Android 浏览器中启动自定义应用

By Felix

Use an <intent-filter> with a <data> element. For example, to handle all links to twitter.com, you'd put this inside your <activity> in your AndroidManifest.xml:

<intent-filter>
    <data android:scheme="http" android:host="twitter.com"/>
    <action android:name="android.intent.action.VIEW" />
</intent-filter>

Then, when the user clicks on a link to twitter in the browser, they will be asked what application to use in order to complete the action: the browser or your application.

Of course, if you want to provide tight integration between your website and your app, you can define your own scheme:

<intent-filter>
    <data android:scheme="my.special.scheme" />
    <action android:name="android.intent.action.VIEW" />
</intent-filter>

Then, in your web app you can put links like:

<a href="my.special.scheme://other/parameters/here">

And when the user clicks it, your app will be launched automatically (because it will probably be the only one that can handle my.special.scheme:// type of uris). The only downside to this is that if the user doesn't have the app installed, they'll get a nasty error. And I'm not sure there's any way to check.