windows - Programly detect if my machine remotely accessed -


i have winservice running in 1 pc, connected server can push command winservcice.

one of command should : isconnectedremotly : should return true if connect machine other machine (remotely), , username (the session username connected to)

how can that?

it depends... if mean "is calling process running under remote session, obtain session id process , get info session check if remote session, like:

dword processid; // filled calling program using getcurrentprocessid() dword sessionid, bytecount; lptstr buffer; if (processidtosessionid(processid, &sessionid))     if (wtsquerysessioninformation(wts_current_server_handle, sessionid,                                    wtsisremotesession, &buffer, &bytecount))         if (wtsquerysessioninformation(wts_current_server_handle, sessionid,                                        wtsusername, &buffer, &bytecount))         {   // copy away user name in buffer             wtsfreememory(buffer);         }

Comments

Post a Comment

Popular posts from this blog

c# - Send Image in Json : 400 Bad request -

from android trying send image class of data, iis webservice. (c#) the problem 400 bad request . the image being encoded base64 . , placed json rest of class elements. my guess base64 not valid inside json. server not understand it. if set string "" , post accepted fine. so question is, how make base64 valid within json array? ( tried url.encode no success). or how should send image android webservice? gson gson = new gson(); string json = gson.tojson(record); // record has param { string base64photo } how big image? i'm pretty sure surpassing iis json size limit (default 4 mb). check http://geekswithblogs.net/frankw/archive/2008/08/05/how-to-configure-maxjsonlength-in-asp.net-ajax-applications.aspx or http://www.webtrenches.com/post.cfm/iis7-file-upload-size-limits good luck!
Read more

javascript - addthis share facebook and google+ url -

i attempting use addthis create simple url shares whichever current page user on when click share button. of right now, have twitter working. when user clicks, able see url populated in tweet box. when user clicks facebook or google+, url doesn't seem show in share box. <ul class="addthis_toolbox addthis_default_style "> <li><a class="addthis_button_facebook"><img src="/img/facebook-social.png" width="20" height="20" border="0" alt="share" /></a></li> <li><a class="addthis_button_twitter"><img src="/img/twitter-social.png" width="20" height="20" border="0" alt="share" /></a></li> <li><a class="addthis_button_google_plusone_share"><img src="/img/google-social.png" width="20" height="20" border="0" alt="s...
Read more

ios - Show keyboard with UITextField in the input accessory view -

i have buttona , when buttona clicked, want keyboard show uitextfield in inputaccessoryview. there way have keyboard show manually , set inputaccessoryview without having uitextfield initially? thanks. you cannot summon keyboard without object can become first responder. there 2 ways work around: subclass uiview , implement uikeyinput protocol in it. example: in .h file: @interface inputobject : uiview<uikeyinput> @property (nonatomic, copy) nsstring *text; @property (nonatomic, strong) uiview *inputaccessoryview; // must override inputaccessoryview , since it's readonly default @end in .m file, implement protocol: - (bool)canbecomefirstresponder { return yes; } - (bool)hastext { return [self.text length]; } - (void)inserttext:(nsstring *)text { nsstring *selftext = self.text ? self.text : @""; self.text = [selftext stringbyappendingstring:text]; [[nsnotificationcenter defaultcenter] postnotificationname:kinputobjec...
Read more