Question

I have a button called as Sync. On clicking on it, I need to display a popOverController which displays a few items in UITableView. I have used the following lines of code in .m file to declare the UIButton in ViewWillAppear function :

UIBarButtonItem *addButton = [[UIBarButtonItem alloc] initWithTitle:@"Sync"
                                                               style:UIBarButtonItemStyleBordered
                                                              target:self
                                                              action:@selector(syncAction)] ;

The code for syncAction is :

- (void)syncAction:(id)sender{


Sync = [[SyncTableViewController alloc] initWithStyle:UITableViewStylePlain];


Sync.syncDelegate = self;
self.SyncTableViewPopover = [[UIPopoverController alloc] 
                                   initWithContentViewController:Sync];               

[self.SyncTableViewPopover presentPopoverFromBarButtonItem:sender 
                                        permittedArrowDirections:UIPopoverArrowDirectionDown animated:YES];


}

But, however, on running the application I am getting the following error ::

[splitViewXXXXViewController syncAction]: unrecognized selector sent to instance 0x6b70660

2012-07-02 15:35:59.549 splitView[895:f803] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[splitViewXXXXViewController syncAction]: unrecognized selector sent to instance 0x6b70660'

I am unable to sort it out. Can someone help me to sort it out ?? Thanks and Regards.

Answer 1

You have to use the @selector(syncAction:)

UIBarButtonItem *addButton = [[UIBarButtonItem alloc] initWithTitle:@"Sync"
                                                           style:UIBarButtonItemStyleBordered
                                                          target:self
                                                          action:@selector(syncAction:)] ;

This will resolve you issue