2014-11-23 68 views
-1

我希望androidStartServiceOnBoot.class应该每次执行,但它只执行一次。AndroidStartServiceOnBoot.class在程序中只被调用一次..但我希望它每次调用

@Override public void onReceive(Context context,Intent intent){PowerManager pm =(PowerManager)context.getSystemService(Context.POWER_SERVICE); PowerManager.WakeLock wl = pm.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK,“Your TAG”); //获取锁 wl.acquire();正从乌尔PHP脚本或任何返回

  //You can do the processing here. 
     Bundle extras = intent.getExtras(); 
     StringBuilder msgStr = new StringBuilder(); 

     if(extras != null && extras.getBoolean(ONE_TIME, Boolean.FALSE)){ 
      //Make sure this intent has been sent by the one-time timer button. 
      msgStr.append("One time Timer : "); 
     } 
     Format formatter = new SimpleDateFormat("hh:mm:ss a"); 
     msgStr.append(formatter.format(new Date())); 
     Intent serviceIntent = new Intent(context, AndroidStartServiceOnBoot.class); 
     context.startService(serviceIntent); 
     Toast.makeText(context, msgStr + "test", Toast.LENGTH_LONG).show(); 
     //Release the lock 
     wl.release(); 
} 
+0

您尝试解析的字符串不是有效的json。 – 2014-11-23 05:15:35

回答

0

响应... 被具有附装作为 1.echo“BR”一个
标签----为PHP 等....(删除这对于PHP) 检查你的响应返回从服务器..应该没有
标记,同时返回来自服务器上的文件的响应

+0

我的PHP脚本..怎么办? <?php $ host =“127.0.0.1”; $ uname =“root”; $ pwd =“”; $ db =“mydata”; ($ host,$ uname,$ pwd)或死(“连接失败”); \t mysql_select_db($ db,$ con)或die(“db selection failed”); \t \t $ mobile = $ _ REQUEST ['mno']; \t \t $ flag ['code'] = 0; \t $ result = mysql_query(“SELECT * FROM tracker where mobile like'$ mobile'”)或die(mysql_error()); \t如果(mysql_num_rows($结果)== 0){ \t \t 如果($ R =请求mysql_query( “插入到跟踪器的值( '$移动')”,$ CON)) \t { \t \t // $ flag [“code”] = 1; \t \t $ code = 1; \t \t \t } \t} \t回波json_encode($代码); \t mysql_close($ con); ?> – 2014-11-23 08:15:29

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