iPhone：檢測用戶自上一次屏幕觸摸以來的不活動/空閑時間是否有人實(shí)現(xiàn)了一個功能，如果用戶在某一段時間內(nèi)沒有觸摸屏幕，您就會采取特定的操作嗎？我想找出最好的辦法。UIApplication中有一個與此相關(guān)的方法：[UIApplication sharedApplication].idleTimerDisabled;如果你有這樣的東西，那就太好了：NSTimeInterval timeElapsed = [UIApplication sharedApplication].idleTimeElapsed;然后，我可以設(shè)置一個計時器，并定期檢查這個值，并在它超過閾值時采取一些操作。希望這能解釋我要找的東西。有沒有人已經(jīng)處理過這個問題，或者對你會怎么做有任何想法？謝謝。
3 回答
互換的青春
TA貢獻(xiàn)1797條經(jīng)驗 獲得超6個贊
- (void)sendEvent:(UIEvent *)event {
[super sendEvent:event];
// Only want to reset the timer on a Began touch or an Ended touch, to reduce the number of timer resets.
NSSet *allTouches = [event allTouches];
if ([allTouches count] > 0) {
// allTouches count only ever seems to be 1, so anyObject works here.
UITouchPhase phase = ((UITouch *)[allTouches anyObject]).phase;
if (phase == UITouchPhaseBegan || phase == UITouchPhaseEnded)
[self resetIdleTimer];
}}- (void)resetIdleTimer {
if (idleTimer) {
[idleTimer invalidate];
[idleTimer release];
}
idleTimer = [[NSTimer scheduledTimerWithTimeInterval:maxIdleTime target:self selector:@selector(idleTimerExceeded) userInfo:
nil repeats:NO] retain];}- (void)idleTimerExceeded {
NSLog(@"idle time exceeded");}int retVal = UIApplicationMain(argc, argv, @"AppDelegate", @"AppDelegate");
- 3 回答
- 0 關(guān)注
- 735 瀏覽
添加回答
舉報
0/150
提交
取消