url-scheme

问题 In my app I want to have two Different URL Schemes. Like One and Two So the user can open my app with: one://something and two://something I am using this: - (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url { } How will the app know if the user types one or two? 回答1: handleOpenURL is deprecated, so if you're targeting iOS 4.2 or later, you should instead use application:openURL:sourceApplication:annotation: In both cases, you will be passed an NSURL, on which you can

问题 We are working on an iOS app and we happen to be using the Gmail app's URL Scheme to compose a new email for the user. We figured out most of it, however we are having some trouble adding new lines in the body of the email. So far our URL looks like this: googlegmail://co?from=username@example.com&to=helpdesk@example.com&subject=TITLE&body=What%20ever%20they%20include%20for%20the%20body We have already tried using %0D%0A , as well as a few other options, however none of those seem to work. Is

问题 I'm noticing strange behavior in Safari recently. I register a url scheme for the my app, and enter myapp:// into Safari. This launches my app immediately. Then I go back to Safari, and enter myapp:// into Safari again, this time it prompts me "Open this page in "myapp"?" Cancel or Open. My app will launch if I tap on open, and subsequent attempts the same alert shows. If I try tapping on cancel, my app will not launch. which is expected. However, if I enter myapp:// into the URL bar again, I

问题 I'm noticing strange behavior in Safari recently. I register a url scheme for the my app, and enter myapp:// into Safari. This launches my app immediately. Then I go back to Safari, and enter myapp:// into Safari again, this time it prompts me "Open this page in "myapp"?" Cancel or Open. My app will launch if I tap on open, and subsequent attempts the same alert shows. If I try tapping on cancel, my app will not launch. which is expected. However, if I enter myapp:// into the URL bar again, I

问题 I am using URL scheme in my iPhone app,from a page I switch user to safari,and a button click from Web page,I am reverting back to app At that time ,some parameters are passed by the web page like myapp://parameter=1 How can I find this parameter from my app ? I have placed a break point in -(BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)ur but while revert back.it is not get called. 回答1: You can use this url host to find out the parameters, where parameters could be of

问题 Is there any way to get all the list of installed application and check a specific bundle/app can be open which is already installed in swift ? 回答1: You can only open an app if you know its URL Scheme just like facebook provides for example. fb:// There is no way to get the list of applications. Can't say anything about jail broken devices. 来源： https://stackoverflow.com/questions/40327664/swift-get-all-installed-application-and-open-specific-one

问题 I have to open my iOS app whenever user open a link of my app in browser. I have already used a like myApp:// but i want to open my app even when user open an http link a from browser. Just like pinterest. Pinterest opens app even if i open a regular http link like this http://www.pinterest.com/pseudoamber/ and using URL scheme as well like this pinterest://www.pinterest.com/pseudoamber/ . My app is opening on myApp://www.myapp.com now i want to open my app when user open an http link like

问题 this is yet another .htaccess question. And I already did my literature review. Would appreciate any help. Requirements: Force HTTPS only for a few URLs. Browser shouldn't say partially encrypted page for SSL pages. I am using CodeIgnitor and tweaked the base_url in config.php to: $config['base_url'] = (isset($_SERVER['HTTPS']) && $_SERVER['HTTPS']=='on') ? 'https://' : 'http://' ; $config['base_url'] .= $_SERVER['HTTP_HOST']; $config['base_url'] .= preg_replace('@/+$@','',dirname($_SERVER[

问题 I have implemented the ability to make a call from clicking on a row of descriptive tableview of my hotel; I used the URL scheme by writing the following code in method "didSelectRowAtIndexPath ": NSString *phoneNumber=element.phone; [NSString *phoneNumberScheme = [NSString stringWithFormat:@"tel:%@ ", phoneNumber]; NSlog(phoneNumberScheme); phoneNumberScheme = [phoneNumberScheme stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; [[UIApplication sharedApplication] openURL:[NSURL

问题 I have implemented the ability to make a call from clicking on a row of descriptive tableview of my hotel; I used the URL scheme by writing the following code in method "didSelectRowAtIndexPath ": NSString *phoneNumber=element.phone; [NSString *phoneNumberScheme = [NSString stringWithFormat:@"tel:%@ ", phoneNumber]; NSlog(phoneNumberScheme); phoneNumberScheme = [phoneNumberScheme stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; [[UIApplication sharedApplication] openURL:[NSURL