I am using this code to run the maps.app from my app.
NSString* urlStr = [NSString stringWithFormat: #"http://maps.google.com/maps?saddr=%f,%f&daddr=%f,%f", s_lat, s_long, d_lat, d_long];
[[UIApplication sharedApplication] openURL: [NSURL URLWithString:urlStr]];
The issue is that I didnt manage to get the current location.
I want something like this:
poi1.url = [NSString stringWithFormat:#"http://maps.google.com/maps?saddr=Current Location&daddr=%f,%f",lat,lon];
Have you tried appending the your label after the position data?
http://maps.google.com/maps?q=37.771008,+-122.41175+(You+can+insert+your+text+here)&iwloc=A&hl=en
Edit: And don't forget to encode your URL correctly!
NSURL* url = [[NSURL alloc] initWithString:[addr stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
Related
I had used following code in my app but it opens new window and shows route fromcurrent location to destination.
code:
NSURL *url = [NSURL URLWithString:urlAddress];
[[UIApplication sharedApplication] openURL:url];
NSString *urlString = [NSString stringWithFormat:#"http://maps.apple.com/maps? daddr=%f,%f&saddr=%f,%f",12.8400,77.6700, 12.9610850,77.604692699999990];
NSString *escapedString = [urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url=[NSURL URLWithString:escapedString];
[[UIApplication sharedApplication]openURL:url];
but my requirement is to show route from current location to destination location without launching any url.
You need to request a directions for example from Google
http://maps.googleapis.com/maps/api/directions/json?origin=%f,%f&destination=%f,%f&sensor=true&mode=%#&departure_time=%ld
And then get the routes from received json and write the path on the map.
I think you could use a WebView and load that url in it.
You can use UIWebView and load the above URL in it
NSString *urlString = [NSString stringWithFormat:#"http://maps.apple.com/maps? daddr=%f,%f&saddr=%f,%f",12.8400,77.6700, 12.9610850,77.604692699999990];
NSString *escapedString = [urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url=[NSURL URLWithString:escapedString];
[webveiew loadRequest : [NSURLRequest requestwithURL:url]];
Use google maps api. Go through their documentation. Its helpful. I had used it in my application.
I've a a pdf placed in my Code files. I want it to open in iBooks, while researching I've found this :
NSString *stringURL = #"itms-books:";
NSURL *url = [NSURL URLWithString:stringURL];
[[UIApplication sharedApplication] openURL:url];
NSString *stringURL = #"itms-bookss:";
NSURL *url = [NSURL URLWithString:stringURL];
[[UIApplication sharedApplication] openURL:url];
How can I use this or what other way i can open my PDF File in iBook?
Regards
Maybe too late but I created a small example where the method explained in andycodes used. I uploaded to Github and you can download it here:
https://github.com/cjimenezpacho/openpdfibooks
Basically is this piece of code in a IBAction and the required delegate methods:
NSURL *url = [[NSBundle mainBundle]
URLForResource: #"iPhoneintro" withExtension:#"pdf"];
self.docController = [UIDocumentInteractionController interactionControllerWithURL:url];
self.docController.delegate = self;
[self.docController presentOpenInMenuFromRect:CGRectZero inView:self.view animated:YES];
Links to Apple documentation:
Class:
https://developer.apple.com/library/ios/documentation/UIKit/Reference/UIDocumentInteractionController_class/Reference/Reference.html
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:#"itms-bookss://%#",self.pathToFile]]];
this does open a local pdf to Ibooks needs 2 s's.
Edit: to be specific on this this will open the local PDF if it is in iBooks already locally on the device.
if you need to add it to iBooks so far as I know you need to use the UIDocumentInteractionController.
Take a look at this Tumblr post explaining how do achieve a similar result using UIDocumentInteractionControllers.
http://andycodes.tumblr.com/post/738375724/ios4-sdk-opening-pdfs-in-ibooks
I'd like to make a button call a phone number entered by the user inside the text field. I have a code but it doesn't work.
NSString * phoneNumber = [NSString stringWithFormat:#"%#%#", #"tel://", phoneNumber.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
Anyone has a similar approach to this? Thanks.
I think it's tel: instead of tel://. See this Apple document. Try giving this a shot:
NSString *pn = [#"tel:" stringByAppendingString:phoneNumber.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:pn]];
See my answer to another question for some sample code to handle cases with invalid input.
Basically you do this:
NSString *cleanedString = [[phoneNumber componentsSeparatedByCharactersInSet:[[NSCharacterSet characterSetWithCharactersInString:#"0123456789-+()"] invertedSet]] componentsJoinedByString:#""];
NSString *escapedPhoneNumber = [cleanedString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:#"tel://%#", escapedPhoneNumber]];
Update: I noticed that the string you created shares the some name ("phoneNumber") as the text field from which you try to get the text. You may want to rename either of those two.
Currently I am doing this to 'get directions':
NSString *googleMapsURLString;
googleMapsURLString = [NSString stringWithFormat:
#"http://maps.google.com/?saddr=%1.6f,%1.6f&daddr=%1.6f,%1.6f",
curLoc.latitude, // Start
curLoc.longitude,
self.hotspot.coordinate.latitude,
self.hotspot.coordinate.longitude];
NSLog(#"Opening URL: %#",googleMapsURLString);
NSURL *url = [NSURL URLWithString:googleMapsURLString];
[[UIApplication sharedApplication] openURL:url];
Does anyone know if there is a parameter to get directions via walking instead of defaulting to driving?
Just add
dirflg=w
to your parameters.
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