How to force Share Intent to open a specific app?

Question

I like share intent, it is perfect to open sharing apps with image and text parameters.

But now i'm researching the way to force share intent to open a specific app from the list, with the paramters given to the share intent.

This is my actual code, it shows the list of sharing apps installed on the phone. Please, can somedone tell me what i should add to the code to force for example official twitter app? and official faccebok app?

Intent sharingIntent = new Intent(Intent.ACTION_SEND);
Uri screenshotUri = Uri.parse("file:///sdcard/test.jpg");  
sharingIntent.setType("image/*");
sharingIntent.putExtra(Intent.EXTRA_TEXT, "body text");
sharingIntent.putExtra(Intent.EXTRA_STREAM, screenshotUri);
startActivity(Intent.createChooser(sharingIntent, "Share image using"));

Thanks

Solution 1

For Facebook

public void shareFacebook() {
        String fullUrl = "https://m.facebook.com/sharer.php?u=..";
        try {
            Intent sharingIntent = new Intent(Intent.ACTION_SEND);
            sharingIntent.setClassName("com.facebook.katana",
                    "com.facebook.katana.ShareLinkActivity");
            sharingIntent.putExtra(Intent.EXTRA_TEXT, "your title text");
            startActivity(sharingIntent);

        } catch (Exception e) {
            Intent i = new Intent(Intent.ACTION_VIEW);
            i.setData(Uri.parse(fullUrl));
            startActivity(i);

        }
    }

For Twitter.

public void shareTwitter() {
        String message = "Your message to post";
        try {
            Intent sharingIntent = new Intent(Intent.ACTION_SEND);
            sharingIntent.setClassName("com.twitter.android","com.twitter.android.PostActivity");
            sharingIntent.putExtra(Intent.EXTRA_TEXT, message);
            startActivity(sharingIntent);
        } catch (Exception e) {
            Log.e("In Exception", "Comes here");
            Intent i = new Intent();
            i.putExtra(Intent.EXTRA_TEXT, message);
            i.setAction(Intent.ACTION_VIEW);
            i.setData(Uri.parse("https://mobile.twitter.com/compose/tweet"));
            startActivity(i);
        }
    }

Solution 2

There is a way more generic to do that, and is not necessary to know the full package name of the application intent.

See this post: How to customize share intent in Android?

Solution 3

100% Working solution

if you want to share something in any app you want or open a url via every action, just use this method:

private void shareOrViewUrlViaThisApp(String appPackageName, String url) {
    boolean found = false;
    Intent intent = new Intent(Intent.ACTION_VIEW);
    intent.setData(Uri.parse(url));
    
    List<ResolveInfo> resInfo = getPackageManager().queryIntentActivities(intent, 0);
    if (!resInfo.isEmpty()){
        for (ResolveInfo info : resInfo) {
            if (info.activityInfo.packageName.toLowerCase().contains(appPackageName) ||
                    info.activityInfo.name.toLowerCase().contains(appPackageName) ) {
            
                intent.setPackage(info.activityInfo.packageName);
                found = true;
                break;
            }
        }
        if (!found)
            return;

        startActivity(Intent.createChooser(intent, "Select"));
    }
}

and simply call:

shareOrViewUrlViaThisApp(<your package name>,<your url>);

This answer was inspired from this.