如何在Cordova Android应用程序中深度链接到特定页面？

Question

I am using launchmyapp plugin to create scheme url for my cordova android app. I have created myapp:// as my scheme and am using intent://scan/#Intent;scheme=myapp://;S.browser_fallback_url=https://play.google.com/store/apps/details?id=org.myapp;end inside an <a> tag to open the app from web browser.

Now when I want to open any specific page (like settings page) of my app using the scheme URL. I tried using myapp://settings as suggested on the plugin's page but it doesn't seem to work. Again the homepage of my app is only getting opened.

I have tried this answer , but it doesn't seem to work. Any help is much appreciated.

Answer 1

I had achieved this using cordova-universal-links-plugin

step 1 : install cordova-universal-links-plugin

step 2 : add below tag in your project's config file

<universal-links>
    <host name="yourappname.com" scheme="http">
        <path event="launchedAppFromLink" url="*" />
    </host>
</universal-links>

step 3 : Subscribe the above event "launchedAppFromLink" on device ready event. by adding below code

universalLinks.subscribe('launchedAppFromLink', 
onApplicationDidLaunchFromLink);

function onApplicationDidLaunchFromLink(eventData) {

  console.log('Did launch app from the link: ' + eventData.url);

}

after this your app link will be like : http://yourappname.com/

step 4 : on clicking above link if your app is installed in device then it will open.

step 5 : to open app specific register on branch.io

step 6 : register your app. After app registration it gives a link which is used to redirect or open the app.

step 7 : after that link just add your page name for example "myappname.app.link/yourapppage.html" and it will open that same page in mobile if your app is installed and if not then it will redirect to playstore with your app link