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Android 启动应用程序从具有多种服务的 BroadcastReceiver 启动

转载 作者:行者123 更新时间:2023-11-30 03:40:16 26 4
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我需要在启动完成时启动 2 个服务。第一个服务正常启动,但第二个服务似乎没有启动。我不知道我是否必须创建两个 BroadcastReceiver 或者一个就足够了。这是我的代码。我已将这两项服务放在一个 BroadcastReceiver 中。拜托,你能告诉我我做错了什么吗?

提前致谢

AndroidManifest.xml:

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.pruebas.appservicelocator"
android:versionCode="1"
android:versionName="1.0"
android:installLocation="internalOnly" >

<uses-sdk
android:minSdkVersion="8"
android:targetSdkVersion="16" />

<!-- Startup service -->
<uses-permission android:name="android.permission.RECEIVE_BOOT_COMPLETED"/>
<!-- GPS -->
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
<!-- UUID -->
<uses-permission android:name="android.permission.READ_PHONE_STATE"/>
<!-- Acceso a web service -->
<uses-permission android:name="android.permission.INTERNET"></uses-permission>

<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" >
<activity
android:name="com.pruebas.appservicelocator.MainActivity"
android:label="@string/app_name" >
<intent-filter>
<action android:name="android.intent.action.MAIN" />

<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>

<service android:name=".Servicio">
<intent-filter>
<action android:name="com.pruebas.appservicelocator.Servicio"/>
</intent-filter>
</service>

<service android:name=".ServicioBD">
<intent-filter>
<action android:name="com.pruebas.appservicelocator.ServicioBD"/>
</intent-filter>
</service>

<receiver android:name=".Recibidor" android:enabled="true" android:exported="true">
<intent-filter>
<action android:name="android.intent.action.BOOT_COMPLETED"/>
<category android:name="android.intent.category.DEFAULT"/>
</intent-filter>
</receiver>
</application>

</manifest>

Recibidor.java:

package com.pruebas.appservicelocator;

import android.content.BroadcastReceiver;
import android.content.Context;
import android.content.Intent;
import android.util.Log;
import android.widget.Toast;

public class Recibidor extends BroadcastReceiver {

@Override
public void onReceive(Context context, Intent intent) {
android.os.Debug.waitForDebugger();
Toast.makeText(context, "Iniciando Recibidor", Toast.LENGTH_LONG).show();
final String TAG = "Recibidor";
Log.i(TAG, "Iniciando Recibidor");

if (intent.getAction().equalsIgnoreCase("android.intent.action.BOOT_COMPLETED")) {
Toast.makeText(context, "Iniciando Intent", Toast.LENGTH_LONG).show();
Log.i(TAG, "Iniciando Intent");

Intent servicio = new Intent();
servicio.setAction("com.pruebas.appservicelocator.Servicio");
context.startService(servicio);

Intent servicioBD = new Intent();
servicio.setAction("com.pruebas.appservicelocator.ServicioBD");
context.startService(servicioBD);

Log.i(TAG, "Iniciando Servicios");
Toast.makeText(context, "Iniciando Servicio", Toast.LENGTH_LONG).show();
}
}

}

“Servicio”服务工作正常,所以我不写代码。如果您需要,请告诉我,我会写出来。

ServicioBD.java:

package com.pruebas.appservicelocator;

import com.pruebas.utils.UsersLocationsDBHelper;

import android.app.Service;
import android.content.Intent;
import android.os.IBinder;
import android.widget.Toast;

public class ServicioBD extends Service{

private UsersLocationsDBHelper locDBHelper = null;
private static String TAG = "ServicioBD";

@Override
public IBinder onBind(Intent intent) {
return null;
}

@Override
public void onCreate() {
super.onCreate();
Toast.makeText(this, "SERVICIOBD ON CREATE", Toast.LENGTH_LONG).show();

}

@Override
public void onDestroy() {
super.onDestroy();
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
startForeground(0, null);

Toast.makeText(this, "SERVICIOBD ON START COMMAND", Toast.LENGTH_LONG).show();


return START_STICKY;
}
}

最佳答案

我发现了错误。

  Intent servicio = new Intent();
servicio.setAction("com.pruebas.appservicelocator.Servicio");
context.startService(servicio);

Intent servicioBD = new Intent();
servicio.setAction("com.pruebas.appservicelocator.ServicioBD");
context.startService(servicioBD);

必须是

  Intent servicio = new Intent();
servicio.setAction("com.pruebas.appservicelocator.Servicio");
context.startService(servicio);

Intent servicioBD = new Intent();
servicioBD.setAction("com.pruebas.appservicelocator.ServicioBD");
context.startService(servicioBD);

关于Android 启动应用程序从具有多种服务的 BroadcastReceiver 启动,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15861585/

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