Question

我正在研究一种游戏，其中通过长按对象本身来设置游戏对象的属性。属性的值由长按手势的持续时间确定。我正在使用UILongPressGestureRecognizer，所以它是这样的：

[gameObjectView addGestureRecognizer:[[UILongPressGestureRecognizer alloc] 
                                       initWithTarget:self action:@selector(handle:)]];

然后是处理函数

- (void)handle:(UILongPressGestureRecognizer)gesture {
  if (gesture.state == UIGestureRecognizerStateEnded) {
    // Get the duration of the gesture and calculate the value for the attribute
  }
}

在这种情况下如何获得长按手势的持续时间？

Answer 1

我很确定手势不会存储此信息供您访问。您只能在其上设置名为minimumPressDuration的属性，该属性是识别手势之前的时间。

使用ios 5（未经测试）的解决方法：

创建一个名为timer：@property (nonatomic, strong) NSTimer *timer;

的NSTimer属性

还有一个柜台：@property (nonatomic, strong) int counter;

然后@synthesize

- (void)incrementCounter {
    self.counter++;
}

- (void)handle:(UILongPressGestureRecognizer)gesture {
    if (gesture.state == UIGestureRecognizerStateBegan) {
         self.counter = 0;
         self.timer = [NSTimer scheduledTimerWithTimeInterval:1 target:self selector:@selector(incrementCounter) userInfo:nil repeats:yes];
    }
    if (gesture.state == UIGestureRecognizerStateEnded) {
        [self.timer invalidate];
    }
}

因此，当手势开始时，启动一个计时器，每秒触发一次增量方法，直到手势结束。在这种情况下，您需要将minimumPressDuration设置为0，否则手势将不会立即开始。然后用计数器做任何你想做的事情！

Answer 2

不需要计时器。你可以用这种方式实现它：

- (void)handleRecognizer:(UILongPressGestureRecognizer *)gesture
{
    static NSTimeInterval pressStartTime = 0.0; //This an be moved out and be kept as a property

    switch ([gesture state])
    {
        case UIGestureRecognizerStateBegan:
            //Keeping start time...
            pressStartTime = [NSDate timeIntervalSinceReferenceDate];
            break; /* edit*/
        case UIGestureRecognizerStateEnded:
        {
            //Calculating duration
            NSTimeInterval duration = [NSDate timeIntervalSinceReferenceDate] - pressStartTime;
            //Note that NSTimeInterval is a double value...
            NSLog(@"Duration : %f",duration);
            break;
        }
        default:
            break;
    }
}

如果您想要获得长按的实际持续时间，也不要忘记在创建时将手势识别器的minimumPressDuration设置为0：
myLongPressGestureRecognizer.minimumPressDuration = 0

Answer 3

到目前为止，面向对象的Cocoa Touch中最干净，最简单的解决方案似乎是UILongPressGesture的子类。这是一个用Swift编写的例子。

    class MyLongPressGesture : UILongPressGestureRecognizer {
        var startTime : NSDate?
    }

    func installGestureHandler() {
            let longPress = MyLongPressGesture(target: self, action: "longPress:")
            button.addGestureRecognizer(longPress)
    }

    @IBAction func longPress(gesture: MyLongPressGesture) {
            if gesture.state == .Began {
                    gesture.startTime = NSDate()
            }
            else if gesture.state == .Ended {
                    let duration = NSDate().timeIntervalSinceDate(gesture.startTime!)
                    println("duration was \(duration) seconds")
            }
    }

如果您想要包括第一次点击的时间，可以在计算持续时间时通过添加回手势来包括它.minimumPressDuration。鉴于在被触发的手势和被调用的.Start处理程序之间可能存在少量（微小的）时间，缺点是它可能不是微秒精确的。但对于绝大多数应用而言并不重要。

Answer 4

请参阅“minimumPressDuration”属性。根据文件：

手指必须按下手势视图的最小时间段   被承认。

[...]

时间间隔以秒为单位。默认持续时间为0.5   秒。

Answer 5

我知道这是一个迟到的答案，但这对我来说在IOS 7＆amp; 8，无需创建计时器。

UILongPressGestureRecognizer *longGR = [[UILongPressGestureRecognizer alloc] initWithTarget:self action:@selector(yourAction)]; // create the recognizer
longGR.minimumPressDuration = 10.0f; //Ten Seconds
longGR.allowableMovement = 50.0f; //Allowable Movement while being pressed
[gameObjectView setUserInteractionEnabled:YES]; //If setting interaction to a non-interactable object such as a UIImageView
[gameObjectView addGestureRecognizer:longGR]; //Add the gesture recognizer to your object

Answer 6

您可以通过 Swift 3.0

中的关注来获取它

逻辑：只需指定按下时间并找到触摸结束时间的差异

代码：

//variable to calculate the press time
static var pressStartTime: TimeInterval = 0.0

func handleRecognizer(gesture: UILongPressGestureRecognizer) -> Double {
    var duration: TimeInterval = 0

    switch (gesture.state) {
    case .began:
        //Keeping start time...
        Browser.pressStartTime = NSDate.timeIntervalSinceReferenceDate

    case .ended:
        //Calculating duration
        duration = NSDate.timeIntervalSinceReferenceDate - Browser.pressStartTime
        //Note that NSTimeInterval is a double value...
        print("Duration : \(duration)")

    default:
        break;
    }

    return duration
}

Answer 7

似乎在新的iOS版本（目前对我来说是10）上，ul{ padding-left: 0; }和.begin状态的长按识别器回调一个接一个地发生，只有在事件发生后结束，只有当您从屏幕上抬起手指时。

这会使两个事件之间的差值以毫秒为单位，而不是您实际搜索的时间。

在此问题解决之前，我决定迅速从头开始创建另一个手势识别器，这就是它的工作原理。

.ended

然后，您可以轻松访问import UIKit.UIGestureRecognizerSubclass class LongPressDurationGestureRecognizer : UIGestureRecognizer { private var startTime : Date? private var _duration = 0.0 public var duration : Double { get { return _duration } } override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent) { startTime = Date() // now state = .begin //state = .possible, if you would like the recongnizer not to fire any callback until it has ended } override func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent) { _duration = Date().timeIntervalSince(self.startTime!) print("duration was \(duration) seconds") state = .ended //state = .recognized, if you would like the recongnizer not to fire any callback until it has ended } }手势识别器的.duration属性。

例如：

LongPressDurationGestureRecognizer

此特定示例未考虑实际发生了多少触摸或其位置。但是您可以轻松地使用它，扩展LongPressDurationGestureRecognizer。

iOS：如何获得长按手势的持续时间？

7 个答案: