如何注册应用程序以响应自定义URL方案打开请求？

Question

Like this:

NSString *stringURL = @"appname://";
NSURL *url = [NSURL URLWithString:stringURL];
[[UIApplication sharedApplication] openURL:url];

I slightly remember it was necessary to write a value-key to Info.plist. How?

Answer 1

Add this to plist .

The app will by called by @"readtext://" url

Answer 2

This seems to be the question that I answered (with screenshots & source code) here .

And I've posted a full walkthrough of how to do this on my blog .