ios UIPanGestureRecognizer pointer position

Question

UIPanGestureRecognizer *panRecognizer = [[UIPanGestureRecognizer alloc] initWithTarget:self action:@selector(pan:)];
[self addGestureRecognizer:panRecognizer];

- (void)pan:(UIPanGestureRecognizer *)gesture {
    NSLog(@"%f", [gesture translationInView:self].x);
}

The above code will log the relative position of my current pan, but how can I get the absolute position for the view I'm in?

I'm simply just wanting to slide a UIImageView to wherever the user's finger is.

Answer 1

translationInView gives you the pan translation (how much x has changed) and not the position of the pan in the view (the value of x). If you need the position of the pan, you have to use the method locationInView.

You can find the coordinates relatively to the view as follows:

- (void)pan:(UIPanGestureRecognizer *)gesture {
    NSLog(@"%f", [gesture locationInView:self].x);
}

Or relatively to the superview:

- (void)pan:(UIPanGestureRecognizer *)gesture {
    NSLog(@"%f", [gesture locationInView:self.superview].x);
}

Or relatively to the window:

- (void)pan:(UIPanGestureRecognizer *)gesture {
    NSLog(@"%f", [gesture locationInView:self.window].x);
}

Answer 2

Swift 5

Use the method .location() that returns a CGPoint value. [documentation]

For example, relative location of your gesture to self.view:

let relativeLocation = gesture.location(self.view)
print(relativeLocation.x)
print(relativeLocation.y)

Answer 3

I think a simple way of something like this is to get the x and y of the touch and tracking it, once it has 2 points (say X:230 Y:122) you set the scroll of the UIScroll view to the x and y.