objective-c - 将TableViewCell推送到另一个ViewController时出错

我试图将数据从TableViewCell传递到另一个ViewController。但是在另一个ViewController中没有显示数据。这是我的代码

-(void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath{


   PeripheralManager *objSelected=[device objectAtIndex:indexPath.row];

   [self prepareForSegue:@"TableDetails" sender:objSelectedDevice];
 }


-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{


   if ([segue.identifier isEqualToString:@"TableDetails"])
   {


   DetailViewController *detail=segue.destinationViewController;
   detail.dataArray=device;
   }

 }

错误信息

nested push animation can result in corrupted navigation bar
2012-10-24 12:01:39.805 [3182:707] nested push animation can result in corrupted navigation bar
2012-10-24 12:01:40.164 [3182:707] Finishing up a navigation transition in an unexpected state. Navigation Bar subview tree might get corrupted.
2012-10-24 12:01:40.167 [3182:707] Finishing up a get navigation transition in an unexpected state. Navigation Bar subview tree might corrupted.

最佳答案

删除您的多余代码只有这样做-

在DetailViewController.h中

@property(nonatomic, retain)NSMutableArray *dataArray;

在DetailViewController.m中

@synthesize dataArray = _dataArray;

现在在TableViewController.m中，只需编写以下内容-

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
   if([segue.identifier isEqualToString:@"TableDetails"])
   {
     DetailViewController *detailViewObject = segue.destinationViewController;
     detailViewObject.dataArray = anyArray;
   }
}

我在这里通过NSMutableArray。