I have an NSURL:开发者_如何转开发 serverCall?x=a&y=b&z=c What is the quickest and most efficient way to get the value of y?
I am using the new NSURL bookmark data API introduced in OS X 10.6 to store an \"alias\" to a file system resource. When I use
I have the following stri开发者_Go百科ng... NSString *googleSearchString = @\"http://www.google.com/search?q=lyrics+%22Tænder+På+Dig%22+%22Jakob+Sveistrup%22\";
I am confused by Apple\'s documentation about the NSURL class开发者_如何学C. In NSURL, they say the following:
Having a problem getting the URL for a resource for some reason:This code is in viewDidLoad, and it\'s worked in other applications, but not here for some reason:
Apple says, NSURL is developed using RFC 1738 (and some othe开发者_Go百科rs). Now RFC 1738 specifies only that an web URL has a scheme and a scheme specific part.
I\'m using ASIHTTP and trying to perform a GET request for a site: NSURL *url = [[NSURL URLWithString:@\"/verifyuser.aspx?user=%@\" relativeToURL:@\"http://domain.com\"],userNameretain];
it may be very easy, but I don\'t seems to find out why is URLWithString: returning nil here. //localisationName is a arbitrary string here
To open Google Maps with directions, i\'m using a formatted NSString inside a NSURL. But it doesn\'t work with [[UIApplication sharedApplication] openURL:nsurl];
I have problem with NSURL and Thailand language for example \"http://www.xyp.com?var=ไทย\" it does\'t work but if you use
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