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django: Call to remote service kills python


django: Call to remote service kills python

By : aVis
Date : October 17 2020, 11:12 PM
With these it helps The developpement server is single-threaded so if you open another connection it hangs the server.
You could try for dev purposes : http://github.com/jaylett/django_concurrent_test_server
code :


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Android remote service doesn't call service methods

Android remote service doesn't call service methods


By : Brian Ford
Date : March 29 2020, 07:55 AM
I wish this help you
I need IPC to control the service from different activities. So I decide to develop a remote service with AIDL.
How to call an Android remote service (IPC) from a Widget / local service?

How to call an Android remote service (IPC) from a Widget / local service?


By : user2618314
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I found the answer to my own question. To make things easier for others, here's the solution:
When doing a remote service, one has to write the AIDL which will be compiled into a sort of stub interface, the implementation of that interface (i.e. the code that is executed when someone calls the remote methods), and a class that extends "Service" which returns the implementation class in the onBind() method. (A normal local service would return null in that method)
code :
public class RemoteService extends Service {
  public IBinder onBind(Intent intent) {
    Log.i("RemoteService", "onBind() called");
    return new RemoteServiceImpl();
  }
  /**
   * The IRemoteInterface is defined through IDL
   */
  public class RemoteServiceImpl extends IRemoteService.Stub {
    public void remoteDetonateBirthdayCake() throws RemoteException {
        //your code here
    }
  };
}
<service android:name="RemoteService"> 
    <intent-filter>
        <action android:name="com.sofurry.favorites.IRemoteService"></action>
</intent-filter>
</service>
IRemoteService mService;
RemoteServiceConnection mConnection = new RemoteServiceConnection();
getApplicationContext().bindService(new Intent(IRemoteService.class.getName()), mConnection, Context.BIND_AUTO_CREATE);
class RemoteServiceConnection implements ServiceConnection {
    public void onServiceConnected(ComponentName className, 
        IBinder service ) {
        mService = IRemoteService.Stub.asInterface(service);
        isBound = true;
    }

    public void onServiceDisconnected(ComponentName className) {
        mService = null;
        isBound = false;
    }
};
mService.remoteDetonateBirthdayCake();
Django's objects.all() kills python

Django's objects.all() kills python


By : donkitjong
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You can try to count objects first, and them iterated with sliced version. Something like:
code :
step = 10
count = MyModel.objects.count()/step
for i in xrange(count):
    for m in MyModel.objects.all()[i*step:(i+1)*step]:
        # doing something with m
How to call many remote WCF service from one central WCF service using Async call?

How to call many remote WCF service from one central WCF service using Async call?


By : Readeo
Date : March 29 2020, 07:55 AM
I wish this helpful for you You can install the "Task Parallel Library for .NET 3.5" from NuGEt and use tasks that way. Then you can wrap you wcfClient.GetSiteInfoBegin and wcfClient.EndSiteInfoBegin methods with Task.Factory.FromAsync.
This is untested, but maybe something like this:
code :
public AboutInfo GetAboutInfo()
{
 AboutInfo aboutInfo = new AboutInfo();

 SiteInfo siteInfo = new SiteInfo()
 {
    ID                = site.ID
   ,Name              = site.Name
   ,DatabaseVersion   = "Unavailable"
 };

  var tasks = new List<Task<SiteInfo>>();
 foreach (Site site in sites)
 {
   try
   {
     string uri = Utilities.CombineUri(site.Uri, "svc/About.svc/ws");
     AboutServiceClient wcfClient = new AboutServiceClient("About");
     wcfClient.Endpoint.Address   = new EndpointAddress(uri);
      tasks.Add(Task<SiteInfo>.Factory.FromAsync(wcfClient.GetSiteInfoBegin, wcfClient.EndSiteInfoBegin, null)
       .ContinueWith(t =>
       {                    
            siteInfo.DatabaseVersion = t.Result.DatabaseVersion.DatabaseVersion;
       }, TaskScheduler.FromCurrentSynchronizationContext()));

   }
   catch (Exception e)
   { //...code stripped out... 
    }
}

Task.WhenAll(tasks.ToArray()).ContinueWith
    ( ts =>
    {
        ts.ForEach( t => aboutInfo.Sites.Add(t.Rrsult); 
    });

return aboutInfo;   
Limiting execution time of a function call in Python kills my Kernel

Limiting execution time of a function call in Python kills my Kernel


By : Christian Schmalisch
Date : March 29 2020, 07:55 AM
it helps some times You're creating 10 signals with SIGALRM handler, meaning you now have 10 exceptions going on at the same time. You may want to instead try:
code :
signal.signal(signal.SIGALRM, signal_handler)
signal.alarm(10)   # Ten seconds

for i in range(10):
    try:
        time.sleep(0.2) #  The function I want to apply
        print("Ok it works")
    except Exception, msg:
        print "Timed out!"
        break
for i in range(10):
    signal.signal(signal.SIGALRM, signal_handler)
    signal.alarm(10)   # Ten seconds
    try:
        time.sleep(i * 2) #  Force it to break,
        print("Ok it works")
    except Exception, msg:
        print "Timed out!"
    signal.alarm(0)
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