ios - 无法识别的选择器发送到实例 UITableView 单元格推送到 TabbarController

Question

我只是尝试将 mid 发送到 customerDetailviewcontroller，然后 segue 首先进入 tabbar Controller 。

有我的prepareforsegue方法

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    if ([segue.identifier isEqualToString:@"showDetail"]) {
        NSIndexPath *indexPath = [self.tableView indexPathForSelectedRow];
        ApiClass *apiClass = [self.customerCards objectAtIndex:indexPath.row];
        CustomerDetailViewController *customerDetailViewController = (CustomerDetailViewController *)segue.destinationViewController;
        customerDetailViewController.mid = apiClass.mid;



    }
}

_mid NSString * @"1642"

那是错误。

Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[UITabBarController setMid:]: unrecognized selector sent to instance 0x126e53580'
*** First throw call stack:
(0x1838d6530 0x1948ac0e4 0x1838dd5f4 0x1838da3ac 0x1837dec4c 0x1000ae2b4 0x1886b401c 0x18820ce14 0x1882ca6dc 0x188166b8c 0x1880d85f0 0x18388ed98 0x18388bd24 0x18388c104 0x1837b91f4 0x18cbdb6fc 0x18814a10c 0x1000e2b08 0x194f2aa08)
libc++abi.dylib: terminating with uncaught exception of type NSException

这是我从 json 获取的数据。所以它不是 nil。

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Answer 1

我弄清楚了。主要问题是我将 segue 推送到标签栏 Controller 。我用这段代码修复了它。

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
if ([segue.identifier isEqualToString:@"showDetail"]) {
    NSIndexPath *indexPath = [self.tableView indexPathForSelectedRow];
    ApiClass *apiClass = [self.customerCards objectAtIndex:indexPath.row];
    UITabBarController *tabBarController = [segue destinationViewController];
    CustomerDetailViewController *customerDetailViewController = (CustomerDetailViewController *)[[tabBarController viewControllers]objectAtIndex:0];
    customerDetailViewController.mid = apiClass.mid;



}