我的应用程序是通过 NetConnection 保存(到服务器)数据。如果网络不可用,我会在本地保存,然后当网络再次可用时发送到服务器。我的问题是检查经常连接互联网所以我尝试了服务功能来检查网络连接。但只打了一次电话。如何解决我的问题。任何人都可以帮助我
提前致谢!
update
package com.android.cdtech;
import android.app.Activity;
import android.content.BroadcastReceiver;
import android.content.Context;
import android.content.Intent;
import android.database.Cursor;
import android.net.ConnectivityManager;
import android.view.View;
class ReceiverName extends Activity {
BroadcastReceiver r = new BroadcastReceiver(){
@Override
public void onReceive(Context context, Intent intent) {
ConnectivityManager cm = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
if (cm == null)
return;
if (cm.getActiveNetworkInfo() != null && cm.getActiveNetworkInfo().isConnected()) {
saveData();
} else {
// Do nothing or notify user somehow
}
}
// code to handle broadcase
private void saveData() {
final saveData dh=new saveData(this); //Class for Creating a Database
webService calService=new webService();
dh.open();
Cursor c = dh.pay();
String text = "";
do {
text = text + " " + "\n" + c.getString(4);
System.out.println(c.getCount());
// Toast.makeText(this,"Name:" +c.getString(1)+c.getString(2)+c.getString(3)+c.getString(4)+"",Toast.LENGTH_LONG).show();
calService.paymentReceipt("PaymentReceipt", c.getString(1), c.getString(2), c.getString(3), c.getString(4), "gf", "0");
}
while (c.moveToNext());
dh.close();
}
};
}
您可以在没有任何计时器的情况下完成此操作,只需注册接收器以侦听连接更改:
<receiver android:name=".ReceiverName" >
<intent-filter >
<action android:name="android.net.wifi.STATE_CHANGE" />
<action android:name="android.net.conn.CONNECTIVITY_CHANGE" />
</intent-filter>
</receiver>
并检查连接是否建立:
public class ReceiverName extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
ConnectivityManager cm = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
if (cm == null)
return;
if (cm.getActiveNetworkInfo() != null && cm.getActiveNetworkInfo().isConnected()) {
// Send here
} else {
// Do nothing or notify user somehow
}
}
}
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