ios - 在矩形框(如开关)内移动按钮(向左/向右)

Question

我正在尝试创建一个控件，用户可以在该控件中触摸并移动框架内的按钮。这是我的代码。

- (void)wasDragged:(UIButton *)button withEvent:(UIEvent *)event
{

   UITouch *touch = [[event touchesForView:button] anyObject];

    // get delta
    CGPoint previousLocation = [touch previousLocationInView:button];
    CGPoint location = [touch locationInView:button];
    CGFloat delta_x = location.x - previousLocation.x;
    CGFloat delta_y = location.y - previousLocation.y;

    // move button
    button.center = CGPointMake(button.center.x + delta_x,
                                button.center.y + delta_y);   


}

我能够移动按钮(通过触摸和拖动)，但是如何限制按钮，以便它只能在矩形框架内向左/向右移动。

Answer 1

也许这种方法会帮助您。我在不久前制作的一个简单的Pong游戏中使用了它。它是用于UIView的，它是乒乓球游戏的弹跳板。我将弹跳板的移动限制在x方向，而不是在屏幕范围之外。

如果不清楚，请写评论，然后尝试解释。

// Method for movement of the bouncing pad. Restricted movement to x-axis inside of bounds.
- (void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event {

    UITouch *aTouch = [touches anyObject];
    CGPoint loc = [aTouch locationInView:self];
    CGPoint prevloc = [aTouch previousLocationInView:self];

    CGRect myFrame = self.frame;

    // Checking how far we have moved from the previous location
    float deltaX = loc.x - prevloc.x;

    // Note that we only update the x-position of the pad to prevent it from moving in the y-direction.
    myFrame.origin.x += deltaX;

    // Making sure that the bouncePad cannot move outside of the screen
    if(myFrame.origin.x < 0){
        myFrame.origin.x = 0;
    } else if (myFrame.origin.x + myFrame.size.width > [UIScreen main Screen].bounds.size.width) {
        myFrame.origin.x = [UIScreen mainScreen].bounds.size.width - myFrame.size.width;
    }

    // Setting the bouncing pad frame to the one with the updated position from the touches moved event.
    [self setFrame:myFrame];

}