Question: Is there a way to implement OpenUrl functionality(as in iOS OpenUrl) in android?

for example,
If a user is redirected from a webpage(in a browser) to "`myapp://main`",
android will launch my app.

Answer: 
the answer is just to add this to manifest inside the activity you want to start :

```
<activity
    android:name="com.dimrix.something.BootActivity"
    android:label="@string/app_name"
    android:screenOrientation="portrait" >
    <intent-filter>
        <action android:name="android.intent.action.MAIN" />
        <category android:name="android.intent.category.LAUNCHER" />
    </intent-filter>

         <intent-filter>
            <data android:scheme="SomeScheme" />
            <data android:scheme="otherScheme" />
            <action android:name="android.intent.action.VIEW" />
            <category android:name="android.intent.category.DEFAULT" />
            <category android:name="android.intent.category.BROWSABLE" />
         </intent-filter>

    </activity>

```

and than if you write inside the browser 

SomeScheme:// 

or

otherScheme:// 

it will start the activity....

you can even send info after the scheme if you want to do action according to the url by getting the intent data in the activity

```
Uri data = getIntent().getData();

```

so you can write

SomeScheme://water

and respond to the "water" as you like with the Uri data .
That can be done with Intent Filters.

See [Allowing Other Apps to Start Your Activity](http://developer.android.com/training/basics/intents/filters.html) in the Android documentation, or answers to [this question](https://stackoverflow.com/questions/5596261/intent-filter-to-launch-my-activity-when-custom-uri-is-clicked).