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我的 iPhone 项目上有一个 Popover:

弹出窗口截图

当我点击一张小图片时,它会出现:

-(void)tapDetected:(UITapGestureRecognizer *)sender
{
    UIImageView *iboImageView = sender.view;
    UITableView *tableView = (UITableView *)iboTableView;
    NSIndexPath *index = [NSIndexPath indexPathWithIndex:iboImageView.tag];
    MovieCell *cell = (MovieCell *)[tableView cellForRowAtIndexPath:[NSIndexPath indexPathForRow:iboImageView.tag inSection:0]];
    [ApplicationManager getInstance].currentMovie=cell.nameLabel.text;

    DemoTableController *controller = [[DemoTableController alloc] initWithStyle:UITableViewStylePlain]; //This is my popover controller (what is displayed inside popover )
    FPPopoverController *popover = [[FPPopoverController alloc] initWithViewController:controller];
    popover.contentSize = CGSizeMake(150,158);
    [popover presentPopoverFromView:cell.iboPopImage];
}

当我按下一个单元格时,我想折叠弹出框。有没有类似的东西self.close

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1 回答 1

3

从 FPPopoverController.h 文件:

/** @brief Dismiss the popover **/
-(void)dismissPopoverAnimated:(BOOL)animated;

/** @brief Dismiss the popover with completion block for post-animation cleanup **/
typedef void (^FPPopoverCompletion)();
-(void)dismissPopoverAnimated:(BOOL)animated completion:(FPPopoverCompletion)completionBlock;

使用这些方法中的任何一种来关闭控制器。

于 2013-10-23T11:50:52.177 回答