如何从UIGestureRecognizer获取UITouch位置

问题描述

我想从UIGestureRecognizer获取UITouch的水龙头位置，但我不知道如何从文档和其他SO问题。你们中有一个能指导我吗？

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?

- (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];

}

固定码在这里 - AXIS

    CGPoint touchLocation = [[CCDirector sharedDirector] convertToGL:[self convertToNodeSpace:[tapRecognizer locationInView:[[CCDirector sharedDirector] openGLView]]]];

推荐答案

您可以使用 locationInView ：在UIGestureRecognizer上的方法。如果您为视图传递nil，此方法将返回窗口中触摸的位置。

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.

- (void)handleTap:(UITapGestureRecognizer *)tapRecognizer
{
    CGPoint touchPoint = [tapRecognizer locationInView: _tileMap]
}

还有一个有用的委托方法 gestureRecognizer：shouldReceiveTouch：。

There is also a helpful delegate method gestureRecognizer:shouldReceiveTouch:. Just make sure to implement and set your tap gesture's delegate to self.

保持对姿势识别器的引用。

Keep a reference to the gesture recognizer.

@property UITapGestureRecognizer *theTapRecognizer;

启动姿势识别器

_theTapRecognizer = [[UITapGestureRecognizer alloc] initWithTarget: self action: @selector(someMethod:)];
_theTapRecognizer.delegate = self;
[someView addGestureRecognizer: _theTapRecognizer];

监听代理方法。

-(BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
    CGPoint touchLocation = [_tileMap convertTouchToNodeSpace: touch];
    // use your CGPoint
    return YES;
}

这篇关于如何从UIGestureRecognizer获取UITouch位置的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！