Launch custom app from URL

Question

Hi i have tried so much to open custom app from url, but i failed , i have this link to open custom app click here

as per link i have done coding, my coding is below

AndroidManifest.xml file....

     <application
    android:icon="@drawable/icon"
    android:label="@string/app_name" >
    <uses-library android:name="com.google.android.maps" />

    <activity
        android:name=".AnimatedScreen"
        android:label="Last Alert Pro"
        android:screenOrientation="portrait" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

     <activity
        android:name=".MapInfo"
        android:label="@string/app_name"
        android:screenOrientation="portrait" >
        <activity
        android:name=".MapInfo"
        android:label="@string/app_name"
        android:screenOrientation="portrait" >
        <intent-filter>
            <data android:scheme="my.lastalertpro.scheme" />
            <action android:name="android.intent.action.VIEW" />
            <category android:name="android.intent.category.DEFAULT" />
            <category android:name="android.intent.category.BROWSABLE" />
        </intent-filter>
    </activity>
    </activity>

Now in coding part html link in browser is like below

    String linkForAndroid = "my.lastalertpro.scheme//" + lon + "/" + lat + "/" + alt;

Now i in Java part the coding for fetching data is

    Uri data = getIntent().getData();
    String scheme = data.getScheme(); // "http"

    if (scheme != null) {
    String host = data.getHost(); // "my.lastalertpro.scheme"
    List<String> params = data.getPathSegments();
     try {
    lon = Double.parseDouble(params.get(0)); // longitude
    } catch (Exception e) {
    lon = 0.0;
        }

Now the problem is that i am getting this link in gmail. when i got mail and try to click on this link the browser list is not displaying and also application is started directly in default browser :(

use the URL also in data tag as android:host="your website(www.google.com)" — Syn3sthete, Oct 15 '12 at 08:48

score 3 · Accepted Answer · answered Oct 15 '12 at 09:48

Finally i found the answer

if you send below link in gmail...

      String linkForAndroid = "my.lastalertpro.scheme//" + lon + "/" + lat + "/" + alt;

Do not know why but it's considered like this way in ANDROID PHONE ---NOT--- ON DESKTOP COMPUTER

  in mail if you got link and click on this the link will be looking like 

  http://my.lastalertpro.scheme/70.77188993/22.28305819/96+m.

so you have to pass intent-filter in AndroidManifest.xml file like this way

   <intent-filter>
      <data android:scheme="http" android:host="my.lastalertpro.scheme" />
      <action android:name="android.intent.action.VIEW" />
      <category android:name="android.intent.category.DEFAULT" />
      <category android:name="android.intent.category.BROWSABLE" />
   </intent-filter>

score 0 · Answer 2 · answered Oct 15 '12 at 08:54

0

Try adding the given below tags along with action_view

<action android:name="android.intent.action.VIEW" />
<category android:name="android.intent.category.DEFAULT" />
<category android:name="android.intent.category.BROWSABLE" />

answered Oct 15 '12 at 08:54

Hitesh Jain

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Launch custom app from URL

2 Answers2