ios 如何从 UIGestureRecognizer 获取 UITouch 位置
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How to get UITouch location from UIGestureRecognizer
提问by Oscar Apeland
I want to get the UITouch location of my tap from UIGestureRecognizer, but I can not figure out how to from looking at both the documentation and other SO questions. Can one of you guide me?
我想从 UIGestureRecognizer 获取我的水龙头的 UITouch 位置，但我无法弄清楚如何查看文档和其他 SO 问题。你们中的一位可以指导我吗？
- (void)handleTap:(UITapGestureRecognizer *)tapRecognizer
{
CCLOG(@"Single tap");
UITouch *locationOfTap = tapRecognizer; //This doesn't work
CGPoint touchLocation = [_tileMap convertTouchToNodeSpace:locationOfTap];
//convertTouchToNodeSpace requires UITouch
[_cat moveToward:touchLocation];
}
FIXED CODE HERE - THIS ALSO FIXES INVERTED Y AXIS
固定代码在这里 - 这也固定倒 Y 轴
CGPoint touchLocation = [[CCDirector sharedDirector] convertToGL:[self convertToNodeSpace:[tapRecognizer locationInView:[[CCDirector sharedDirector] openGLView]]]];
回答by MJN
You can use the locationInView:method on UIGestureRecognizer. If you pass nil for the view, this method will return the location of the touch in the window.
您可以使用locationInView:UIGestureRecognizer 上的方法。如果为视图传递 nil，则此方法将返回触摸在窗口中的位置。
- (void)handleTap:(UITapGestureRecognizer *)tapRecognizer
{
CGPoint touchPoint = [tapRecognizer locationInView: _tileMap]
}
There is also a helpful delegate method gestureRecognizer:shouldReceiveTouch:. Just make sure to implement and set your tap gesture's delegate to self.
还有一个有用的委托方法gestureRecognizer:shouldReceiveTouch:。只需确保实现并将您的点击手势的委托设置为 self。
Keep a reference to the gesture recognizer.
保留对手势识别器的引用。
@property UITapGestureRecognizer *theTapRecognizer;
Initiailze the gesture recognizer
初始化手势识别器
_theTapRecognizer = [[UITapGestureRecognizer alloc] initWithTarget: self action: @selector(someMethod:)];
_theTapRecognizer.delegate = self;
[someView addGestureRecognizer: _theTapRecognizer];
Listen for delegate methods.
侦听委托方法。
-(BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
CGPoint touchLocation = [_tileMap convertTouchToNodeSpace: touch];
// use your CGPoint
return YES;
}
回答by user3108511
In Swift:
在斯威夫特：
func handleFrontTap(gestureRecognizer: UITapGestureRecognizer) {
print("tap working")
if gestureRecognizer.state == UIGestureRecognizerState.Recognized
{
print(gestureRecognizer.locationInView(gestureRecognizer.view))
}
}
回答by user3352358
Try this:
尝试这个：
-(void) didMoveToView:(SKView *)view{
oneFingerTap = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(oneTapDetected:)];
oneFingerTap.numberOfTapsRequired=1;
oneFingerTap.numberOfTouchesRequired=1;
[view addGestureRecognizer:oneFingerTap];
}
-(void)oneTapDetected:(UITapGestureRecognizer *)recognizer{
NSLog(@"one tap detec");
tapPositionOneFingerTap = [oneFingerTap locationInView:self.view];
NSLog(@"%f, %f",tapPositionOneFingerTap.x,tapPositionOneFingerTap.y);
}
This prints the coordinates of each tap in your console.
这会打印控制台中每个点击的坐标。
回答by user3352358
Apple Docssay
苹果文档说
UIGestureRecognizer
- (NSUInteger)numberOfTouchesThe number of UITouch objects in a private arraymaintained by the receiver.
UIGestureRecognizer
- (NSUInteger)numberOfTouches接收者维护的私有数组中 UITouch 对象的数量。
So you shouldn't access them.
所以你不应该访问它们。
Using the value returned by this method in a loop, you can ask for the location of individual touches using the
locationOfTouch:inView:method.
在循环中使用此方法返回的值，您可以使用该
locationOfTouch:inView:方法询问单个触摸的位置。
