iphone - 如何获得任意控制 objective-c 的屏幕触摸位置？

Question

您好，我想获取任何控件或发生触摸的任何位置的触摸位置/点。

为此，我已经实现了这个，但我没有得到正确的接触点。

   // Create gesture recognizer, notice the selector method
UITapGestureRecognizer *oneFingerTwoTaps =
[[[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(oneFingerTwoTaps)] autorelease];
oneFingerTwoTaps.delegate=self;
// Set required taps and number of touches
[oneFingerTwoTaps setNumberOfTapsRequired:1];
[oneFingerTwoTaps setNumberOfTouchesRequired:1];
[[self view] addGestureRecognizer:oneFingerTwoTaps];

和

 - (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch {

    CGPoint point= [touch locationInView:touch.view];
        NSLog(@"Point - %f, %f",point.x,point.y);
    NSLog(@"Touch");
    return NO; // handle the touch
}

当我试图点击任何 UIButton、UIImage、UITableView 时，它没有给我正确的点击点，我做错了什么吗？请帮我。谢谢你。

Answer 1

您的代码打印触摸发生在 View 中的位置。因此，如果您触摸中间的 50x100 大小的按钮，它将打印“Point 25.0, 50.0”。

如果要找到UIScreen的触摸位置，就得进行值转换:

 - (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch {

    CGPoint point = [touch locationInView:touch.view];
    CGPoint pointOnScreen = [touch.view convertPoint:point toView:nil];
    NSLog(@"Point - %f, %f", pointOnScreen.x, pointOnScreen.y);
    NSLog(@"Touch");
    return NO; // handle the touch
}

或者直接获取窗口(屏幕)空间中的坐标:

 - (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch {

    CGPoint pointOnScreen = [touch locationInView:nil];
    NSLog(@"Point - %f, %f", pointOnScreen.x, pointOnScreen.y);
    NSLog(@"Touch");
    return NO; // handle the touch
}