Getting url with Intent Android

Question

Im trying to get the URL that was clicked in order to open the app. Im just having trouble finding where I can get that information. I have a manifest which opens app when clicked. The link would be "http://host.com/file.html?param=1&param2=2" which Im trying to give to the app to handle.

 <intent-filter> <data android:scheme="http" android:host="host.com" android:pathPrefix="/file.html" /> <action android:name="android.intent.action.VIEW" /> <category android:name="android.intent.category.BROWSABLE" /> <category android:name="android.intent.category.DEFAULT" /> </intent-filter>

EDIT

@Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); Intent intent = getIntent(); Uri uri = intent.getData(); try { url = new URL(uri.getScheme(), uri.getHost(), uri.getPath()); } catch (MalformedURLException e) { // TODO Auto-generated catch block e.printStackTrace(); } }

Ben Siver · Accepted Answer · 2015-02-09 22:04:29Z

8

You should be able to get the intent data in your activity with the following:

Uri uri = this.getIntent().getData(); URL url = new URL(uri.getScheme(), uri.getHost(), uri.getPath());

answered Feb 9, 2015 at 22:04

Ben Siver

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2 Comments

William McCarty Over a year ago

I added my current OnCreate method up top. still isn't seeming to work though.

William McCarty Over a year ago

My onCreate Method seems to be not being called, from my understanding it will be called when the system calls onCreate and it will redirect to the onCreate I made by overriding it.

ror · Accepted Answer · 2019-10-11 18:33:16Z

Basically once you obtain your Uri (that is, getIntent().getData()), you can read your complete url as uri.toString(). Assuming your android:host is valid and set, you can also get actual query data by calling uri.getEncodedQuery().

Here is more details from working sample, if needed:

This is my manifest setup:

 <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="boo.acme.com" android:scheme="http" /> </intent-filter>

This is how I read Uri:

Uri uri = this.getIntent().getData();

Assuming user clicked url "http://boo.acme.com/?ll=43.455095,44.177416", uri.toString() produces "http://boo.acme.com/?ll=43.455095,44.177416" and uri.getEncodedQuery() produces "ll=43.455095,44.177416".

Hope that helps!

Dr. Abhishek · Accepted Answer · 2019-02-22 20:19:47Z

Uri uri = this.getIntent().getData(); try { URL url = new URL(uri.getScheme(), uri.getHost(), uri.getPath()); } catch (MalformedURLException e) { }

You will have to Surround with try/catch.

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