How to simply open directory/folder?

Question

DONT mark this as duplicate, before reading what I need.

I have seen many similar topics, but in none of them I've found solution. I need the simplest thing: In my application I have button "View Media Files". After clicking that button, i need to be opened (with built-in File Explorer) this directory - SD_CARD/my_folder where are media files (and I want to click any of them and they should be opened in default Media player)..

I have used all suggested answers on SO , like this:

    Intent intent = new Intent(Intent.ACTION_VIEW);
    Uri mydir = Uri.parse("/sdcard/Recorder_Videos");
    intent.setDataAndType(mydir, "*/*");
    startActivity(intent);

but all they do: after clicking button, it opens "Choose File" menu: (Where I cant still play media files when clicking)

Answer 1

The solution (not complete) I have found, was that I was missing file:// prefix. Here is my solution (however, it shows all kinds of applications on first view):

public void openFolder(String location)
{
    // location = "/sdcard/my_folder";
    Intent intent = new Intent(Intent.ACTION_VIEW);
    Uri mydir = Uri.parse("file://"+location);
    intent.setDataAndType(mydir,"application/*");    // or use */*
    startActivity(intent);
}

p.s. Strange and surprising, but there doesnt exist the standard definition of "File Browser" in stock Android systems (unless you install 3rd party "File Explorer").. That's why "resource/folder" mime-type doesnt work by default..

However, let's say a simple truth. File-Browser is a SIMPLE and ESSENTIAL part of any OS system. And it's quite disrespectful from Android, saying that it's not their job, and throwing the responsiblity to 3rd party apps.

Answer 2

You can use type DocumentsContract.Document.MIME_TYPE_DIR which works on several devices and launches File Explorer. You can refer this SO for more details.