android-Monodroid意向服务的BroadcastReciever不触发OnReceive
作者:互联网
好的,所以我一直在尝试为monodroid essentially going off of this source material运行一个玩具意图服务,并且遇到了一些问题,并将其归结为广播接收器.我怀疑ACTION_RESP字符串还有更多.如果没有别的话,我当然可以忍受,也许是对它的确认,或者是设置接收机可以看到的动作的正确方法.
所以,这就是我到目前为止所拥有的…
我的专心服务
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using Android.App;
using Android.Content;
using Android.OS;
using Android.Runtime;
using Android.Views;
using Android.Widget;
using Android.Util;
namespace Mono_For_Android_Intent_Service
{
[Service]
public class TestIntentService : IntentService
{
public static String PARAM_IN_MSG = "imsg";
public static String PARAM_OUT_MSG = "omsg";
#region Constructor
public TestIntentService()
: base("TestIntentService")
{
}
#endregion
protected override void OnHandleIntent(Intent intent)
{
string msg = intent.GetStringExtra(PARAM_IN_MSG);
//Do Stuff
string result = msg + " " + DateTime.Now.ToString();
Intent BroadcastIntent = new Intent(this,typeof(Mono_For_Android_Intent_Service.MainActivity.MyBroadcastReceiver));
BroadcastIntent.SetAction(Mono_For_Android_Intent_Service.MainActivity.MyBroadcastReceiver.ACTION_RESP);
BroadcastIntent.AddCategory(Intent.CategoryDefault);
BroadcastIntent.PutExtra(PARAM_OUT_MSG,result);
SendBroadcast(BroadcastIntent);
}
}
}
主要活动和广播接收器
using System;
using Android.App;
using Android.Content;
using Android.Runtime;
using Android.Views;
using Android.Widget;
using Android.OS;
namespace Mono_For_Android_Intent_Service
{
[Activity(Label = "Mono_For_Android_Intent_Service", MainLauncher = true, Icon = "@drawable/icon")]
public partial class MainActivity : Activity
{
private MyBroadcastReceiver Receiver;
protected override void OnCreate(Bundle bundle)
{
base.OnCreate(bundle);
// Set our view from the "main" layout resource
SetContentView(Resource.Layout.Main);
IntentFilter filter = new IntentFilter(MyBroadcastReceiver.ACTION_RESP);
filter.AddCategory(Intent.CategoryDefault);
Receiver = new MyBroadcastReceiver();
RegisterReceiver(Receiver, filter);
var button = FindViewById(Resource.Id.start);
button.Click += (s,e)=>
{
Intent msgIntent= new Intent(this,typeof(TestIntentService));
msgIntent.PutExtra(TestIntentService.PARAM_IN_MSG,"Me message, arh");
Toast.MakeText(this, "Starting intent service!", ToastLength.Short).Show();
StartService(msgIntent);
};
TextView txtView = FindViewById(Resource.Id.status);
}
public class MyBroadcastReceiver : BroadcastReceiver
{
public static readonly string ACTION_RESP = "Mono_For_Android_Intent_Service.Intent.Action.MESSAGE_PROCESSED";
public override void OnReceive(Context context, Intent intent)
{
Toast.MakeText(context, "Firing On Receive!", ToastLength.Short).Show();
var activity = (MainActivity)context;
TextView txtView = activity.FindViewById(Resource.Id.status);
txtView.Text = intent.GetStringExtra(TestIntentService.PARAM_OUT_MSG).ToString();
Toast.MakeText(context, "On Receive complete!", ToastLength.Short).Show();
}
}
}
}
>我做错了吗?我靠近吗?我想以为我已经很近了,但是如果我真的在timbucktoo的话,我一定想知道…
>如果是这样,发送和接收该广播的正确方法是什么?
解决方法:
如果要在系统中注册广播接收器,则需要添加[BroadcastReceiver]属性.
样品:
https://github.com/xamarin/monodroid-samples/blob/master/ApiDemo/App/OneShotAlarm.cs
标签:intentservice,xamarin-android,broadcastreceiver,android 来源: https://codeday.me/bug/20191101/1986847.html