How to get URI from an asset File?

Question

I have been trying to get the URI path for an asset file.

uri = Uri.fromFile(new File("//assets/mydemo.txt"));

When I check if the file exists I see that file doesn't exist

File f = new File(filepath);
if (f.exists() == true) {
    Log.e(TAG, "Valid :" + filepath);
} else {
    Log.e(TAG, "InValid :" + filepath);
}

Can some one tell me how I can mention the absolute path for a file existing in the asset folder

Heterogenesis

Answer 1

There is no "absolute path for a file existing in the asset folder". The content of your project's assets/ folder are packaged in the APK file. Use an AssetManager object to get an InputStream on an asset.

For WebView, you can use the file Uri scheme in much the same way you would use a URL. The syntax for assets is file:///android_asset/... (note: three slashes) where the ellipsis is the path of the file from within the assets/ folder.

Viperish

The reason why I wanted an absolute path was that I wanted a URI for the file. – Heterogenesis

file://android_asset/..., where ... is the relative path within your project's assets/ directory. – Viperish

I tried that and i get it still as an invalid path. I have verified the apk is containing the txt file within its asset folder – Heterogenesis

Here is a sample project using file://android_asset/, though not for a Uri: github.com/commonsguy/cw-advandroid/tree/master/WebView/GeoWeb1 – Viperish

I can confirm that Commons' tip works at least for the purpose of showing resources in a WebView. – Puri

@Viperish it looks like you missed a '/' in your URI. It should be file:///. – Molt

I tried this as new File("file:///android_asset/sounds/beep.mid");, but result file doesn't exists. Any ideas? P.S. file is in assets/sounds/ folder. – Caribou

Is it right that ´file:///android_asset/...´ works only in WebView and is not possible to get InputStream using URI/URL/Uri outside WebView? – Kimi

@Lubbo: AFAIK, the only thing that pays attention to file:///android_asset/ is WebView. It's possible that ContentResolver and openInputStream() works with it, but I do not recall that it does. The syntax in that earlier comment definitely will not work. – Viperish

for a quick note android_asset is a not your asset directory's name, its same no mater what is your asset directory's name,file:///android_asset/ this part will be same no mater what is your asset directory's name – Legalize

Answer 2

The correct url is:

file:///android_asset/RELATIVEPATH

where RELATIVEPATH is the path to your resource relative to the assets folder.

Note the 3 /'s in the scheme. Web view would not load any of my assets without the 3. I tried 2 as (previously) commented by CommonsWare and it wouldn't work. Then I looked at CommonsWare's source on github and noticed the extra forward slash.

This testing though was only done on the 1.6 Android emulator but I doubt its different on a real device or higher version.

EDIT: CommonsWare updated his answer to reflect this tiny change. So I've edited this so it still makes sense with his current answer.

Kallick

Doesn't work anymore String fileName = "file:///android_asset/file.csv"; System.out.println(new File(fileName).exists()); // prints false – Vaal

@Vaal the constructor for File that accepts a string is expecting a path not a URI. I haven't coded for Android for a few years and things may have changed. I'm unsure if you can access the embedded assets folder using a normal java File class. I seem to remember that you have to use an AssetManager. Files in the assets folder are placed on the device in the read-only bundled APK and compressed (APKs are actually zip files), thus more work to read it back. Try something like new File(new URI(filename)) – Kallick

Answer 3

Finally, I found a way to get the path of a file which is present in assets from this answer in Kotlin. Here we are copying the assets file to cache and getting the file path from that cache file.

@Throws(IOException::class)
    fun getFileFromAssets(context: Context, fileName: String): File = File(context.cacheDir, fileName)
            .also {
               if (!it.exists()) {
                it.outputStream().use { cache ->
                    context.assets.open(fileName).use { inputStream ->
                            inputStream.copyTo(cache)
                    }
                  }
                }
            }

Get the path to the file like:

val filePath =  getFileFromAssets(context, "fileName.extension").absolutePath

Belshin

Notice that if the fileName have directory dir/file. It will crash with path the cache/dir/file not exist. Also I think better have a checking if(!it.exist) inside the also block, then it will not copy the file every time even the file is already on the cache – Tolan

@ShylendraMadda, @Yeung, if(!it.exist) thows an error unresolved reference. any solution ? – Fetus

Check that file exists in your cache with that file name? @Fetus – Belshin

@ShylendraMadda, I know filename only. Can we load it with script without knowing actual uri – Fetus

I think if(!it.exist) should put before it.outputStream()... – Tolan

@Tolan Updated, Thanks – Belshin

@SamChen Glad that it helped you – Belshin

@Shailendra Madda Can you help me with this: https://mcmap.net/q/48973/-android-open-assets-image-file-with-intent-and-fileprovider-shows-size-0/3466808, would be very appreciated! – Rebuke

for someone if have some dir before the file name, like -> ("raw/filename.ext" ), inside if (!it.exists()) block befor the it.outputStream().use { cache -> ... add this code to create directory if is not exist --> val fileDir = File( context.cacheDir, File.separator.toString() + "raw" ) fileDir.mkdir() – Bosch

Answer 4

Please try this code working fine

 Uri imageUri = Uri.fromFile(new File("//android_asset/luc.jpeg"));

    /* 2) Create a new Intent */
    Intent imageEditorIntent = new AdobeImageIntent.Builder(this)
            .setData(imageUri)
            .build();

Prepay

Answer 5

Be sure ,your assets folder put in correct position.

