nsurl

I need to load an image from a url and set it inside an UIImageView; the problem is that I don't know the exact size of the image, then how can I show the image correctly? Just use the size property of UIImage, for example: NSURL *url = [NSURL URLWithString:path]; NSData *data = [NSData dataWithContentsOfURL:url]; UIImage *img = [[UIImage alloc] initWithData:data]; CGSize size = img.size; In swift: var url = NSURL.URLWithString("http://www.example.com/picture.png") var data = NSData(contentsOfURL : url) var image = UIImage(data : data) image.size // if you need it In swift regarding using

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 NSString* webName = [localisationName stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; NSString* stringURL = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%@,Montréal,Communauté-Urbaine-de-Montréal,Québec,Canadae&output=csv&oe=utf8&sensor=false&key=", webName]; NSURL* url = [NSURL URLWithString:stringURL]; You need to escape the non-ASCII characters in your hardcoded URL as well: //localisationName is a arbitrary string here

I have an app where the user can choose an image either from the built-in app images or from the iphone photo library. I use an object Occasion that has an NSString property to save the imagePath . Now in the case of the built-in app images I do get the file name as an NSString an save in the [occasion imagePath] . But in the 2nd case where the user picks an image form the photo library I get an NSURL which I want to convert to an NSString to be able to save it in [occasion imagePath ]. Is it possible to convert the NSURL to an NSString ? Randall In Objective-C: NSString *myString = myURL

Apple with iOS 10 has deprecated openURL: for openURL:option:completionHandler If I have: [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.google.com"]]; How it will become? options:<#(nonnull NSDictionary *)#> in detail [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.google.com"] options:<#(nonnull NSDictionary<NSString *,id> *)#> completionHandler:nil]; Thanks Update options:@{} For empty dictionary with no key and value http://useyourloaf.com/blog/querying-url-schemes-with-canopenurl/ Write like this. Handle completionHandler

I am trying to convert a String to NSURL and my code for that is Below: var url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitude),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US" println("This is String: \(url)") var remoteUrl : NSURL? = NSURL(string: url) println("This is URL: \(remoteUrl)") And console prints something like this: This is String: https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997|-34.4356434,150.8858692|-34.4250728,150.89314939999997|-34

I have converted the code for making a phone call from Objective-C to Swift, but in Objective-C, we can set the type of the URL that we like to open (e.g. telephone, SMS, web) like this: @"tel:xx" @"mailto:info@example.es" @"http://stackoverflow.com" @"sms:768number" The code in Swift is: UIApplication.sharedApplication().openURL(NSURL(string : "9809088798") I read that have not released any scheme parameter for tel: , but I don't know if Swift can detect if the string is for making a phone call, sending email, or opening a website. Or may I write: (string : "tel//:9809088798") ? I am pretty