Basically, i'm trying to program a "tweet this" button from inside my application. Depending on their spot in the application, they can click the tweet button and it'll shoot them out to Safari with a tweet message that changes depending on where they are.
In order to create URLs, I have to escape the query string that I want to put in the NSUrl object. So, I do this:
NSString* escapedTweet = [@"Some String With Spaces" stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding];
After concatenating the base, my url comes out "http://www.twitter.com/home/?status=Some&20String%20With%20Spaces" - looked at it in the debugger and this is definitely the value (as expected). Now, I create my URL and launch safari:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:escapedUrlString]];
But here's where the issue comes up: OpenUrl appears to be escaping my percent signs, so the actual URL that Safari goes to is "http://www.twitter.com/home/?status=Some%2520String%2520With%2520Spaces", which is obviously a problem since twitter creates the status message as "Some%20String%20With%20Spaces".
NSUrl will NOT allow me to create a URL with spaces in it, so i'm completely lost as to how to get my URLs to just include a %20. Has anyone else run into this issue or found a workaround?
I'm running on an iPad with an up to date OS, not sure if it's the same issue on the iPhone.
Edit: In a nutshell, how do I get openUrl to open http://www.twitter.com/home/?status=Some%20Url%20With%20Spaces without escaping my percent signs and creating a URL like http://www.twitter.com/home/?status=Some%2520Url%2520With%2520Spaces?