NSURL object value is null - ios

Question

I am generating a URL as given by string parameters but url gives null value.
For generating URL I have implemented the following code .
NSURL *url = [[NSURL alloc] init];
NSString *strURL = #"ftp://Administrat:ABC(R%-#TRDOP#xx.xx.xx.xx/arrows.png";
url = [NSURL URLWithString:[NSString stringWithString:strURL]];
NSLog(#"URL :: %#",url);
Thanks

Answer 1

You need to escape the special characters in the url query string.
Use:
NSString *strURL = #"ftp://www.jerox.com/Administrator:#123#TRDOP#%$/arrows.png";
strURL =[strURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url = [NSURL URLWithString:strURL];
NSLog(#"URL :: %#",url);
Also I need to mention some mistakes in your code:
No need to allocate and init the NSURL object, because you are again assigning another object to that pointer
No need of using stringWithString: there

Answer 2

Make use of the following code.
NSString *sURL = #"ftp://www.jerox.com/Administrator:#123#TRDOP#%$/arrows.png";
NSURL *url = [[NSURL alloc] initWithString:[sURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
NSLog(#"URL :: %#",url);

Answer 3

From your posted code sample, here is the problem:
NSString *strURL = #"ftp://Administrat:ABC(R%-#TRDOP#xx.xx.xx.xx/arrows.png";
Look carefully at the authority component (username, password and host). An # symbol is used in URLs to separate username & password from the host. Because your password contains an # character, it must be percent escaped. The percent encoding of # is %40, giving you instead the code:
NSString *strURL = #"ftp://Administrat:ABC(R%-%40TRDOP#xx.xx.xx.xx/arrows.png";
You really ought to be escaping other URL-specific characters in there too, like the lone % symbol.