I am trying to access google maps' forward geocoding service from my iphone app.
When i try to make an NSURL from a string with a pipe in it I just get a nil pointer.
NSURL *searchURL = [NSURL URLWithString:#"http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false"];
I dont see any other way in the google api to send bounds coordinates with out a pipe.
Any ideas about how I can do this?
Have you tried replacing the pipe with %7C (the URL encoded value for the char |)?
As stringByAddingPercentEscapesUsingEncoding is deprecated, you should use stringByAddingPercentEncodingWithAllowedCharacters.
Swift answer:
let rawUrlStr = "http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
let urlEncoded = rawUrlStr.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())
let url = NSURL(string: urlEncoded)
Edit: Swift 3 answer:
let rawUrlStr = "http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
if let urlEncoded = rawUrlStr.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
let url = NSURL(string: urlEncoded)
}
If you want to be safe for whatever weird characters you will put in the future, use stringByAddingPercentEscapesUsingEncoding method to make the string "URL-Friendly"...
NSString *rawUrlStr = #"http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
NSString *urlStr = [rawUrlStr stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *searchURL = [NSURL URLWithString:urlStr];
Related
I have the following code - note it has to objects with temp, but I will explain.
NSString *temp = _passedOnURL;
NSString *temp = #"http://google.com"; //I comment the one out that I do not use.
NSLog(#"TEMP - %#", temp);
NSURL *feedURL = [NSURL URLWithString:temp];
NSLog(#"FEED URL - %#", feedURL);
The _passedOnURL is a string with the contents passed from a Segue.
Now when I use the 1st temp, the FEED URL returns (null), but when I Log Temp it is still there, so somehow the NSURL does not read the string.
When I hardcode the string with the second temp - there is no issue.
In my mind there is no difference for the NSURL when it is reading the NSString yet, it seems to behave different.
Is there any reason for this??
EDIT
When I do the following code I have no issues:
_passedOnURL = #"http://www.google.com";
so I really have no explanation for this???
try escaping it : [NSURL URLWithString: [temp stringByAddingPercentEscapesUsingEncoding: NSASCIIStringEncoding]]
It seems you have an invalid url string stored in temp. Not every string can be converted to a url but the valid url. Invalid chars and format will lead a nil object after +URLWithString:. So would you let us know what is stored in temp when you try this?
According to the doc for URLWithString:
Parameters
URLString
The string with which to initialize the NSURL object. Must be a URL
that conforms to RFC 2396. This method parses URLString according to
RFCs 1738 and 1808.
Return Value
An NSURL object initialized with URLString. If the string was
malformed, returns nil.
So my guess is that your _passedOnURL is not a valid URL.
I would do a NSLog on your _passedOnURL to check if you are getting the string correctly from the other segue.
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://
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 this situation where I need to make a URL with a string that contains ".
For instance:
NSString *myStringWithURL = [NSString stringWithFormat:
#"http://myurl.com/setParam.php?setName=
{\"name\":\"%#\"} ", name];
// myStringWithURL will return the desired URL but it is a String.
// of course I need it as URL in order to make the request.
NSURL *targetURL = [NSURL URLWithString : myStringWithURL];
//This will return nil.
So the question is how to get my URL to not return nil?
The part {"name":"My Name"} should be a JSON but this seemed simpler than using a NSDictionary and then using the parser of JSON and so on.
I thank you very much for your help!!
use stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding to convert your string
another issue where I seem to have found an solution for ObjC but not MonoTouch.
I want a NSUrl from an URL (as string).
The string may contain whitespace and backslashes.
Why is NSUrl returning null for such string, even though these are valid urls in a browser?
For example:
NSUrl foo = NSUrl.FromString(#"http://google.com/search?\query");
foo == null
Any suggestions?
[NSURL URLWithString:[googlSearchString stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding]];
Referenced by
Stack Overflow....
You probably need to process the string first with
stringByAddingPercentEscapesUsingEncoding:
... so that it can be processed a valid URL.
URLWithString:
Creates and returns an NSURL object initialized with a provided string.
(id)URLWithString:(NSString *)URLString
Parameters
URLString
The string with which to initialize the NSURL object. Must be a URL that conforms to RFC 2396. This method parses URLString according to RFCs 1738 and 1808. (To create NSURL objects for file system paths, use fileURLWithPath:isDirectory: instead.)
Return Value
An NSURL object initialized with URLString. If the string was malformed, returns nil.
NSString * urlString = #"http://example/newcase/path/fileNames";
urlString = [urlString stringByAddingPercentEscapesUsingEncoding:NSISOLatin1StringEncoding];
NSURL * url = [NSURL URLWithString:urlString];
NSLog(#"URL: %#", url);