设置UIStoryboardSegue子类的属性

Question

我创建了一个UIStoryboardSegue的子类，以便通过可选的动画实现模态轮回。

（的UIStoryboardSegue子类）： .H

#import <UIKit/UIKit.h> 

@interface ModalSegue_OptionalAnimation : UIStoryboardSegue 
@property (readwrite) BOOL withAnimation; 
@end 

.M

#import "ModalSegue_OptionalAnimation.h" 

@implementation ModalSegue_OptionalAnimation 

-(void) perform{ 
    BOOL _withAnimation_Va = self.withAnimation; 
    [[[self sourceViewController] navigationController] pushViewController:[self destinationViewController] animated:_withAnimation_Va]; 
} 

@end 

但是我不确定现在如何调用从外面这个属性。

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender { 
    if([[segue identifier] isEqualToString:@"segue_qVC_to_cVC_checkAnswer"]) { 
     CheckAnswerViewController *cVC = [segue destinationViewController]; 

     if(segue_QVC_ISEXAM) { 
      //Something like this: 
      //segue.withAnimation = NO; 
      //Settings the property to NO is like 'I dont animation when performing the segue' 
     } 
    .... 

在我的故事板中，我已经将刚刚创建的类设置为自定义。

Answer 1

尝试这样：

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender { 
if([[segue identifier] isEqualToString:@"segue_qVC_to_cVC_checkAnswer"]) { 
    CheckAnswerViewController *cVC = [segue destinationViewController]; 

    if(segue_QVC_ISEXAM) { 
     ModalSegue_OptionalAnimation *customSegue = (ModalSegue_OptionalAnimation *)segue; 
     customSegue.withAnimation = NO; 

     //Something like this: 
     //segue.withAnimation = NO; 
     //Settings the property to NO is like 'I dont animation when performing the segue' 
    } 
....