detect

问题 Some android user using auto-clicker apps available on play-store to click multiple time. So I want to detect/block user who tries to use auto-clicker.Is there any way in android to detect such users?If yes, then how we can detect those users? 回答1: There is no API to detect if an autoclicker is running. All autoclickers use accessibility services to emulate clicks, and there is an API that allows you to detect if any accessibility service is running. The problem is, those services also

问题 I program in recent years to Android and I wonder something: How to detect the presence of headphones? There is a method: isWiredHeadsetOn() but it doesn't work. I've tried that but it doesn't work: AudioManager am = (AudioManager)getSystemService(AUDIO_SERVICE); Log.i("am.isWiredHeadsetOn()", am.isWiredHeadsetOn() + ""); if (am.isWiredHeadsetOn()) { //code } Thank you (and sorry if I made spelling mistakes, I am French) 回答1: @Phil answer at https://stackoverflow.com/a/19102661/4758255 is a

来源： https://stackoverflow.com/questions/20868030/detect-all-oracle-versions-installed

来源： https://stackoverflow.com/questions/48019043/python-detect-android

问题 We are developing a Web-App, which launches on Desktop and on tablets (iPad, Android or a surface). Now we are building our own keyboard for number inputs. When you set the focus on an input field with a mousclick, the costum keyboard opens correct. But when you set the focus to the input with a touched click (tablet), the default keyboard opens also. Our idea is, to detect, if there was a mouse-click or a touched click. If it's a touched click, we can set the readonly="true" property to the

问题 I am using android-beacon-library to discover nearby Beacons and it is work fine and very reliable. Is it possible to use the same library to detect the Bluetooth signal of an Android device? I couldn't find any good reliable method to discover the nearby Bluetooth devices, that's why I want to use this library. 回答1: The library is optimized to detect a very specific kind of Bluetooth LE device -- a beacon as the library name would suggest. It is not designed to detect more general non-beacon

问题 I have tried KeyDown and NSEvent , but they require a NSWindow object to be active. My hope is that I can put an app on the status bar and alert the user when it has pressed CapsLock , even if the user is in any other app. My app idea doesn't have a window for settings or anything else, is just an indicator. Can it even be done? I am guessing the AppDelegate but can't figure out how to make it receive modifier events. Any help really appreciated! I have looked on stack overflow for a

问题 I have this problem today, i saw this solution: How to detect my application is idle in c#? I tried that, but my form is covered with userControls and other elements, and the mouseover or keydown events are only fired in the margins of those elements. Is there a better way? 回答1: Hacking together a solution with timers and mouse events is unnecessary. Just handle the Application.Idle event. Application.Idle += Application_Idle; private void Application_Idle(object sender, EventArgs e) { // The

问题 I am running the script to automate test cases and having this unique problem. I have detected and used IDs of the elements for click etc. purpose. However, all of a sudden these ids have changed and the script works no more. Another weird thing is those IDs are same as in script when inspected in Chrome but different in Firefox driver browser. Firebug for test driver: - <p class="description" onclick="selectElementTextListForIE(this,'tile29', 'tile19');selectElementTextList(this,'tile29', ''

问题 I have a matrix [1 2 3 6 7 1 2 1] and would like to remove mirror imaged pairs..i.e. output would be either: [1 2 3 6 7 1] or [3 6 7 1 2 1] Is there a simple way to do this? I can imagine a complicated for loop, something like (or a version which wouldn't delete the original pair..only the duplicates): for i=1:y var1=(i,1); var2=(i,2); for i=1:y if array(i,1)==var1 && array(i,2)==var2 | array(i,1)==var2 && array(i,2)==var1 array(i,1:2)=[]; end end end thanks 回答1: How's this for simplicity - A