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how to get an uri of an image resource in android

Posted by: admin March 11, 2020 Leave a comment

Questions:

I need to open an intent to view an image as follows:

Intent intent = new Intent(Intent.ACTION_VIEW);
Uri uri = Uri.parse("@drawable/sample_1.jpg");
intent.setData(uri);
startActivity(intent);

The problem is that Uri uri = Uri.parse("@drawable/sample_1.jpg"); is incorrect.

How to&Answers:

The format is:

"android.resource://[package]/[res id]"

[package] is your package name

[res id] is value of the resource ID, e.g. R.drawable.sample_1

to stitch it together, use

Uri path = Uri.parse("android.resource://your.package.name/" + R.drawable.sample_1);

Answer:

Here is a clean solution which fully leverages the android.net.Uri class via its Builder pattern, avoiding repeated composition and decomposition of the URI string, without relying on hard-coded strings or ad hoc ideas about URI syntax.

Resources resources = context.getResources();
Uri uri = new Uri.Builder()
    .scheme(ContentResolver.SCHEME_ANDROID_RESOURCE)
    .authority(resources.getResourcePackageName(resourceId))
    .appendPath(resources.getResourceTypeName(resourceId))
    .appendPath(resources.getResourceEntryName(resourceId))
    .build();

Answer:

public static Uri resourceToUri(Context context, int resID) {
        return Uri.parse(ContentResolver.SCHEME_ANDROID_RESOURCE + "://" +
                context.getResources().getResourcePackageName(resID) + '/' +
                context.getResources().getResourceTypeName(resID) + '/' +
                context.getResources().getResourceEntryName(resID) );
    }

Answer:

For those having error, you may be entering the wrong package name. Just use this method.

public static Uri resIdToUri(Context context, int resId) {
    return Uri.parse(Consts.ANDROID_RESOURCE + context.getPackageName()
                     + Consts.FORESLASH + resId);
}

Where

public static final String ANDROID_RESOURCE = "android.resource://";
public static final String FORESLASH = "/";

Answer:

You want the URI of the image resource, and R.drawable.goomb is an image resource. The Builder function creates the URI that you are asking for:

String resourceScheme = "res";
Uri uri = new Uri.Builder()
  .scheme(resourceScheme)
  .path(String.valueOf(intResourceId))
  .build();

Answer:

Based on the answers above I want to share a kotlin example on how to get a valid Uri for any resource in your project. I think it’s the best solution because you don’t have to type any strings in your code and risk typing it wrongly.

    val resourceId = R.raw.scannerbeep // r.mipmap.yourmipmap; R.drawable.yourdrawable
    val uriBeepSound = Uri.Builder()
        .scheme(ContentResolver.SCHEME_ANDROID_RESOURCE)
        .authority(resources.getResourcePackageName(resourceId))
        .appendPath(resources.getResourceTypeName(resourceId))
        .appendPath(resources.getResourceEntryName(resourceId))
        .build()