NSURL with string

I have problem with NSURL. I am trying to create NSURL with string

code

    NSString *prefix = (@"tel://1234567890 ext. 101");
    NSString *dialThis = [NSString stringWithFormat:@"%@", prefix];
    NSURL *url = [[NSURL alloc] initWithString:dialThis];
    NSLog(@"%@",url);

also tried

    NSURL *url = [NSURL URLWithString:dialThis];

but it gives null . what is wrong ?

Thanks..

Your problem is the unescaped spaces in the URL. This, for instance, works:

    NSURL *url = [NSURL URLWithString:@"tel://1234567890x101"];

Edit: As does this..

    NSURL *url2 = [NSURL URLWithString:[@"tel://1234567890 ext. 101"
        stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

Before passing any string as URL you don't control, you have to encode the whitespace:

 NSString *dialThis = [prefix stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
 // tel://1234567890%20ext.%20101

As a side note, iOS is not going to dial any extension. The user will have to do that manually.

From Apple URL Scheme Reference: Phone Links:

To prevent users from maliciously redirecting phone calls or changing the behavior of a phone or account, the Phone application supports most, but not all, of the special characters in the tel scheme. Specifically, if a URL contains the * or # characters, the Phone application does not attempt to dial the corresponding phone number.

Im not sure the "ext." in phone number can be replce by what value? but you can try like this,

NSString *prefix = [NSString stringWithString: @"tel://1234567890 ext. 101"];

    NSString *dialThis = [NSString stringWithFormat:@"%@", prefix];

    NSURL *url = [NSURL URLWithString:[dialThis stringByReplacingOccurrencesOfString:@" ext. " withString:@"#"]];
// it might also represent by the pause symbol ','.

you can go to find the ext. is equivalent to what symbol in the phone, then replace it.

but dunno it can be work in actual situation or not....

As with iOS 9.0,

stringByAddingPercentEscapesUsingEncoding:

has been deprecated.

Use the following method for converting String to NSURL.

let URL = "URL GOES HERE"
let urlString = URL.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLFragmentAllowedCharacterSet())

If you've got something you think should be a URL string but know nothing about how URL strings are supposed to be constructed, you can use NSURL's URLWithDataRepresentation:relativeToURL: method. It parses the URL string (as bytes in an NSData) and percent-encodes characters as needed. Use the NSUTF8StringEncoding for best results when converting your NSString to NSData.

NSURL *url = [NSURL URLWithDataRepresentation:[@"tel:1234567890 ext. 101" dataUsingEncoding:NSUTF8StringEncoding] relativeToURL:nil];
NSLog(@"%@",url);

creates a URL with the string 1234567890%20ext.%20101

It attempts to do the right thing. However, for best results you should find the specification for the URL scheme you using and follow it's syntax to create your URL string. For the tel scheme, that is https://www.rfc-editor.org/rfc/rfc3966.

P.S. You had "tel://" instead of "tel:" which is incorrect for a tel URL.

Try this one, It works for me....

NSString *prefix = (@"tel://1234567890 ext. 101");
NSString *dialThis = [NSString stringWithFormat:@"%@", prefix];
NSURL *url = [NSURL URLWithString:[queryString stringByReplacingOccurrencesOfString:@" " withString:@"%20"]];
NSLog(@"%@",url);

Make an extension for use in any part of the project as well:

extension String {
    var asNSURL: NSURL! {
        return NSURL(string: self)
    }
}

From now you can use

let myString = "http://www.example.com".asNSURL

or

myString.asNSURL