openurl

QDesktopServices::openUrl with selecting specified file in explorer

半世苍凉 提交于 2019-12-05 13:23:48
In most coding programs, you can right click on the item and click show in explorer and it shows the file in explorer with the item selected. How would you do that in Qt with QDesktopServices? (or any way to do it in QT) ernie you can use this method to select file on Windows or MacOS ,if you want select on linux you can find a way in QtCreator sources. void select(const QString& path){ #if defined(Q_OS_WIN) const QString explorer = "explorer"; QStringList param; if (!QFileInfo(path).isDir()) param << QLatin1String("/select,"); param << QDir::toNativeSeparators(path); QProcess::startDetached

Waze doesn't load navigation from Swift

北城以北 提交于 2019-12-05 06:32:54
I integrated Waze into my Swift app, but when I click on the button, Waze opens but nothing happens with the navigation. I juste see the app and that's all, instead of launching the navigation. Here is the code: @IBAction func openWazeAction(_ sender: Any) { // open waze if UIApplication.shared.canOpenURL(URL(string: "waze://")!) { let urlStr = String(format: "waze://ul?ll=%f,%f&navigate=yes", (selectedBorne?.location?.x)!, (selectedBorne?.location?.y)!) print(urlStr) UIApplication.shared.open(URL(string: urlStr)!) } else { UIApplication.shared.open(URL(string: "http://itunes.apple.com/us/app

NSWidgetExtensionContext openURL Swift

妖精的绣舞 提交于 2019-12-05 06:10:47
I have been attempting to implement a button to open my iOS app from its widget. I realize this issue has been beaten to death on the forums but I cannot find explanation with the specific error I am receiving. Perhaps some of you more experienced iOS developers can shed some light on this. I am developing an update to one of my iOS apps for iOS 10 using XCode 8.1 and Swift 2. Code for my widget's button: URL scheme added to the widget's info.plist: The runtime error I receive when pressing the OpenApp button: AppWidget[11872:3577323] __55-[_NCWidgetExtensionContext openURL:completionHandler:]

How to link to Update page for our app

半城伤御伤魂 提交于 2019-12-04 22:29:29
We prompt users to upgrade their app if they're running an outdated version.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 current version as a querystring parameter in the openURL call or is there another way to make sure the

Possible to open iPhone Safari and target a specific window?

匆匆过客 提交于 2019-12-04 19:19:09
I have an app that uses OAuth to authenticate, which means the user must be directed to the website to authorize the application. I prefer to make it obvious to the user that they are using a standard browser to authorize at the original site rather than just using a web view to show the content within my app. However, every time they are directed to Safari (via openURL), it launches a new Safari window and once Safari has 8 windows open, it simply fails to work at all. I'm wondering if there is an equivalen to the "target" attribute of an anchor tag in html which would allow me to cause all

Story link flow using SSO facebook - openURL not being called

老子叫甜甜 提交于 2019-12-04 04:42:11
问题 I have implemented Single-Sign-On in my iOS 4.3 app successfully. Now I want to publish a link to the users facebook wall so that when his/her friends that also own the app clicks the link they should be redirected to my app. The way I interpret http://developers.facebook.com/docs/mobile/ios/build/#deeplink , is that one of the delegate functions - (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url, (pre iOS 4.2) - (BOOL)application:(UIApplication *)application openURL:

Open URL Schemes in IOS

余生颓废 提交于 2019-12-04 01:44:34
问题 I have 2 apps, which are meant for different purpose, where I should not allow user to use both apps in same device. How can I check whether other app installed or not? I think I can do it using Open URL as following by putting app bundle id which is not working, I am stuck to get url for my app EX : "fb://" if ([[UIApplication sharedApplication] canOpenURL:@"My App URL"]) { //App installed } else{ //App Not installed } 回答1: You have 2 Apps.Now you want to open First App from the Second App.I

How to use [[UIApplication sharedApplication] openURL:] open other app?

我怕爱的太早我们不能终老 提交于 2019-12-03 19:42:14
问题 I followed http://iosdevelopertips.com/cocoa/launching-your-own-application-via-a-custom-url-scheme.html instruction to open app1(GlassButton) within app2(FontTest). The open method of FontTest as following: -(void)open { BOOL res = [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"glassbutton://"]]; if (res) { [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"glassbutton://"]]; } } The value of "res" is "YES", but nothing happen after openURL method be

Link to app manage subscriptions in app store

我的梦境 提交于 2019-12-02 15:31:30
Currently with In app purchase the only way to cancel an auto-renewing subscription is to do the following with the device: Settings > Store > View my account > Manage my subscription Is it possible programmatically to link directly to the Manage my subscription page in the app store? I know I can open the app store via something like NSURL *url = [NSURL URLWithString:@"itms-apps://itunes.apple.com"]; [[UIApplication sharedApplication] openURL:url]; I have seen other apps do this but I can't seem to figure out how. The new and official way (according to WWDC 2018 Session 705 ) is the following

Xamarin iOS get status back of a phone call

泪湿孤枕 提交于 2019-12-02 09:54:19
问题 Is there a way to get the call status back after a call is made from an app. I am using following to make a call NSUrl url = new NSUrl("tel://" + phoneStr); UIApplication.SharedApplication.OpenUrl(url); It shows a pop up with "Cancel" and "Call" button. If user cancels it stays back in the app, if user clicks call it initiates a call. I would like to get the click action if user made a call or canceled it. Is there a way to get that status 回答1: On iOS 10+ We can use CXCallObserver to capture