Imray

You can put it wherever you want and define the path in your build.gradle file as follows sourceSets { main { assets.srcDirs = ['src/main/assets'] } } – Abatement

How to get path of hello.html ? please add – Unclog

Answer 6

Works for WebView but seems to fail on URL.openStream(). So you need to distinguish file:// protocols and handle them via AssetManager as suggested.

Familiar

Answer 7

InputStream is = getResources().getAssets().open("terms.txt");
String textfile = convertStreamToString(is);
    
public static String convertStreamToString(InputStream is)
        throws IOException {

    Writer writer = new StringWriter();
    char[] buffer = new char[2048];

    try {
        Reader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"));
        int n;
        while ((n = reader.read(buffer)) != -1) {
            writer.write(buffer, 0, n);
        }
    } finally {
        is.close();
    }

    String text = writer.toString();
    return text;
}

Diazotize

Answer 8

Try this out, it works:

InputStream in_s = 
    getClass().getClassLoader().getResourceAsStream("TopBrands.xml");

If you get a Null Value Exception, try this (with class TopBrandData):

InputStream in_s1 =
    TopBrandData.class.getResourceAsStream("/assets/TopBrands.xml");

Parkman

If not working , please Try this one : InputStream in_s1 = TopBrandData.class.getResourceAsStream("/assets/TopBrands.xml"); – Parkman

Answer 9

try this :

Uri uri = Uri.parse("android.resource://"+getPackageName()+"/"+R.raw.cat); 

I had did it and it worked

Googins

Actually it works. Check ContentResolver.openAssetFileDescriptor – Novelty

Worked. Thanks a lot. Lol, I just looking for get URI internal file // CREATE RAW FOLDER INSIDE RES, THEN NAME FILE CONTAIN ONLY a..z0..9 character, NO UPPERCASE !!! – Palpate

Answer 10

Since it's actually data in the APK file and not on the emulator, you won't be able to find an "absolute path". I've eventually implemented in this pretty straightforward way :

private fun getUriFromAsset(context: Context, assetFileName: String): Uri? {
    val assetManager = context.assets
    var inputStream: InputStream? = null
    var outputStream: FileOutputStream? = null
    var tempFile: File? = null

    return try {
        inputStream = assetManager.open(assetFileName)
        tempFile = File.createTempFile("temp_asset", null, context.cacheDir)
        outputStream = FileOutputStream(tempFile)

        inputStream.copyTo(outputStream)

        Uri.fromFile(tempFile)
    } catch (e: IOException) {
        e.printStackTrace()
        null
    } finally {
        inputStream?.close()
        outputStream?.close()
        tempFile?.deleteOnExit()
    }
}

Faris

Answer 11

Yeah you can't access your drive folder from you android phone or emulator because your computer and android are two different OS.I would go for res folder of android because it has good resources management methods. Until and unless you have very good reason to put you file in assets folder. Instead You can do this

try {
      Resources res = getResources();
      InputStream in_s = res.openRawResource(R.raw.yourfile);

      byte[] b = new byte[in_s.available()];
      in_s.read(b);
      String str = new String(b);
    } catch (Exception e) {
      Log.e(LOG_TAG, "File Reading Error", e);
 }

Crown

Answer 12

If you are okay with not using assets folder and want to get a URI without storing it in another directory, you can use res/raw directory and create a helper function to get the URI from resID:

internal fun Context.getResourceUri(@AnyRes resourceId: Int): Uri =
    Uri.Builder()
        .scheme(ContentResolver.SCHEME_ANDROID_RESOURCE)
        .authority(packageName)
        .path(resourceId.toString())
        .build()

Now if you have a mydemo.txt file under res/raw directory you can simply get the URI by calling the above helper method

context.getResourceUri(R.raw.mydemo)

Reference: https://stackoverflow.com/a/57719958

Rosemaria

Answer 13

Worked for me Try this code

   uri = Uri.fromFile(new File("//assets/testdemo.txt"));
   String testfilepath = uri.getPath();
    File f = new File(testfilepath);
    if (f.exists() == true) {
    Toast.makeText(getApplicationContext(),"valid :" + testfilepath, 2000).show();
    } else {
   Toast.makeText(getApplicationContext(),"invalid :" + testfilepath, 2000).show();
 }

Eccrinology