假设我有3个视图控制器标记为"A","B"和"C".现在,"A"是窗口的rootViewController,当点击按钮时它以模态方式呈现"B".在"B"中,当按下一个按钮时,它应该被"A"解雇,然后"A"将立即以模态方式呈现C.如何做到这一点？这是我希望实现这一目标的代码,但我没有成功.
在"A"viewController中,我声明了一个属性,用于在"B"viewController被"A"解除时,在要调用的头文件中保存一个块.
@property (nonatomic, copy) void (^presentZapLaunch)(void);
这是"A"viewController呈现"B"的现有方法
-(void)presentNextViewCon
{
CYCGestureZapZapViewController *gestureViewCon = [[CYCGestureZapZapViewController alloc]init];
if (!self.presentZapLaunch) {
__weak CYCZapZapViewController *weakRefCon = self;
self.presentZapLaunch = ^{
CYCZapZapViewController *preventWeakRefCon = weakRefCon;
CYCZapZapLaunchViewController *zapLaunch = [[CYCZapZapLaunchViewController alloc]init];
NSLog(@"Called");
[preventWeakRefCon presentViewController:zapLaunch animated:YES completion:nil];
};
}
[self presentViewController:gestureViewCon animated:YES completion:nil];
}
这是由"A"解雇的"B"解雇方法,"A"应立即出现"C"
-(void)presentNextViewCon
{
NSLog(@"Hello");
[self.presentingViewController dismissViewControllerAnimated:self completion:^{[(CYCZapZapViewController *)self.presentingViewController presentZapLaunch];}];
}
*请注意,我使用"A"视图控制器作为窗口的rootViewController,"A"以模式方式呈现"B"视图控制器.所有"A","B"和"C"都是视图控制器.