i have this idiotic Question .
Is there any way open an another ios Application inside a view controller of a app.
I know we can open a another Ios app if we have url schema and the calling application goes to background and called application goes to background.
Is there any way ?? i just got a dream ,so am asking can we do it??
It is not possible to do it with the iPhone SDK. You can do this though by using Private APIs.
You can make it work on jailbroken devices.
But if you want to call native social apps like facebook or twitter you should use this code.
It works and apple approves it.
NSURL *fbNativeAppURL = [NSURL URLWithString:#"fb://"];
if ( [[UIApplication sharedApplication] canOpenURL:fbNativeAppURL])
{
[[UIApplication sharedApplication] openURL:fbNativeAppURL];
}
or
NSURL *twNativeAppURL = [NSURL URLWithString:#"twitter://"];
if ( [[UIApplication sharedApplication] canOpenURL:twNativeAppURL])
{
[[UIApplication sharedApplication] openURL:twNativeAppURL];
}
Related
I am working on a map application in my iPhone app.
I have a button go.
When the user clicks this button in this method I want to check if user has installed the waze application on his iphone. If yes then navigate to waze application otherwise open iPhone's default map app.
Try to do this way :
NSString *wazeAppURL = #"waze://";
NSString *mapsAppURL = #"maps://";
BOOL canOpenURL = [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:wazeAppURL]];
NSString *url = canOpenURL ? wazeAppURL : mapsAppURL;
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:url]];
Here, canOpenURL allows you to test if the Waze app is installed on your iPhone. if iPhone can open the url waze:// it means you already have the app and it will launch it. Otherwise it will launch the default Maps app. Safari won't be called.
To open an app you need to call
BOOL canOpenURL = [[UIApplication sharedApplication]
canOpenURL:[NSURL URLWithString:#"app://"]];
if ( canOpenUrl ) [[UIApplication sharedApplication]
openURL:[NSURL URLWithString:url]];
To find all the url, go to this page: http://handleopenurl.com/
For waze in particular, http://handleopenurl.com/scheme/waze
hope this helps.
Note that on iOS you can also navigate to Google Maps -- and pass along the query string or geopoint. Here's one example of navigating to a specific geopoint:
if (self.mapView.userLocation.location) {
NSString *urlAsString = [NSString stringWithFormat:#"comgooglemaps://?q=%f,%f", self.mapView.userLocation.location.coordinate.latitude, self.mapView.userLocation.location.coordinate.longitude];
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:urlAsString]]) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlAsString]];
}
}
Just a suggestion to enhance the user experience.
In my app, I have integrated native map app for showing directions between two points on button action. I am using the following code.
- (IBAction)showRoute
{
NSString *urlString = [NSString stringWithFormat:#"http://maps.apple.com/maps? daddr=%#,%#&saddr=%f,%f",d_latitude,d_longitude, s_latitude, s_longitude];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlString]];
}
But the problem is when click the button it first redirects to safari and then maps app - is there any way to miss out the safari step ?
In which iOS have you tried.? I tried on iOS 6 simulator works fine, opens the map app directly. In iOS simulator it opened safari has it doesn't have map app on simulator.
I have tried the below code on button action, its similar the one you used.
NSString* url = [NSString stringWithFormat:#"http://maps.apple.com/maps?daddr=%f,%f&saddr=Current%%20Location", 40.7142, 74.0064];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:url]];
I am able to call from my iphone application by using below code:
NSString *phoneNumber = #"tel://1234567890";
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
Now, i want to know how to return to my application back when the call ends ?
UIWebView *phoneCallWebview = [[UIWebView alloc] init];
// [self.view addSubview:phoneCallWebview];
NSURL *callURL = [NSURL URLWithString:[NSString stringWithFormat:#"tel:%#", 9238928399]];
[phoneCallWebview loadRequest:[NSURLRequest requestWithURL:callURL ]];
As far as I'm aware, such interaction is impossible since your application has been demoted to background, and all UI interaction has been delegated to the Phone app, and the user.
I found this SO question
End call, don't return to app automatic in iphone 3.1 version
Which pointed to an article on apple dev forums
https://devforums.apple.com/message/128046 (dev account required)
Which says it was a change in iOS 3.1 but a "workaround" is
use UIWebView to open the tel: url, after the call, your app will relaunch, but you get the annoying do you want to make this call alert.
I have't verified this works as described, just thought I'd point it out
From iOS 5, use below...
NSString *phoneNumber = [#"telprompt://" stringByAppendingString:#"12345678"];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
Just use telprompt:// instead of tel://
telprompt will prompt the user first, and when call ends,it will go back to your application.
NSString *myNumber = [#"telprompt://" stringByAppendingString:txtMobileNo.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:myNumber]];
I can make a call from my app by use this APIs.
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"tel:XXXXXX"]];
I would like to return to my app where I left after the users ends the call. Is that possible?
Try this:
UIWebView *callingWebview;
[callingWebview loadRequest:[NSURLRequest requestWithURL:]];
no it's not possible
As title says I'd like to know how to restart my iPhone app after doing this:
[[UIApplication sharedApplication] openURL:[NSURL UrlWithString:#"tel://0123456789"]]
It seems pretty simple as I saw many topics also talking about restoring the very state of the application when openURL is called, but I can't find how to simply restart the app when the calling is finished.
Is it supposed to be the default behavior? As for me, the iPhone opens Favorites after call is finished, I don't know why.
You can't. Starting an app is solely user's responsibility - which I consider a good thing.
check the discussion here: https://devforums.apple.com/message/128046#128046
create a UIWebView to load the phone url like this:
UIWebView *webview = [[UIWebView alloc] initWithFrame:[UIScreen mainScreen].applicationFrame];
[webview loadRequest:[NSURLRequest requestWithURL:url]];
just use
[[UIApplication sharedApplication] openURL:[NSURL UrlWithString:#"telprompt://0123456789"]]
It will return to the app after call finished
You can't restart an app after a phone call, as your app has terminated and your code is no longer being run.
If you want to restart after the user visits a webpage, you can put a link with a custom scheme in that webpage, and register it with your app. The user can then tap the link to open your app again.