openurl

问题 I need to send a pre-filled message from my app to the fb messenger app. I followed the post SO Link. The custom URL just opens the fb messenger app but shows no change. Below is the code that I have tried: [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"fb-messenger://contacts"]]; [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"fb-messenger://share"]]; The above code just opens the fb messenger app, it does not navigate to contacts page or the compose

问题 I am trying to open link in Safari browser from inside my iOS app. Its a hybrid app so I made changes in iOS native side (which I am totally unfamiliar with) I have tried this code from below posts: Open target="_blank" links outside of UIWebView in Safari NSLog(@"open safari"); // NSLog(urlSafari); // [[UIApplication sharedApplication] openURL:urlSafari]; [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[request.URL.absoluteString substringFromIndex:7]]]; //[NSURL

问题 We prompt users to upgrade their app if they're running an outdated version. When users tap our update button, I use openURL with an address like itms://itunes.apple.com/us/app/our-app-title/id12345?mt=8 to load the App Store app to the listing for our app. With that method, however, the resulting screen has a button labeled "Open" not "Update." If users open the App Store app first, then navigate to our app's listing (or go to the update tab), the button is labeled "Update." Can I pass the

问题 I am not even sure how to define the problem but here it goes. I have an application that uses Facebook SDK for user login. I followed the Facebook authorization tutorial. I am not 100% sure how it works but this part in my "AppDelegate.m" seems important. - (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation { return [FBSession.activeSession handleOpenURL:url]; } So far so good. Now I want to implement

问题 My app freezes after calling openURL on iOS 9.3 builds 13E233 & 13E234. I tried using dispatch_after but that does not fix the issue. Here's the code, nothing special. + (void)someMethod:(UIView *)senderView { [Utility showLoadingHUDWithText:nil inView:senderView]; [[SomeClient sharedClient] someNetworkAPI:^(id result) { [Utility hideAllHUDsForView:senderView]; NSDictionary *dict = (NSDictionary *)result; NSString *someString = dict[@"someKey"]; NSURL *url = [NSURL URLWithString:someString];

问题 with iOS9 application openURL:sourceApplication:annotation is deprecated func application(application: UIApplication, openURL url: NSURL, sourceApplication: String?, annotation: AnyObject?) -> Bool { return FBSDKApplicationDelegate.sharedInstance().application(application, openURL: url, sourceApplication: sourceApplication, annotation: annotation) } so now we should use; func application(app: UIApplication, openURL url: NSURL, options: [String : AnyObject]) -> Bool {} FBSDKApplicationDelegate

问题 I have an app that will be using both Facebook and Instagram API's, both of which require me to use this delegate method: - (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation { return [PFFacebookUtils handleOpenURL:url]; } and this is the code provided by instagram: -(BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id

问题 I just upgraded from XCode 6.4 to Xcode 7 GM and get started to change the code to be compliant with Swift 2. I could not come over the following errors. The project is a keyboard extension and the snippet is from the containing app. let s = "https://itunes.apple.com/" UIApplication.sharedApplication().openURL(NSURL(string : s)!) Error 1: 'sharedApplication()' is unavailable: Use view controller based solutions where appropriate instead. Error 2: 'openURL' is unavailable. This might be

问题 This question already has answers here : Access a particular field in arbitrarily nested JSON data [duplicate] (3 answers) Closed last year . I go through this part: How do I extract the data from that URL? I only want to print out the "networkdiff": 58954.60268219 . from urllib import urlopen url = urlopen('http://21.luckyminers.com/index.php?page=api&action=getpoolstatus&api_key=8dba7050f9fea1e6a554bbcf4c3de5096795b253b45525c53562b72938771c41').read() print url This is what the API display

问题 This question already has answers here : Access a particular field in arbitrarily nested JSON data [duplicate] (3 answers) Closed last year . I go through this part: How do I extract the data from that URL? I only want to print out the "networkdiff": 58954.60268219 . from urllib import urlopen url = urlopen('http://21.luckyminers.com/index.php?page=api&action=getpoolstatus&api_key=8dba7050f9fea1e6a554bbcf4c3de5096795b253b45525c53562b72938771c41').read() print url This is what the API display