Trying to convert a string to NSURL and this is not happening.
barcodeTextLabel.text = foundCode.barcodeString;
urlToGrab = [NSString stringWithFormat:#"%#", foundCode.barcodeString]; // foundCode.barcodeString is an NSString
urlToGrab shows the following "error invalid CFStringRef"
This is how you create an NSURL from an NSString:
NSURL *url = [NSURL URLWithString:#"http://www.google.com"];
You can use following for creating the file path to url.
NSURL *yourURL = [NSURL fileURLWithPath:#"/Users/xyz/Desktop/abc.sqlite"];
If foundCode.barcodeString is the string you want as your URL, then (like the above answer) use the NSURL class method URLWithString:(NSString *).
Your code should look like:
NSURL urlToGrab = [NSURL URLWithString:foundCode.barcodeString];
Where is your error coming into to play? The way your code is, urlToGrab is an instance of NSString. I would imagine you would get an error like you described if you tried to make an HTTP request on an NSString rather than NSURL.
Swapnali patil's answer works, but I will add an explanation.
You will get a nil if the format of NSString doesn't fit file criteria for NSURL (file:///xxx/file.ext).
My needs were with loading a JPG image via URL file path to nsdata; NSURL * u=[[NSURL alloc] initWithString:fpath] returned nil, but NSURL *yourURL = [NSURL fileURLWithPath:fpath] as in mentioned answer worked. A URL for files will be file:///users/xxx/pic.jpg format and give disk access. NSURL * u=[[NSURL alloc] initWithString:(NSString*) ] will also give nil object if nsstring is web URL but if missing http://
Related
When trying to load a request I do like this:
NSString *urlStrings = [NSLocalizedStringFromTable(#"kPicturesURL", #"urls", nil) stringByAppendingPathComponent:#"Articles/a.pdf"];
NSURL *url = [NSURL URLWithString:urlStrings];
Inside of "urls" I have this key:
/* Service pictures directory */
"kPicturesURL" = "http://192.168.2.104/myApp/Pictures";
And I get this result: (The first line is "urlStrings" and the second is the "url")
(NSString *) $6 = 0x092a8f60 http:/192.168.2.104/myApp/Pictures/Articles/a.pdf
2012-11-22 09:18:20.093 NPE[7680:c07] Couldn't issue file extension for path: /192.168.2.104/myApp/Pictures/Articles/a.pdf
I've tried those questions:
NSString and NSUrl not converting properly
NSURL not getting allocatd with NSString
Pass NSString into NSURL URLWithString ?
NSString to NSURL ?
NSString to NSURL
Non of those worked, what seems to be the problem?
Thanks!
Maybe instead of:
http:/192.168.2.104/myApp/Pictures/Articles/a.pdf
This:
http://192.168.2.104/myApp/Pictures/Articles/a.pdf
So, using one of the following did worked eventually.
This one:
NSString *stringURL = [url absoluteString];
When you have a filesystem path, you can have a ".." removed (and the previous path component removed as well) by using the stringByResolvingSymlinksInPath selector. How can I achieve the same thing for a URL? For example I start out with, say:
www.example.com/themes/themeA/../common/assetA.png
Which I need converted to:
www.example.com/themes/common/assetA.png
For a URL use the NSURL method:
- (NSURL *)standardizedURL
Returns a new URL that points to the same resource as the original URL and is an absolute path.
Example:
NSString *s = #"www.example.com/themes/themeA/../common/assetA.png";
NSURL *u = [NSURL URLWithString:s];
NSURL *su = [u standardizedURL];
NSLog(#"su: %#", su);
NSLog output:
su: www.example.com/themes/common/assetA.png
What about the following?
NSString* resolved_url
= [[[NSURL URLWithString: #"www.example.com/themes/themeA/../common/assetA.png"] standardizedURL] absoluteString];
If you want NSURL instead of NSString, remove call to absoluteString.
I try to display an image with a URL source in an iOS application, but it doesn't show up.
The url of the image is live example path.
When escaping this string using the following Objective-C code:
NSString *url= [(NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (CFStringRef)originalpath, NULL, CFSTR("øæ"), kCFStringEncodingUTF8) autorelease];
the result is (with encoding of øæ) : live xml path
All my files where the URLs are stored use text encoding (UTF-8).
How do I escape the URL in a right way, such that the image will be displayed?
don't go for ascii encoding try it
NSString *URLString = [yourImagepath stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url = [NSURL URLWithString:URLString];
I hope it will help you out.
I believe you need to escape your url-string, before using it as an url...
NSString *URLString = [yourImagepath stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding];
NSURL *url = [NSURL URLWithString:URLString];
Had the same problem on iOS with German umlauts. The NSURL category NSURL+IFUnicodeURL solved the problem for me, don't get scared by the amount of C code in there, it's well-hidden :)
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
I Have tried
NSString *alertTxt =textfieldName.text;
NSLog(#"Name: %#",alertTxt);
NSURL *urlAddress = [[NSURL alloc ] initWithString: #"%#/login/index.php",alertTxt];
error : too many argument to function initstring ..
the user will giv the url address to the alertText field and i hav to embed in to the webkit may i know the process how to make it !!!
Where im making mistake kindly help me out in this
Thanks
-initWithString: can only take a complete string, not a format string. To use -initWithString: you need to complete the string first.
NSString* urlString = [NSString stringWithFormat:#"%#/login/index.php", alertTxt];
NSURL* urlAddress = [[NSURL alloc] initWithString:urlString];
BTW, it may be more efficient to use -stringByAppendingString:.
NSString* urlString = [alertTxt stringByAppendingString:#"/login/index.php"];