The recommended encoding scheme to use is UTF-8.
A space is assigned number 32, which is 20 in hexadecimal. When you see “%20,” it represents a space in an encoded URL, for example, http://www.example.com/products%20and%20services.html.
Since URLs often contain characters outside the ASCII set, the URL has to be converted into a valid ASCII format. URL encoding replaces unsafe ASCII characters with a "%" followed by two hexadecimal digits. URLs cannot contain spaces. URL encoding normally replaces a space with a plus (+) sign or with %20.
addingPercentEncoding(withAllowedCharacters:) Returns a new string made from the receiver by replacing all characters not in the specified set with percent-encoded characters.
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.
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];
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.
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!!!
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With