当我的应用设置为不在后台运行时，是否可以在URL方案中获取参数？

Question

I am trying to open my app using URL Scheme with a parameter on it, my problem is, is it possible to get the parameter when the app is set to not run in the background.

SampleApp-info.plist

  <key>UIApplicationExitsOnSuspend</key>
        <true/>

AppDelegate.m

- (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation;
{

    NSLog(@"url recieved: %@", url);
    NSLog(@"scheme: %@", [url scheme]);
    NSLog(@"query string: %@", [url query]);
    NSLog(@"host: %@", [url host]);

    return YES;
}

When UIApplicationExitsOnSuspend is set to true and try to open my app from Safari, it's not creating any logs but when I remove the key it is working well. I have to restart my app to load the first viewcontroller and suspending it on the background is the best way, I think...

Can anyone help me on this? or a better way so I can reload my app starting in the first viewcontroller? Newbie here.

I really appreciate any help. Thank you and regards.

Answer 1

UIApplicationExitsOnSuspend Settings

You have to set UIApplicationExitsOnSuspend to false . URL Scheme Callback will not work with UIApplicationExitsOnSuspend set to true . If you want to have a simular behavior like UIApplicationExitsOnSuspend set to true just add the following code

- (void)applicationDidEnterBackground:(UIApplication *)application {
    [super applicationDidEnterBackground:application];
    exit(0);
}

First you have to make an entry in in Info.plist

eg entry your URL Scheme. In this case (in the picture) the callback will be blurry:// the identifier should be normally unique

Callback Implementation

For Pre iOS 9 swift

optional func application(_ application: UIApplication,
            handleOpenURL url: NSURL) -> Bool {
    NSLog("URL is %@", url)
    return true
}

For Pre iOS 9 objective-c

- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url {
    // handler code here
    NSLog(@"Url is %@, url);
    return YES;
}

- (BOOL)application:(UIApplication *)application
            openURL:(NSURL *)url
  sourceApplication:(NSString *)sourceApplication
         annotation:(id)annotation
{
    // handler code here
    NSLog(@"Url is %@, url);
    return YES;
}

IOS 9 objective-c

- (BOOL)application:(UIApplication *)app
        openURL:(NSURL *)url
        options:(NSDictionary<NSString *,
                         id> *)options {
    NSLog(@"Url is %@", url);
    return YES;
}

IOS 9 - Swift

optional func application(_ app: UIApplication,
              openURL url: NSURL,
              options options: [String : AnyObject]) -> Bool {
   NSLog("URL is", NSURL)
   return true
}

Test the callback

Just open the weburl in the example case blurry://test with the mobile safari on your Handy