ios - 侧滑菜单

Question

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

if ([segue isKindOfClass:[SWRevealViewControllerSegueSetController class]]) {


    SWRevealViewControllerSegueSetController *swSegue = (SWRevealViewControllerSegueSetController*) segue;
    swSegue.performBlock = ^(SWRevealViewControllerSegueSetController* rvc_segue, UIViewController* svc,UIViewController* dvc){

    UINavigationController* navController = (UINavigationController*)self.revealViewController.frontViewController;
        [self.revealViewController setFrontViewPosition:FrontViewPositionLeft animated:YES];
    };

}

}

我调用它是为了在单击 TableView 中的特定单元格时调用我的不同 UI，但它们给我一个错误“在类型 SWRevealViewC ontrollerSegueSetController 上找不到属性 performblock”

求助!!!提前致谢。

Answer 1

是的，我有同样的错误

但是在您的 SWRevealViewController.h 文件的最后一节中，删除这段代码中的注释

@interface SWRevealViewControllerSegue : UIStoryboardSegue     

@property (nonatomic, strong) void(^performBlock)(
     SWRevealViewControllerSegue* segue, UIViewController* svc,
     UIViewController* dvc );

@end

与最后一节的 SWRevealViewController.m 文件相同，从此代码中删除此注释

@implementation SWRevealViewControllerSegue  // DEPRECATED

-(void)perform {
    if ( _performBlock )
        _performBlock( self, self.sourceViewController,self.destinationViewController ); 
}

@end

我认为这会在你的项目中起作用