Android：如何按资源ID检索文件名和资源扩展名

Question

I have the following:

getResources().getResourceEntryName(resourceId);

The problem is, that it retrieves only the file name without the extension.

For example, if I have the following res/drawable/pic.jpg, the

getResources().getResourceEntryName(resourceId);

is returning the value "pic". The extension .jpg is missing.

Answer 1

To get "res/drawable/pic.jpg" you could use this:

    TypedValue value = new TypedValue();
    getResources().getValue(resourceId, value, true);
    // check value.string if not null - it is not null for drawables...

    Log.d(TAG, "Resource filename:" + value.string.toString());
    // ^^ This would print res/drawable/pic.jpg

Source: android/frameworks/base/core/java/android/content/res/Resources.java

Answer 2

You can do so

    Field[] fields = R.raw.class.getFields();
    for (int count = 0; count < fields.length; count++) {

        // Use that if you just need the file name
        String filename = fields[count].getName();
        Log.i("filename", filename);
        int rawId = getResources().getIdentifier(filename, "raw", getPackageName());
        TypedValue value = new TypedValue();
        getResources().getValue(rawId, value, true);
        String[] s = value.string.toString().split("/");
        Log.i("filename", s[s.length - 1]);
    }

Answer 3

Short answer: You can't.

Another way to do this, would be to put your graphics inside the assets folder. Then you can access the Files directly, without your App needing any permission.

You can, for example, do so in your Activity:

AssetManager am = this.getApplicationContext().getAssets()
InputStream is = am.open(foldername+"/"+filename)
Bitmap myNewImage = BitmapFactory.decodeStream(is);

I hope that this will accomplish what you had in mind.

---

UPDATE : it seems it is indeed possible, see Aleksandar Stojiljkovic's answer instead.

Answer 4

This should be the best solution:

 TypedValue value = new TypedValue();
    getResources().getValue(resourceId, value, true); 
String resname = value.string.toString().substring(13, value.string.toString().length());

resname = "pic.jpg"