iOS : How to do proper URL encoding? - ios

Question

I'm unable to open a URL into UIWebView so I've seached & found that I need to encode URL, so I tried to encode it but, I've facing problem in URL encoding : My URL is http://somedomain.com/data/Témp%20Page%20-%20Open.html (It's not real URL).
I'm concerned with %20 that I tried to replace using stringByReplacingOccuranceOfString:#"" withString:#"" , it give me the URL I wanted like http://somedomain.com/data/Témp Page - Open.html However its not opening in UIWebView but amazingly it opens in Safari & FireFox perfect. Even I open unencoded URL its automatically converts and open the page I'm looking for.
I've google for URL encoding & it points me to different results I already checked but no results help me out!! I tried different functions answers in different URL encoding question but it just changed all special characters and make my URL like, http%3A%2F%2Fsomedomain.com%2Fdata%2FT... which can't open in UIWebView and even in any browser.
It gives the following Error Log in UIWebView delegate
- (void)webView:(UIWebView *)webView didFailLoadWithError:(NSError *)error { }
Error Code : 101
& Description : Error Domain=WebKitErrorDomain Code=101 "The operation couldn’t be completed. (WebKitErrorDomain error 101.)" UserInfo=0x6e4cf60 {}

Answer 1

The answer #Dhaval Vaishnani provided is only partially correct. This method treats the ?, = and & characters as not to be encoded, since they're valid in an URL. Thus, to encode an arbitrary string to be safely used as a part of an URL, you can't use this method. Instead you have to fall back to using CoreFoundation and CFURLRef:
NSString *unsafeString = #"this &string= confuses ? the InTeRwEbZ";
CFStringRef safeString = CFURLCreateStringByAddingPercentEscapes (
NULL,
(CFStringRef)unsafeString,
NULL,
CFSTR("/%&=?$#+-~#<>|\\*,.()[]{}^!"),
kCFStringEncodingUTF8
);
Don't forget to dispose of the ownership of the resulting string using CFRelease(safeString);.
Also, it seems that despite the title, OP is looking for decoding and not encoding a string. CFURLRef has another, similar function call to be used for that:
NSString *escapedString = #"%32%65BCDEFGH";
CFStringRef unescapedString = CFURLCreateStringByReplacingPercentEscapesUsingEncoding (
NULL,
(CFStringRef)escapedString,
CFSTR(""),
kCFStringEncodingUTF8
);
Again, don't forget proper memory management.

Answer 2

I did some tests and I think the problem is not really with the UIWebView but instead that NSURL won't accept the URL because of the é in "Témp" is not encoded properly. This will cause +[NSURLRequest requestWithURL:] and -[NSURL URLWithString:] to return nil as the string contains a malformed URL. I guess that you then end up using a nil request with -[UIViewWeb loadRequest:] which is no good.
Example:
NSLog(#"URL with é: %#", [NSURL URLWithString:#"http://host/Témp"]);
NSLog(#"URL with encoded é: %#", [NSURL URLWithString:#"http://host/T%C3%A9mp"]);
Output:
2012-10-02 12:02:56.366 test[73164:c07] URL with é: (null)
2012-10-02 12:02:56.368 test[73164:c07] URL with encoded é: http://host/T%C3%A9mp
If you really really want to borrow the graceful handling of malformed URLs that WebKit has and don't want to implement it yourself you can do something like this but it is very ugly:
UIWebView *webView = [[[UIWebView alloc]
initWithFrame:self.view.frame]
autorelease];
NSString *url = #"http://www.httpdump.com/texis/browserinfo/Témp.html";
[webView loadHTMLString:[NSString stringWithFormat:
#"<script>window.location=%#;</script>",
[[[NSString alloc]
initWithData:[NSJSONSerialization
dataWithJSONObject:url
options:NSJSONReadingAllowFragments
error:NULL]
encoding:NSUTF8StringEncoding]
autorelease]]
baseURL:nil];

Answer 3

The most straightforward way is to use:
NSString *encodedString = [rawString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
iDhaval was close, but he was doing it the other way around (decoding instead of encoding).
Anand's way would work, but you'll most likely have to replace more characters than spaces and new lines. See the reference here:
http://en.wikipedia.org/wiki/Percent-encoding#Percent-encoding_reserved_characters
Hope that helps.

Answer 4

It's very simple to encode the URL in iPhone. It is as following
NSString* strURL = #"http://somedomain.com/data/Témp Page - Open.html";
NSURL* url = [NSURL URLWithString:[strURL stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
It's a perfect way to encode the URL, I am using it and it's perfectly work with me.
Hope it will help you!!!

Answer 5

This may useful to someone who's reach to this question for URL encoding, as my question likely different which has been solved and accepted, this is the way I used to do encoding,
-(NSString *)encodeURL:(NSString *)urlString
{
CFStringRef newString = CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (CFStringRef)urlString, NULL, CFSTR("!*'();:#&=+#,/?#[]"), kCFStringEncodingUTF8);
return (NSString *)CFBridgingRelease(newString);
}

Answer 6

You can try this
NSString *url = #"http://www.abc.com/param=Hi how are you";
NSString* encodedUrl = [url stringByAddingPercentEscapesUsingEncoding:
NSASCIIStringEncoding];

Answer 7

I think this will work for you
[strUrl stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet]
the Native method for URL Encoding.

Answer 8

Swift 4.x
let originalString = "https://www.somedomain.com/folder/some cool file.jpg"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
print(escapedString!)

Answer 9

You probably need to break the URL down into it's constituent parts and then URL encode the host and path but not the scheme. Then put it back together again.
Create an NSURL with the string and then use the methods on it such as host, scheme, path, query, etc to pull it apart. Then use CFURLCreateStringByAddingPercentEscapes to encode the parts and then you can put them back together again into a new NSURL.

Answer 10

can you please Try this out.
//yourURL contains your Encoded URL
yourURL = [yourURL stringByReplacingOccurrencesOfString:#" " withString:#"%20"];
yourURL = [yourURL stringByReplacingOccurrencesOfString:#"\n" withString:#""];
NSLog(#"Keyword:%# is this",yourURL);
I am not sure,but I have solved using this in my case.
Hope this will solve yours.