我有一个名为TVOutViewController(.h& .m)的视图控制器,它应该处理我的外部屏幕。如何“告诉”View Controller这样做?
我已经做了什么:
NSLog(@"Current Number of screens: %i", [[UIScreen screens] count]);
if([[UIScreen screens]count] > 1) {
CGSize maxSize;
UIScreenMode *maxScreenMode;
for(int i = 0; i < [[[[UIScreen screens] objectAtIndex:1] availableModes]count]; i++)
{
UIScreenMode *current = [[[[UIScreen screens]objectAtIndex:1]availableModes]objectAtIndex:i];
if(current.size.width > maxSize.width)
{
maxSize = current.size;
maxScreenMode = current;
}
}
UIScreen *externalScreen = [[UIScreen screens] objectAtIndex:1];
externalScreen.currentMode = maxScreenMode;
现在我的阵列中有一个外部屏幕(并且已识别)。但是如何在此屏幕上添加(例如)标签?
有没有这样的方式:
Screen Handled by the TVOutViewController = TheExternalScreen //Pseudocode
[Screen Handled by the TVOutViewController addSubview: aLabel]; //Pseudocode
谢谢!
答案 0 :(得分:0)
知道了。必须定义UIWindow GLOBALLY !
答案 1 :(得分:0)
查看此示例代码: https://github.com/quellish/AirplayDemo
这几乎可以满足您的需求。