小编Fer*_*ndo的帖子

我正在创建一个必须显示超过1个通知的应用.我正在查找信息,我在课堂上使用此代码:

Notification noti = new NotificationCompat.Builder(this)

我可以在栏上显示多个通知,但问题是当我点击它时只显示第一个.我认为它是因为挂起的intent获得的id总是相同的,但是我发送了一个不同的id.

通知活动:单击按钮时显示新通知.

import android.app.Activity;
import android.app.Notification;
import android.app.NotificationManager;
import android.app.PendingIntent;
import android.content.Intent;
import android.os.Bundle;
import android.support.v4.app.NotificationCompat;
import android.util.Log;
import android.view.View;

public class NotificationActivity extends Activity {
int notificationID = 1;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_notification_example);
}

public void onClick(View v){
    displayNotification();
}

protected void displayNotification(){
    Intent i = new Intent(this, NotificationView.class);
    i.putExtra("notificationID", notificationID);

    PendingIntent pendingIntent = PendingIntent.getActivity(this, 0, i, 0);
    NotificationManager nm = (NotificationManager)getSystemService(NOTIFICATION_SERVICE);

    CharSequence ticker ="Display";
    CharSequence contentTitle = …

我正在寻找有关如何在Android上使用HttpPost方法发送信息的信息,我总是看到:

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(posturl);

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("Name","Var1"));
params.add(new BasicNameValuePair("Name2","Var2"));

httppost.setEntity(new UrlEncodedFormEntity(params));    
HttpResponse resp = httpclient.execute(httppost);
HttpEntity ent = resp.getEntity();

问题是我无法做到这一点,因为我必须连接到接收XML格式的String的资源.

关于如何在不使用字符串的情况下仅发送字符串的任何想法 List<nameValuePair>