繁体   English   中英

是否可以从另一个类调用performSegueWithIdentifier?

[英]Is it possible to call a performSegueWithIdentifier from another class?

我只是建立一个类来正确管理我的数据库和JSON请求。 现在的问题是,我该如何执行搜寻?

这是我认为的代码:

- (IBAction)loginClick:(id)sender
{
    NSString *post = [NSString stringWithFormat:@"username=test&password=test"];
    [[DataManagement sharedManager] WebServiceLogin:post];
}
- (void) showTypeView
{
    [self performSegueWithIdentifier:@"showTypeView" sender:nil];
}

在我的课上:

    -(void)connectionDidFinishLoading:(NSURLConnection *)connection
    {
...
                switch ([[response valueForKey:@"success"] intValue])
                {
                    case 0:
                    {
                        NSLog(@"error: %@ error Description: %@", [response valueForKey:@"success"], [response valueForKey:@"error_message"]);
                        break;
                    }
                    case 1:
                    {
                        LoginViewController *showView = [LoginViewController new];
                        [showView showTypeView];
                        break;
                    }
                    default:
                        break;
                }
...
    }

启动时出现错误:**

*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Receiver (<LoginViewController: 0x165afd30>) has no segue with identifier 'showTypeView''
*** First throw call stack:
(0x2592e2eb 0x250fadff 0x29e2b037 0xe1819 0xdb64f 0x25f64de1 0x25f64d99 0x25f64e8d 0x25e261ef 0x25edf04f 0xa77cab 0xa7f835 0x25e171e3 0x258415f9 0x25e170cb 0x25e16f95 0x25e16e29 0x258f1257 0x258f0e47 0x258ef1af 0x25841bb9 0x258419ad 0x26abbaf9 0x29b2dfb5 0xe3ea9 0x254f4873)
libc++abi.dylib: terminating with uncaught exception of type NSException

**

如果您使用的是segueWithIdentifier,则需要已经在Storyboard中内置了segue,并将其正确标记为“ showTypeView”。 否则,您应该使用导航控制器来推送视图控制器,或者使用self presentViewController来显示模式视图控制器。

编辑:基于Larme的注释,您可以像这样构建一个委托:

// In your class.h file
@property (weak, nonatomic)id<SegueDelegate> delegate;

// In class.m file
LoginViewController *showView = [LoginViewController new];
self.delegate = showView;
[self.delegate segue];


// In LoginViewController.h
@protocol SegueDelegate
-(void)segue;
@end

@interface LoginViewController: UIViewController <SegueDelegate>
-(void)segue;
@end

// In LoginViewController.m
@implementation LoginViewController
-(void)segue
{
    [self performSegueWithIdentifier:@"showTypeView" sender:nil];
}
@end

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM