2014-05-10 6 views
-1

기존 계정 중 하나를 선택합니다. viewController이 선택한 계정을 전달하고 싶습니다. 하지만 난이 오류가 -> 여기ACAccount 인스턴스로 보낸 인식 할 수없는 선택기 (NSInvalidArgumentException)

내 코드 "인식 할 수없는 선택기 인스턴스로 전송"

선택 계정 tableView에 :

-(void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath{ 
    AccountsTableViewCell *selectedRow = [self.accountTable cellForRowAtIndexPath:indexPath]; 
    choosenAccount = selectedRow.twitAccount; 
    [self performSegueWithIdentifier:@"selectAccount" sender: nil]; 

} 

SEGUE :

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender 
{ 
    if ([segue.identifier isEqualToString:@"selectAccount"]) { 
     TwitterFollowersViewController *vc = (TwitterFollowersViewController *) segue.destinationViewController; 
     vc.twitterAccount = choosenAccount; 
    } 
} 

ViewController.h 파일 :

@interface TwitterFollowersViewController : UIViewController 
     @property (retain, nonatomic) ACAccount *twitterAccount; 
@end 

LOG :

2014-05-10 23:54:13.297 tweet++[10143:60b] -[UIViewController setTwitterAccount:]: unrecognized selector sent to instance 0xa45d870 
2014-05-10 23:54:13.300 tweet++[10143:60b] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[UIViewController setTwitterAccount:]: unrecognized selector sent to instance 0xa45d870' 
*** First throw call stack: 
(
    0 CoreFoundation      0x0194e1e4 __exceptionPreprocess + 180 
    1 libobjc.A.dylib      0x016cd8e5 objc_exception_throw + 44 
    2 CoreFoundation      0x019eb243 -[NSObject(NSObject) doesNotRecognizeSelector:] + 275 
    3 CoreFoundation      0x0193e50b ___forwarding___ + 1019 
    4 CoreFoundation      0x0193e0ee _CF_forwarding_prep_0 + 14 
    5 tweet++        0x000071ac -[TwitterAccountsViewController prepareForSegue:sender:] + 268 
    6 UIKit        0x008f1efa -[UIStoryboardSegueTemplate _perform:] + 156 
    7 UIKit        0x004ae41c -[UIViewController performSegueWithIdentifier:sender:] + 72 
    8 UIKit        0x0f20f8fc -[UIViewControllerAccessibility(SafeCategory) performSegueWithIdentifier:sender:] + 63 
    9 tweet++        0x00006fff -[TwitterAccountsViewController tableView:didSelectRowAtIndexPath:] + 287 
    10 UIKit        0x004779a1 -[UITableView _selectRowAtIndexPath:animated:scrollPosition:notifyDelegate:] + 1513 
    11 UIKit        0x00477b14 -[UITableView _userSelectRowAtPendingSelectionIndexPath:] + 279 
    12 UIKit        0x0047c10e __38-[UITableView touchesEnded:withEvent:]_block_invoke + 43 
    13 UIKit        0x003ab0aa ___afterCACommitHandler_block_invoke + 15 
    14 UIKit        0x003ab055 _applyBlockToCFArrayCopiedToStack + 403 
    15 UIKit        0x003aae76 _afterCACommitHandler + 532 
    16 CoreFoundation      0x0191636e __CFRUNLOOP_IS_CALLING_OUT_TO_AN_OBSERVER_CALLBACK_FUNCTION__ + 30 
    17 CoreFoundation      0x019162bf __CFRunLoopDoObservers + 399 
    18 CoreFoundation      0x018f4254 __CFRunLoopRun + 1076 
    19 CoreFoundation      0x018f39d3 CFRunLoopRunSpecific + 467 
    20 CoreFoundation      0x018f37eb CFRunLoopRunInMode + 123 
    21 GraphicsServices     0x01bab5ee GSEventRunModal + 192 
    22 GraphicsServices     0x01bab42b GSEventRun + 104 
    23 UIKit        0x0038df9b UIApplicationMain + 1225 
    24 tweet++        0x000076ad main + 141 
    25 libdyld.dylib      0x04fdc701 start + 1 
) 
libc++abi.dylib: terminating with uncaught exception of type NSException 
+1

storyboard의 클래스 이름을 확인하면 UIViewController가 아니라 TwitterFollowersViewController라고 생각합니다. – Pawan

+0

Google "인식 할 수없는 선택기". 몇 가지 참조 문헌을 연구하십시오. –

+0

(prepareForSegue의'vc'는 TwitterFollowersViewController가 아닙니다.) –

답변

1

체크 스토리 보드에서 클래스 이름, 그것보다는 TwitterFollowersViewController을 자신의 UIViewController 생각합니다.

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