How to implement my very own URI scheme on Android

AndroidAndroid IntentBrowserUriIntentfilter

Android Problem Overview


Say I want to define that an URI such as:

myapp://path/to/what/i/want?d=This%20is%20a%20test

must be handled by my own application, or service. Notice that the scheme is "myapp" and not "http", or "ftp". That is precisely what I intend: to define my own URI schema globally for the Android OS. Is this possible?

This is somewhat analogous to what some programs already do on, e.g., Windows systems, such as Skype (skype://) or any torrent downloader program (torrent://).

Android Solutions


Solution 1 - Android

This is very possible; you define the URI scheme in your AndroidManifest.xml, using the <data> element. You setup an intent filter with the <data> element filled out, and you'll be able to create your own scheme. (More on intent filters and intent resolution here.)

Here's a short example:

<activity android:name=".MyUriActivity">
	<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:scheme="myapp" android:host="path" />
	</intent-filter>
</activity>

As per how implicit intents work, you need to define at least one action and one category as well; here I picked VIEW as the action (though it could be anything), and made sure to add the DEFAULT category (as this is required for all implicit intents). Also notice how I added the category BROWSABLE - this is not necessary, but it will allow your URIs to be openable from the browser (a nifty feature).

Solution 2 - Android

Complementing the @DanielLew answer, to get the values of the parameteres you have to do this:

URI example: myapp://path/to/what/i/want?keyOne=valueOne&keyTwo=valueTwo

in your activity:

Intent intent = getIntent();
if (Intent.ACTION_VIEW.equals(intent.getAction())) {
  Uri uri = intent.getData();
  String valueOne = uri.getQueryParameter("keyOne");
  String valueTwo = uri.getQueryParameter("keyTwo");
}

Solution 3 - Android

I strongly recommend that you not define your own scheme. This goes against the web standards for URI schemes, which attempts to rigidly control those names for good reason -- to avoid name conflicts between different entities. Once you put a link to your scheme on a web site, you have put that little name into entire the entire Internet's namespace, and should be following those standards.

If you just want to be able to have a link to your own app, I recommend you follow the approach I described here:

https://stackoverflow.com/questions/2430045/how-to-register-some-url-namespace-myapp-app-start-for-accessing-your-progra/2430468#2430468

Solution 4 - Android

Another alternate approach to Diego's is to use a library:

https://github.com/airbnb/DeepLinkDispatch

You can easily declare the URIs you'd like to handle and the parameters you'd like to extract through annotations on the Activity, like:

@DeepLink("path/to/what/i/want")
public class SomeActivity extends Activity {
  ...
}

As a plus, the query parameters will also be passed along to the Activity as well.

Solution 5 - Android

As the question is asked years ago, and Android is evolved a lot on this URI scheme.
From original URI scheme, to deep link, and now Android App Links.

Android now recommends to use HTTP URLs, not define your own URI scheme. Because Android App Links use HTTP URLs that link to a website domain you own, so no other app can use your links. You can check the comparison of deep link and Android App links from here

Now you can easily add a URI scheme by using Android Studio option: Tools > App Links Assistant. Please refer the detail to Android document: https://developer.android.com/studio/write/app-link-indexing.html

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionpunnieView Question on Stackoverflow
Solution 1 - AndroidDan LewView Answer on Stackoverflow
Solution 2 - AndroidDiego PalomarView Answer on Stackoverflow
Solution 3 - AndroidhackbodView Answer on Stackoverflow
Solution 4 - AndroidChristianView Answer on Stackoverflow
Solution 5 - AndroidWeidian HuangView Answer on Stackoverflow