如何链接URL以启动我的Android应用程序？

Question

In my app, I want to code a share feature to allow the user to share his/her post.

Now let's say my friend shared his post and I got the link. How can I get that link to launch in my app? And how do I choose which activity it goes to? And how do I get the link so I know which post it's linking to?

I think this is done with intents, but I have no idea how to do it.

Answer 1

Try this, hope it will work for you

String url = "http://www.xample.com";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i); 

Answer 2

Add an intent-filter for your Activity (in the Manifest of your app):

<intent-filter>
    <action android:name="android.intent.action.VIEW" />
    <category android:name="android.intent.category.DEFAULT" />
    <category android:name="android.intent.category.BROWSABLE" />
    <data android:host="example.com"
    android:pathPattern=".*"
    android:scheme="http" />
</intent-filter>

In this example, your app will handle all links which starts with http://example.com/

Then, in your Activity, this way you can read the URL which was used to call your app:

Uri uri = getIntent().getData();

This will return something like this: http://example.com/link.php?post=1

Use the different methods of uri to get the required information.