我的 iPhone 程序使用 URL 架构并成功工作，要做到这一点只需 休息本教程。好的，当程序启动时，我使用该函数
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions;
，并且 launchOptions 包含我需要的所有数据。这个问题就解决了。但是，当我使用 URL 模式并且我的程序打开时。好的，程序打开并显示其中的最后一个窗口。但我想去另一个地方。
问题：我如何知道该程序已从具有 URL 架构的 Web 浏览器重新激活？我必须使用什么功能？我没有找到任何可以解决它的方法。
my iPhone program use a URL schema and work successful, to do this is only fallow this tutorial. Ok, when the program start, i use the function
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions;
And the launchOptions have all data that I need. This problem is solved. BUT, and when I use the URL schema and my program IS OPEN. Ok, the program open and show the last windows that was in it. But I want to go to another place.
The question: How can I know the program was reactivated from a web browser with a URL Schema? What function I have to use? I didn't found any that can solve it.
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是的...正确的功能是这个！！！！ （感谢onnoweb）
这有点晦涩，为什么不使用“wasOpenedByURL”。呸..我必须使用对象中的所有函数才能知道它的作用。
YES... The correct function is this!!!! (Thanks onnoweb)
It's a little obscure, why not use "wasOpenedByURL". Bah.. I have to use all function in a object to know what it do.