ios - 反向滑动 UIButton 数组

Question

在我的应用程序中，我有很多 UIButton 堆叠在一起。我想使用滑动手势来回浏览它们。在 viewDidLoad 中我有:

self.buttonArray = [[NSMutableArray alloc] initWithObjects:map, boardmembers, facebook, twitter, lessonbooks, schedule, nil];
    map.hidden = NO;
    boardmembers.hidden = facebook.hidden = twitter.hidden = lessonbooks.hidden = schedule.hidden = YES;

要从右向左滑动，并推进它们，我有:

 UIButton *currentVisibleButton = [_buttonArray firstObject];
    UIButton *nextVisibleButton = [_buttonArray objectAtIndex:1];

    [_buttonArray removeObject:currentVisibleButton];
    [_buttonArray addObject:currentVisibleButton];

    currentVisibleButton.hidden = YES;
    nextVisibleButton.hidden = NO;

我在使反向工作时遇到问题，我可以在其中来回移动。我该怎么做？

Answer 1

我不确定我是否完全理解您的代码，但是您发布的内容的类似逆序是:

UIButton *currentVisibleButton = [_buttonArray lastObject];
UIButton *nextVisibleButton = _buttonArray[_buttonArray.count-2];

[_buttonArray removeObject:currentVisibleButton];
[_buttonArray insertObject:currentVisibleButton atIndex:0];

但是将数组用作环的更好方法是保留游标。保持当前可见索引的状态。

@property(assign, nonatomic) NSInteger cursor;

- (void)cursorLeft {
    self.cursor = (self.cursor+1 == _buttonArray.length)? 0 :  self.cursor+1;
}

- (void)cursorRight {
    self.cursor = (self.cursor == 0)? _buttonArray.length-1 :  self.cursor-1;
}

- (UIView *)viewAtCursor {
    return (UIView *)_buttonArray[self.cursor];
}

确保数组中索引为零的按钮是可见的，而其他所有按钮都是隐藏的。现在，无需对阵列进行任何搅动，您就可以像这样四处滑动。

// swipe left
self.viewAtCursor.hidden = YES;
[self cursorLeft];
self.viewAtCursor.hidden = NO;

// swipe right
self.viewAtCursor.hidden = YES;
[self cursorRight];
self.viewAtCursor.hidden = NO;