How to set file or url "open with" programmatically?

Question

When the file or url open, Android ask to user how to open with. My application need to set this option. I cannot completed to finish making application if no solution. Please tell me how to do.

Answer 1

If you share a file with Intent you can first create your Intent openWith object (but not run it yet). Then list all available activities for it like this:

PackageManager pm = (Activity)this.getPackageManager();
List<ResolveInfo> resInfo = pm.queryIntentActivities(openWith, 0);
for (int i = 0; i < resInfo.size(); i++) {
    ResolveInfo ri = resInfo.get(i);
    String packageName = ri.activityInfo.packageName;
    String label = (String) ri.loadLabel(pm);
    String activity = ri.activityInfo.name;
    // choose needed packageName and activity here
}

Then add selected packageName and activity pair to the intent for "open with" action:

intent.setComponent(new ComponentName(packageName, activity));

and run it.

Answer 2

You want to implement ability open files/urls at your app, right? Here is doc https://developer.android.com/training/basics/intents/filters.html So your 'viewer' activity must have intent filter like this

<intent-filter>
    <action android:name="android.intent.action.VIEW" />
    <category android:name="android.intent.category.DEFAULT"/>
    <data android:mimeType="*/*" />
</intent-filter>
<intent-filter>
    <action android:name="android.intent.action.VIEW" />
    <category android:name="android.intent.category.DEFAULT"/>
    <data android:scheme="http|https" />
</intent-filter>