openurl

I´m searching for a opportunity to open a url in java. URL url = new URL("http://maps.google.at/maps?saddr=4714&daddr=Marchtrenk&hl=de"); InputStream is = url.openConnection().getInputStream(); BufferedReader reader = new BufferedReader( new InputStreamReader( is ) ); String line = null; while( ( line = reader.readLine() ) != null ) { System.out.println(line); } reader.close(); I found that way. I added it in my program and the following error occurred. The method openConnection() is undefined for the type URL (by url.openConnection()) What is my problem? I use a tomcat-server with servlets, .

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 what to set for sourceApplication parameter (String) Muthu Ram func application(app: UIApplication,

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)annotation { return [self.instagram handleOpenURL:url]; } As you can see, this is a problem because I can only

When launching an app from a URL, there is a distinction between whether the URL is freshly launching an app, or if it's resuming an app that has been put into a suspended state. My question is, how do I debug the process of a "fresh launch" from a URL? Hitting "Run" in Xcode automatically opens the app. Then, I have to suspend the app to reach Safari and access my test site. But if I close my suspended app, Xcode is no longer attached to it and I'm unable to debug. Lian van der Vyver These steps can be followed on the device. Founded at this link . Run the app from Xcode to install it on your

This question already has an answer here: Access a particular field in arbitrarily nested JSON data [duplicate] 3 answers 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 as a result from print url command: { "getpoolstatus": { "version": "1.0.0", "runtime": 16.618967056274, "data": { "pool

In iOS 6 or 7, the app exit to the home screen when I call UIApplication openUrl with a url of itms-services://XXXX to install a new version of my app (using enterprise deployment with ipa files). In iOS 8, this is no longer the case. Now the app continue running just as nothing has happened, but if I go the home screen, I can see my app icon grayed out, with a downloading pie chart about 66% completed and the text "Downloading..." below. If I now wait for a while (less than a minute), the application is installed correctly and I can start my app again. Has anyone else experienced this

I'm currently developing an App, that needs to open a browser to display a webpage. To do that i use the [UIApplication sharedApplication] openURL method with an url. In iOS 6 this works perfectly, but in iOS 7 it freezes the app for 10+ seconds, then opens the browser and all is good. This happens using ad hoc provisioning. Someone on the internet commented that this was a known problem, however, that one comment was all i could find regarding this problem. I noticed the same problem when calling -[UIApplication openUrl:] from the Application Delegate didReceiveRemoteNotification: or

How to open a url in system default browser by using Swift as programming language and OSX as plattform. I found a lot with UIApplication like UIApplication.sharedApplication().openURL(NSURL(string: object.url)) but this works just on iOS and not on OSX And the Launch Services , I found has no examples for swift and there is a lot deprecated for OSX 10.10 Any help welcome - thanks. Swift 3 or later import Cocoa let url = URL(string: "https://www.google.com")! if NSWorkspace.shared.open(url) { print("default browser was successfully opened") } Connor Knabe For MacOS, you can use this: let url =