InetAddress.getByName("10.2.2.15");
s=new ServerSocket(r);
while(true)
{
Socket client = s.accept();
in = new BufferedReader(new InputStreamReader(client.getInputStream()));
String line=in.readLine();
cIn=null;
while(line!=null){cIn=cIn.concat(line);}
}//while
} catch (IOException e) {
e.printStackTrace();
}
try {
s.close();
in.close();
}
catch (IOException e) {
e.printStackTrace();
}
return cIn;
}//end inBackground
//@SuppressWarnings("null")
protected void onPostExecute(String... cIn)
{
}//onpost execute
}//server class
public class Client extends AsyncTask<String, String, String>
{
PrintWriter out = null;
BufferedReader in=null;
String sIn=null;
//Server s1=new Server();
//int q=s1.r;
TelephonyManager tel = (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE);
String portStr = tel.getLine1Number().substring(tel.getLine1Number().length() - 4);
int q = Integer.parseInt(portStr);
Socket socket;
@Override
protected String doInBackground(String... params) {
try
{
//q=8080;
InetAddress byIpAsName1=InetAddress.getByName("10.0.2.2");
socket = new Socket(byIpAsName1, q);
out = new PrintWriter(socket.getOutputStream(), true);
in = new BufferedReader(new InputStreamReader(socket.getInputStream()));
String line=in.readLine();
sIn=null;
while(line!=null){sIn=sIn.concat(line);}
}
catch (IOException e) {
e.printStackTrace();
}//catch
return sIn;
}//in background
protected void onPostExecute(String... sIn)
{
String c=null;
final TextView tv = (TextView) findViewById(R.id.textView1);
c=c.concat(sIn[0]);
tv.setText(c);
}
}
}//main class
最佳答案
在你的 logcat 中,重要的是这一行:
03-16 23:12:23.434: E/AndroidRuntime(571): java.lang.SecurityException: Requires READ_PHONE_STATE: Neither user 10040 nor current process has android.permission.READ_PHONE_STATE.
这表示为了运行您的代码,您需要 android manifest.xml 中的 READ_PHONE_STATE 权限。
将此行添加到 list 中,在 <application>
之外标记但在 <manifest>
内标签。
<uses-permission android:name="android.permission.READ_PHONE_STATE"></uses-permission>
如果这不能解决问题,问题可能与 this answer 有关.
关于android - 使用 tcp 套接字连接 android 模拟器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9745288/