摘要: 关于形如 Presenting view controllers on detached view controllers is discouraged 的处理

使用模态跳转时，Xcode有时候会出现如下警告

Presenting view controllers on detached view controllers is discouraged <>

这样的警告代码，如果你认为你的层次之间没有问题（其实就是层次问题。present出来的模态窗口，禁止再使用present 来弹出其它的子窗口）只要把self直接模态跳转页面改成从根控制器跳转即可
解决方法：
把

ViewController *viewController = [[ViewController alloc] init];
[self presentViewController:viewController animated:YES completion:nil];

改成

AppDelegate *delegate = (AppDelegate *)[UIApplication sharedApplication].delegate;
ViewController *viewController = [[ViewController alloc] init];
[delegate.window.rootViewController presentViewController:viewController animated:YES completion:nil];

即可

方法一：就是通过上面的方法，先找到appdelegate然后获取window属性

KLViewController *rootVc = (KLViewController *)[(AppDelegate *)[UIApplication sharedApplication].delegate window].rootViewController;

这种方法除了找根控制器还能找appdelegate类里的定义的任何属性，如：

- (void) openOrcloseTheList {
    
    AppDelegate *tempAppDelegate = (AppDelegate *)[[UIApplication sharedApplication] delegate];
    if (tempAppDelegate.theListVC.closed) {
        [tempAppDelegate.theListVC openListView];
    } else {
        [tempAppDelegate.theListVC closeListView];
    }
}

方法二：直接通过keyWindow来找根控制器，更直接

KLViewController *rootVc = (KLViewController *)[UIApplication sharedApplication].keyWindow.rootViewController;

补充：

YZLoginViewController *yzLoginVc = [YZLoginViewController new];

//找根控制器方法

//1.找到appdelegate然后获取window属性
AppDelegate *delegate = (AppDelegate *)[UIApplication sharedApplication].delegate;
[delegate.window.rootViewController presentViewController:yzLoginVc animated:YES completion:nil];

//2.直接通过keyWindow来找根控制器
YZYuQingViewController *rootVc = (YZYuQingViewController *)[UIApplication sharedApplication].keyWindow.rootViewController;
[rootVc presentViewController:yzLoginVc animated:YES completion:nil];