For example:
"http://www.youtube.com" --> ".com"
"http://www.google.com.gr" --> ".com.gr"
"https://made.in.china.chinesewebsite.닷컴" --> ".닷컴"
The host of a URL is a dead-end.
NSURL *URL;
[URL host];
Get the host of the url, break it up at the . delimiter, and take the last component.
- (NSString *) tldOfURL: NSURL *theURL; {
NSString *host = theURL.host;
NSArray *parts = [host componentsSeparatedByString: #"."];
if parts.count < 2 {
return nil;
}
return parts.lastObject;
}
At first my answer was to use NSURLComponents, as in most cases that class does all the heavy lifting for you, but it looks like it doesn't have a mechanism for extracting the TLD.
Note that the above will return nil if the host name does not contain at least one period (.) and will return an empty string if the host ends with a period. Both of these cases should be treated as errors.
P.S. I'm starting to get rusty at Objective-C syntax. Hard to believe, given how many years I's spent writing Objective-C!
My code:
NSString *urlString = [NSString stringWithFormat:#"%#%#%#%#", #"http://api.search.live.net/json.aspx?Appid=am hiding the appid &query=",text,#"&sources=web&web.offset=",offvalue];
NSURL *url = [NSURL URLWithString:urlString];
NSData *data = [NSData dataWithContentsOfURL:url];
The JSON response:
SearchResponse = {
Errors = (
{
Code = 1002;
HelpUrl = "http://msdn.microsoft.com/en-us/library/dd251042.aspx";
Message = "Parameter has invalid value.";
Parameter = "SearchRequest.AppId";
Value = " am hiding the value";
}
);
Query = {
SearchTerms = iphone;
};
Version = "2.2";
};
}
What wrong am I doing here.Can anyone please rectify my query??How to use version 2.2 of the bing API
What is the value of 'text', is this one word or multiple words? You'll need to perform URL_Encode on the parameter 'text' in order to have a valid URL.
See Objective-C and Swift URL encoding
Be sure only to have the URL_Encoding on the text-object and not on the whole URL, otherwise "http://" will be encoded as well, resulting in an invalid URL as well
Eg a space should be %20 , you can verify this by adding a NSLog of the URL
It looks like you entered the wrong AppID judging from.
Parameter = "SearchRequest.AppId";
Make sure that the AppID matches the appID that they assigned your app with the service.
Also (and this sometimes randomly causes issues for me) make sure you put a / at the end of the URL String.
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 {}
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!!!
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);
}
You can try this
NSString *url = #"http://www.abc.com/param=Hi how are you";
NSString* encodedUrl = [url stringByAddingPercentEscapesUsingEncoding:
NSASCIIStringEncoding];
I think this will work for you
[strUrl stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet]
the Native method for URL Encoding.
Swift 4.x
let originalString = "https://www.somedomain.com/folder/some cool file.jpg"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
print(escapedString!)
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.
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.
I have some code to send a url to a remote server. If I do not encode the url, it works perfectly. But if I encode the url, it does not work. So I am pretty sure something is not right with the way I encode the url query string.
Here is my code:
// URL TO BE SUBMITTED.
NSString *urlString =
#"http://www.mydomain.com/test.php?";
// NOW CREATE URL QUERY STRING
NSString *unencoded_query_string =
#"name=%#&user_id=%#&person_name=%#&person_email=%#&privacy=%#";
// PUT PREVIOUSLY SET VALUES INTO THE QUERY STRING
NSString *unencoded_url_with_params =
[NSString stringWithFormat:unencoded_query_string, business , user_id , name , email , privacy_string];
// ENCODE THE QUERY STRING
NSString *escapedString = (__bridge_transfer NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(__bridge CFStringRef)unencoded_url_with_params,
NULL,
(CFStringRef)#"!*'();:#&=+$,/?%#[]",
kCFStringEncodingUTF8);
// NOW APPEND URL TO QUERY STRING
NSString *full_encoded_url_string =
[urlString stringByAppendingString: escapedString];
and then I send this string to the server, and the server does have the correct request file invoked, but isn't able to read the parameters.
Would anyone know what I doing incorrectly here? I am using arc by the way.
Thanks!
I think you probably want to escape each param, not the entire request. Basically you want to escape ampersands, spaces etc that show up in your get variables. Your encoded URL probably looks like this:
http://www.mydomain.com/test.php?name%3DPeter%20Willsey%26user_id%3DUSERID%26person_name%3DPeter%20Willsey%26person_email%3Dpeter%40test.com%26privacy%3D1
and it should look like this:
http://www.mydomain.com/test.php?name=Peter%20Willsey&user_id=25&person_name=Peter%20Willsey&person_email=peter%40test.com&privacy=1
Using the Twitter Search API I have attempted to download JSON data from a specific twitter user. If I use the recommended schema for a from:username twitter recommends http://search.twitter.com/search.json?q=%3Ausername
For example if I want to query tweets FROM #NBCNews, twitter recommends the following:
http://search.twitter.com/search.json?q=%3ANBCNews
In my code I make an NSURLRequest for this JSON data but the NSString that is instantiated somehow has an odd string of numbers and letters where the "%3A" portion of the URL was.
I have made 2 similar strings to test, *path and *workingPath. *path is a call to return all tweets from the user:NBCNews and *workingPath is a call to return all tweets containing NBCNews.
- (void)loadQuery {
NSString *path = [NSString stringWithFormat:#"http://search.twitter.com/search.json?rpp=%d&q=%3ANBCnews",
RESULTS_PERPAGE];
NSString *workingPath = [NSString stringWithFormat:#"http://search.twitter.com/search.json?rpp=%d&q=NBCnews",
RESULTS_PERPAGE,self];
path = [path stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSLog(#"path --> %#", path);
NSLog(#"path --> %#", workingPath);
I have included the output in the debug console where you can see the difference.
Pending breakpoint 1 - ""TwitterSearchViewController.m":334" resolved
2012-01-06 22:01:11.785 emsguide[7286:11603] path --> http://search.twitter.com/search.json?rpp=100&q=0X1.A000040DA604P-1025NBCnews
2012-01-06 22:01:11.786 emsguide[7286:11603] workingPath --> http://search.twitter.com/search.json?rpp=100&q=NBCnews
Current language: auto; currently objective-c
Here the string "0X1.A000040DA604P-1025" replaced %3A in my URL.
Anyone seen this or have thoughts?
Despite what the search.twitter.com result renders, the code does not need %3A and the URL scheme works with from:username
changed:
http://search.twitter.com/search.json?rpp=%d&q=%3ANBCnews
to
http://search.twitter.com/search.json?rpp=%d&q=from:NBCnews