i have two buttons defined. one of them when clicked must open a compose mail screen and the other when clicked must call. i have this defined as below. but when the button is pressed, it does not open either
-(IBAction) phoneButtonPressed:(id) sender{
NSString *phoneNumber = [[NSString alloc] initWithString:#"4216483330"];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
}
-(IBAction) mailButtonPressed:(id) sender{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"mailto:askalibrarian#mail.pitt.edu?subject=ULS Library"]];
}
For the phone#, the url should be prefixed with "tel:" like this:
NSString *phoneNumber = [[NSString alloc] initWithString:#"tel:4216483330"];
For the mail, the problem is you have a space in the url string (in "ULS Library") which needs to be escaped before giving it to NSURL:
NSString *urlString = #"mailto:askalibrarian#mail.pitt.edu?subject=ULS Library";
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[urlString
stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]]];
See the Apple URL Scheme Reference.
Related
I have a variable agencyWebsite and a label that should open the website when clicked with this method:
- (void)website1LblTapped {
NSURL *url = [NSURL URLWithString:self.agencyWebsite];
[[UIApplication sharedApplication] openURL:url];
}
I get a warning in the compiler saying:
Incompatible pointer types sending UILabel* to parameter of type NSString*
And the app crashes when the link is clicked. Any suggestions?
Edit: Here is what I am doing to make the label clickable
UITapGestureRecognizer* website1LblGesture = [[UITapGestureRecognizer alloc] initWithTarget:self action:#selector(website1LblTapped)];
// if labelView is not set userInteractionEnabled, you must do so
[self.agencyWebsite setUserInteractionEnabled:YES];
[self.agencyWebsite addGestureRecognizer:website1LblGesture];
What i used to get it working
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:#"http://%#", self.agencyWebsite.text]];
If agencyWebsite is of type UILabel*, you need to access its text property instead of passing the object itself to URLWithString:.
- (void)website1LblTapped {
NSURL *url = [NSURL URLWithString:self.agencyWebsite.text];
[[UIApplication sharedApplication] openURL:url];
}
Calling self.agencyWebsite will return your UILabel* object, whereas self.agencyWebsite.text will return a NSString* object containing the text from the label.
i am working on signup feature. In this feature when the user create account successfully. i am asking him or her to activate his account.
i want to open the mail application of iphone if user say yes.
now my question is simple how to open mail application from my own application?
#define URLEMail #"mailto:sb#sw.com?subject=title&body=content"
NSString *url = [URLEMail stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding ];
[[UIApplication sharedApplication] openURL: [NSURL URLWithString: url]];
Try this out.
-(void)launchMailAppOnDevice
{
NSString *recipients = #"mailto:myemail#gmail.com?subject=subjecthere";
NSString *body = #"&body=bodyHere";
NSString *email = [NSString stringWithFormat:#"%#%#", recipients, body];
email = [email stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:email]];
}
stringByAddingPercentEscapesUsingEncoding and openURL are deprecated.
Now use this:
#define URLEMail #"mailto:sb#sw.com?subject=title&body=content"
NSString * encodedString = [URLEMail stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLFragmentAllowedCharacterSet]];
UIApplication *application = [UIApplication sharedApplication];
[application openURL:[NSURL URLWithString: encodedString] options:#{} completionHandler:nil];
Ahoy!
The long and short of it is; you can't.
You can create an email compose view for the purpose of sending emails (see MFMailComposeViewController), but you cannot open applications arbitrarily without a purpose.
See this previous post for clarification: Launch an app from within another (iPhone)
Really though, it's not much effort for the user to close your app and open Mail so I wouldn't worry too much about it anyway.
The page I want to open using twitter app:
https://twitter.com/#!/PAGE
To open twitter app I use the following code:
NSURL *urlApp = [NSURL URLWithString: [NSString stringWithFormat:#"%#", #"twitter://https://twitter.com/#!/PAGE"]];
[[UIApplication sharedApplication] openURL:urlApp];
But this code doesn't seem to work as expected, I got only twitter app launched without the page which i want to show.
You are looking for the following url:
twitter:///user?screen_name=PAGE
Note that Twitter is not installed on all devices. You should check the result of openURL method. If it fails, open the page in Safari with regular url.
The following code opens twitter page on twitter app if it is already installed, otherwise opens twitter on safari:
NSURL *twitterURL = [NSURL URLWithString:#"twitter://user?screen_name=username"];
if ([[UIApplication sharedApplication] canOpenURL:twitterURL])
[[UIApplication sharedApplication] openURL:twitterURL];
else
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"http://www.twitter.com/username"]];
Don't forget to replace 'username' with your name.
I know its quite a late response to this question and I agree that, Murat's answer is absolutely correct.
Simply add a check as follows:
NSURL *urlApp = [NSURL URLWithString: [NSString stringWithFormat:#"%#", #"twitter:///user?screen_name=PAGE]];
if ([[UIApplication sharedApplication] canOpenURL:urlApp]){
[[UIApplication sharedApplication] openURL:urlApp];
}
I hope this helps someone. Cheers!! :)
This is the full code required in Swift. I am using Swift 4 but i believe it is the same for Swift 3.
let Username = "YOUR_USERNAME_HERE"
let appURL = NSURL(string: "twitter:///user?screen_name=\(Username)")!
let webURL = NSURL(string: "https://twitter.com/\(Username)")!
let application = UIApplication.shared
if application.canOpenURL(appURL as URL) {
application.open(appURL as URL)
} else {
// if Twitter app is not installed, open URL inside Safari
application.open(webURL as URL)
}
Don't forget to add the Info keys needed to use canOpenURL:
#Alexey: If you just want to know how to launch twitter from your application do this:
NSURL *urlApp = [NSURL URLWithString: [NSString stringWithFormat:#"%#", #"twitter://"]];
if ([[UIApplication sharedApplication] canOpenURL:urlApp]){
[[UIApplication sharedApplication] openURL:urlApp];
}else{
UIAlertView *appMissingAlertView = [[UIAlertView alloc] initWithTitle:#"Twitter App Not Installed!" message:#"Please install the Twitter App on your iPhone." delegate:nil cancelButtonTitle:nil otherButtonTitles:#"Ok",nil];
[appMissingAlertView show];
[appMissingAlertView release];
}
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.
I am new to iPhone developemnt.
NSString myUrl = #"www.google.com";
I need to know how to create a hyperlink for the above NSString variable.
In the MFMailComposeViewController I need to use like below
[mailViewController setMessageBody:myUrl isHTML:YES];
Please help me out.
Thanks for any help.
The easiest way is to make the UILabel into a UIButton, style it (use Custom type to get rid of button look). Then connect to an Action that opens safari.
The action should do this:
NSURL *url = [[[ NSURL alloc ] initWithString: #"http://www.example.com" ] autorelease];
[[UIApplication sharedApplication] openURL:url];