Open an image using URI in Android's default g

Question

I have extracted image uri, now I would like to open image with Android's default image viewer. Or even better, user could choose what program to use to open the image. Something like File Explorers offer you if you try to open a file.

Answer 1

Ask myself, answer myself also:

startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("content://media/external/images/media/16"))); /** replace with your own uri */

It will also ask what program to use to view the file.

Answer 2

If your app targets Android N (7.0) and above, you should not use the answers above (of the "Uri.fromFile" method), because it won't work for you.

Instead, you should use a ContentProvider.

For example, if your image file is in external folder, you can use this (similar to the code I've made here) :

File file = ...;
final Intent intent = new Intent(Intent.ACTION_VIEW)//
                                    .setDataAndType(VERSION.SDK_INT >= VERSION_CODES.N ?
                                                    android.support.v4.content.FileProvider.getUriForFile(this,getPackageName() + ".provider", file) : Uri.fromFile(file),
                            "image/*").addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);

manifest:

<provider
    android:name="android.support.v4.content.FileProvider"
    android:authorities="${applicationId}.provider"
    android:exported="false"
    android:grantUriPermissions="true">
    <meta-data
        android:name="android.support.FILE_PROVIDER_PATHS"
        android:resource="@xml/provider_paths"/>
</provider>

res/xml/provider_paths.xml :

<?xml version="1.0" encoding="utf-8"?>
<paths>
    <!--<external-path name="external_files" path="."/>-->
    <external-path
        name="files_root"
        path="Android/data/${applicationId}"/>
    <external-path
        name="external_storage_root"
        path="."/>
</paths>

If your image is in the private path of the app, you should create your own ContentProvider, as I've created "OpenFileProvider" on the link.

Answer 3

Try use it:

Uri uri =  Uri.fromFile(entry);
Intent intent = new Intent(android.content.Intent.ACTION_VIEW);
String mime = "*/*";
MimeTypeMap mimeTypeMap = MimeTypeMap.getSingleton();
if (mimeTypeMap.hasExtension(
    mimeTypeMap.getFileExtensionFromUrl(uri.toString())))
    mime = mimeTypeMap.getMimeTypeFromExtension(
        mimeTypeMap.getFileExtensionFromUrl(uri.toString()));
intent.setDataAndType(uri,mime);
startActivity(intent);

Answer 4

I use this it works for me

Intent intent = new Intent();
intent.setType("image/*");
intent.setAction(Intent.ACTION_GET_CONTENT);
startActivityForResult(Intent.createChooser(intent,
"Select Picture"), 1);

Answer 5

A much cleaner, safer answer to this problem (you really shouldn't hard code Strings):

public void openInGallery(String imageId) {
  Uri uri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI.buildUpon().appendPath(imageId).build();
  Intent intent = new Intent(Intent.ACTION_VIEW, uri);
  startActivity(intent);
}

All you have to do is append the image id to the end of the path for the EXTERNAL_CONTENT_URI. Then launch an Intent with the View action, and the Uri.

The image id comes from querying the content resolver.

Answer 6

My solution

Intent intent = new Intent();
intent.setAction(Intent.ACTION_VIEW);
intent.setDataAndType(Uri.fromFile(new File(Environment.getExternalStorageDirectory().getPath()+"/your_app_folder/"+"your_picture_saved_name"+".png")), "image/*");
context.startActivity(intent);