使用prepareForSegue从标签中获取标记/ id

Question

我有一组UILabel的，并希望将所选项目的id放在标签中。就像这样：

UILabel *miII = [[UILabel alloc] initWithFrame:CGRectMake(530, 0, 25, 25)];
miII.tag=item.id;

下面我可以在其中设置destinationViewController的destinationViewController属性。我遇到的问题是如何从UILabel访问标记？还是有更好的方法来做这件事？我已经在评论中包含了我的经验，并且不使用UITableView。

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    NSLog(@"prepareForSegue: %@", segue.identifier);
    ItemDetailViewController *myVC = [segue destinationViewController];

    //[myVC setItemId:12]; //  <-- hard-coding this works
    [myVC setItemId:sender.view.tag];   // this doesn't work

}

- (void)tapRecognized:(UIGestureRecognizer *)sender
{
    NSLog(@"that tap was recognized with %d", sender.view.tag); // <-- this works
    [self performSegueWithIdentifier: @"ItemSegue" sender: self];


}

thx预先

Answer 1

sol很简单，但基于其他Q和A的，有点神秘。我做了以下几点：

@interface ListViewController ()
{
@private

int _itemId;
}

.....


-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    // NSLog(@"prepareForSegue: %@", segue.identifier);
    // NSLog(@"prepareForSegue: %@", sender);

    ItemDetailViewController *myVC = [segue destinationViewController];
    [myVC setItemId:_itemId];  // obviously need property for itemId on ItemDetailViewController

}

- (void)tapRecognized:(id)sender
{
    NSLog(@"that tap was recognized with %d", [(UIGestureRecognizer *)sender view].tag);
    _itemId=[(UIGestureRecognizer *)sender view].tag;

在ItemDetailiViewController.h。

@interface ItemDetailViewController : UIViewController

@property (nonatomic) int itemId;