- [UIPopoverController dealloc] 在 popover 仍然可见时达到

Question

我有一个类 (ViewOpenAppointments)，我在其中创建和显示 UIPopover。这是在我的 .h 文件中定义弹出框的代码：

@interface ViewOpenAppointments : UIView  {

}

@property (nonatomic, retain) UIPopoverController *popoverController;
-(void)createOpenAppointmentsPopover: (UIButton *) obViewOpenAppts;
@end

我检查了代码，如果弹出框可见，则将其关闭。这是代码：

    //  create popover
UIViewController* popoverContent = [[UIViewController alloc] init];
//    UIView *popoverView = [[UIView alloc] initWithFrame:CGRectMake(0, 0, 650, 416)];
ViewOpenAppointments *popoverView = [[ViewOpenAppointments alloc]     initWithFrame:CGRectMake(0, 0, 650, 416)];

popoverView.backgroundColor = [UIColor whiteColor];
popoverContent.preferredContentSize = CGSizeMake(650.0, 416.0);

//  create the popover controller
popoverController = [[UIPopoverController alloc]  initWithContentViewController:popoverContent];
popoverController.delegate = (id)self;
[popoverController setPopoverContentSize:CGSizeMake(650, 416) animated:NO];

if ([popoverController isPopoverVisible]) {
    [popoverController dismissPopoverAnimated:YES];
}

[popoverController presentPopoverFromRect:CGRectMake(650, 416, 10, 50) inView: obViewOpenAppts
                 permittedArrowDirections:UIPopoverArrowDirectionUp animated:YES];

问题是关闭弹出框的代码永远不会被命中，这意味着它不可见。但我仍然收到错误消息（在问题标题中描述）。

我做错了什么？

Answer 1

这是一个完整的popover管理示例：

@interface ViewController () <UIPopoverControllerDelegate>
@property (nonatomic, strong) UIPopoverController* currentPop;
@end

@implementation ViewController     

-(IBAction)doPopover1:(id)sender {
    Popover1View1* vc = [[Popover1View1 alloc] initWithNibName:@"Popover1View1" bundle:nil];

    UIPopoverController* pop = [[UIPopoverController alloc] initWithContentViewController:vc];
    self.currentPop = pop;
    [pop presentPopoverFromBarButtonItem:sender
                permittedArrowDirections:UIPopoverArrowDirectionAny
                                animated:YES];
    pop.passthroughViews = nil;
    // make ourselves delegate so we learn when popover is dismissed
    pop.delegate = self;
}

- (void)popoverControllerDidDismissPopover:(UIPopoverController *)pc {
    self.currentPop = nil;
}

通过检查self.currentPop，您可以确保您不会同时呈现两个弹出框（无论如何都是非法的）。