I'm trying to open up the vine app when you tap a vine url from my app. Doing
NSURL *url = [NSURL URLWithString:#"vine://http://vine.co/v/biTaEEwdq2n?1"];
[[UIApplication sharedApplication] openURL:url];
Will indeed open up the app, but not navigate to the video. Any idea how this is done? Is it possible?
After looking at the Javascript on vine.co, I've found that the correct scheme for a post is : vine://post/<#id#> . So, let's try this vine://post/biTaEEwdq2n
It works for me.
Try using vine://v/biTaEEwdq2n or vine://video/biTaEEwdq2n as the URL.
The URL Scheme for the vine profile page is vine://user/USERID
Ex:
vine://user/126850376482649253
Related
I'm creating a facebook game and I want to have a button which let user open my facebook page, however, when I open my page url:
https://www.facebook.com/Spiritbomb.co
It just opens the native Facebook app but does not redirect to my page.
I tried this url on Safari and it works just fine.
Anyone experienced with this could help me please?
Try using NSURL
NSUrl *aUrl = [NSURL URLWithString:#"your facebook url"];
[[UIApplication sharedApplication] openURL:url];
I found out that this type of url doesn't work
https://www.facebook.com/Spiritbomb.co
(What I mean by "doesn't work" is that open this URL in an iOS app will only launch the FB native app, but does not redirect you to the desired Page)
You need to find your Facebook Page ID and use this URL: https://www.facebook.com/YOUR_FB_PAGE_ID
In my case, this URL works: https://www.facebook.com/690543054295905
How to find your FB Page ID: Go to your FB Page, click Edit Page -> Update Page Info
Then scroll to the bottom of the page, you'll find your FB page ID there.
Thanks for all the help, guys.
I am having an app in which I am opening my Facebook Like page.
I am using this Link for that.
But when I try to open this It doesn't go to my Facebook Like page.
It opens up my app with Logo but there is no like page for my app.
It says " This app is not available for your phone."
I have passed the correct AppId from Facebook.
Do I need to set up any additional details from Facebook Developers Guide?
Please help me. Any help will be appreciated.
For my application, I am using this code and it works pretty well. Seems like you are doing something wrong.
NSURL *url = [NSURL URLWithString:#"fb://profile/268179536658143"];
if ([[UIApplication sharedApplication] canOpenURL:url])
{
[[UIApplication sharedApplication] openURL:url];
}
EDIT: You were taking App ID, not the fan page ID. Please correct it.
I'm building an iOS app with the Facebook SDK.
One of the things my customer wants is a button that launches the Facebook app and shows their Facebook page. Is there some way to do this, perhaps using a URL?
Thanks.
iPhone has a bunch of URL schemes such as:
NSURL *url = [NSURL URLWithString:#"fb://<insert function here>"];
[[UIApplication sharedApplication] openURL:url];
Check this site out for more:
http://wiki.akosma.com/IPhone_URL_Schemes#Facebook
Also check out this SO post for more info:
https://stackoverflow.com/a/10416399/738262
Is there anyway to use an iOS URL Scheme to open the YouTube to a particular user's profile I've tried youtube://user/myusername but that did not seem to work.
To open a Youtube page in Youtube app (if it's installed on the device)
you can check whether the device can open the page:
// URL scheme for youtube app
NSString *youtubeURL = #"youtube://www.youtube.com/user/";
// Page name(or channel name)
NSString *youtubePageName = #"YourPageName";
// Check if the device can open in Youtube app or not.
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString: [youtubeURL stringByAppendingString:youtubePageName]]]){
// Open in Youtube app
[[UIApplication sharedApplication] openURL:[NSURL URLWithString: [youtubeURL stringByAppendingString:youtubePageName]]];
}else{
// If device cannot open in youtube app, open the page in browser.
[[UIApplication sharedApplication] openURL:[NSURL URLWithString: #"http://www.youtube.com/user/YourPageName"]];
}
This solution worked for me
iOS 9 edit: Target app's URL scheme must be added to info.plist under LSApplicationQueriesSchemeskey.
Here is a related post: iOS 9 not opening Instagram app with URL SCHEME
This forum post's lack of responses indicates there probably isn't a way to do this currently: YouTube iOS URL Scheme for Channels
But you may be able to request that feature on YouTube's iOS app feedback page. And in the meantime, I'd recommend just linking to users' HTML YouTube pages.
No, few weeks ago I was looking for the same, but i didn't found anything. Instead I've opted for use a UIWebView to show the youtube channel or using the openURL: UIApplication's method to open the channel with Safari.
If you want to save some time, you can use the easy-to-implement TSMiniWebBrowser's control at this Github page.
Here is the youtube Channel Url scheme, try to load it in UIWebView and it will works.
http://m.youtube.com/#/user/channel_name for e.g http://m.youtube.com/#/user/whartonmagazine
I'm working on something cool and I want to link to my Google+ profile in the iOS app.
Has anyone found out how to use the mgc:// URI scheme that the Google+ app on iOS provides to open a specific profile the way you can do it with fb://, twitter:// and almost every other account-based iOS app?
In fact it's really simple. Check your URL in a web browser and replace http:// or https:// with gplus://
Example :
If the URL in your browser is https://plus.google.com/u/0/100711776131865357077 then you open the profile in the app with the following code.
NSURL *gPlusUrl = [NSURL URLWithString:#"gplus://plus.google.com/u/0/100711776131865357077"];
if ([[UIApplication sharedApplication] canOpenURL:gPlusUrl]) {
[[UIApplication sharedApplication] openURL:gPlusUrl];
}
I know this is an old post, but I found the solution here (just search "google"). The URL scheme is gplus://. I have tested this myself and verified that it worked. Usage example:
NSURL *gPlusUrl = [NSURL URLWithString:#"gplus://"];
if ([[UIApplication sharedApplication] canOpenURL:gPlusUrl]) {
[[UIApplication sharedApplication] openURL:gPlusUrl];
}