iphone - 如何在 UITapGestureRecognizer 中的 @selector 中传递参数？

Question

我的表头 View 部分有这个:

    UITapGestureRecognizer *tapGesture = [[UITapGestureRecognizer alloc]initWithTarget:self action:@selector(sectionHeaderTapped:)];

我想在方法 sectionHeaderTapped 中传递部分编号，以便我可以识别哪个部分被点击。

我的方法实现如下所示:

-(void)sectionHeaderTapped:(NSInteger)sectionValue {
    NSLog(@"the section header is tapped ");    
}

我怎样才能实现这个目标？

Answer 1

方法sectionHeaderTapped应具有以下签名之一:

- (void)sectionHeaderTapped:(UITapGestureRecognizer *)sender;
- (void)sectionHeaderTapped;

您必须使用点击的坐标找出被点击的单元格。

-(void)sectionHeaderTapped:(UITapGestureRecognizer *)gestureRecognizer
{
    CGPoint tapLocation = [gestureRecognizer locationInView:self.tableView];
    NSIndexPath *tapIndexPath = [self.tableView indexPathForRowAtPoint:tapLocation];
    UITableViewCell* tappedCell = [self.tableView cellForRowAtIndexPath:tapIndexPath];
}

您可能可以使用该方法获取节标题。但将不同的手势识别器附加到每个节标题可能会更容易。

- (UIView*)tableView:(UITableView*)tableView viewForHeaderInSection:(NSInteger)section
{
    // ...
    UITapGestureRecognizer *tapGesture = [[UITapGestureRecognizer alloc]initWithTarget:self action:@selector(sectionHeaderTapped:)];
    [headerView addGestureRecognizer:tapGesture];
    return headerView;
}

然后

-(void)sectionHeaderTapped:(UITapGestureRecognizer *)gestureRecognizer
{
    UIView *headerView = gestureRecognizer.view;
    // ...
}