ios - 知道 UITapGestureRecognizer 从 UIView 发送什么标签

Question

我不知道我的观点有什么标签。我正在尝试为我的 UITapGestureRecognizers 使用一种方法，然后从 View 中获取 tag。但它不起作用。我试过这两个:

-(void)playGame:(UITapGestureRecognizer *)gr{ // Method that is getting called off my TapGestureRecognizer
    [self performSegueWithIdentifier:@"GameSegue" sender:self]; // Calling the segue
}

-(void)performSegueWithIdentifier:(NSString *)identifier sender:(id)sender{
    UITapGestureRecognizer *tap = (UITapGestureRecognizer *)sender; // Try casting it to an GR
    NSLog(@"%i",tap.view.tag);

    UIView *view = (UIView *)sender; // Try casting it to an view for the tag
    NSLog(@"%i",view.tag);
}

但没有任何效果。我怎么解决这个问题？转换为 UIView 只会给我 self 而不是用于获取标签的 UIView。所以我得到 Tag is not recognized for that instance。

Answer 1

您应该发送 gesture recogniser而不是发送 self .使用 gr而不是 self因为你应该通过 gr作为发件人和来自 gr您可以使用 sender.view.tag 获取 View 标签

-(void)playGame:(UITapGestureRecognizer *)gr{ // Method that is getting called off my TapGestureRecognizer
    [self performSegueWithIdentifier:@"GameSegue" sender:gr]; // Calling the segue
}


-(void)performSegueWithIdentifier:(NSString *)identifier sender:(id)sender{

     //In case of multiple segue its good practice to check identifier
     if([identifier isEqualToString:@"GameSegue"]){

        UITapGestureRecognizer *tap = (UITapGestureRecognizer *)sender; // Try casting it to an GR
       //This will give your tag
       NSLog(@"%i",tap.view.tag);  
   }

}

传递 self意味着你正在通过整个 view或 viewController在任何情况下self被使用。但是作为gr包含添加它的 View ，因此它会给你正确的标签。在你的代码中 performSegueWithIdentifier这条线 UITapGestureRecognizer *tap = (UITapGestureRecognizer *)sender;将获得 View 或其他东西，因此您试图将错误转换为特征识别器。