Develop Reference

How can I get file path on raw resources in Android Studio?

I have been using this tutorial to make some face detections on picture. The problem is when I getting the file path that used on java
String xmlFile = "E:/OpenCV/facedetect/lbpcascade_frontalface.xml";
CascadeClassifier classifier = new CascadeClassifier(xmlFile);
How can I translate on android studio. I try put my lbpcascade_frontalface.xml on raw resources. CascadeClassifier is a class that opencv library provided. The only problem is they only loaded string path (on xmlfile).
this is my code.
String pathtoRes = getRawPathAtempt2(context);
CascadeClassifier cascadeClassifier = new CascadeClassifier();
cascadeClassifier.load(pathtoRes);
I translated to some method like this.
public String getRawPathAtempt2(Context context) {
return "android.resource://" + context.getPackageName() + "/raw/" + "lbpcascade_frontalface.xml";
}
I get the asertions error by opencv that tell me the file is null. Thats mean I had been wrong when I used file path on my method. how can I get file path on raw resources? help me please I have been stuck for several days now

how can i get file path on raw resources?
You can't. There is no path. It is a file on your development machine. It is not a file on the device. Instead, it is an entry in the APK file that is your app on the device.
If your library supports an InputStream instead of a filesystem path, getResources().openRawResource() on a Context to get an InputStream on your raw resource.

This is how i solved the problem with #CommonWare Answer
i'm using input stream to get file
is = context.getResources().openRawResource(R.raw.lbpcascade_frontalface);
so make the file will be opened and i will make a point to the File class
File cascadeDir = context.getDir("cascade", Context.MODE_PRIVATE);
mCascadeFile = new File(cascadeDir, "lbpcascade_frontalface.xml");
and then i make a search to the path file using getAbsolutePath like this
cascadeClassifier.load(mCascadeFile.getAbsolutePath());
take a look at my full code
try {
is = context.getResources().openRawResource(R.raw.lbpcascade_frontalface);
File cascadeDir = context.getDir("cascade", Context.MODE_PRIVATE);
mCascadeFile = new File(cascadeDir, "lbpcascade_frontalface.xml");
os = new FileOutputStream(mCascadeFile);
byte[] buffer = new byte[4096];
int bytesRead;
while ((bytesRead = is.read(buffer)) != -1) {
os.write(buffer, 0, bytesRead);
}
is.close();
os.close();
cascadeClassifier.load(mCascadeFile.getAbsolutePath());
} catch (IOException e) {
Log.i(TAG, "face cascade not found");
}

Uri video = Uri.parse("android.resource://com.test.test/raw/filename");
Using this you can access the file in raw folder, if you want to access the file in asset folder use this URL...
file:///android_asset/filename
Make sure you don't add the extension to the filename. E.g. "/movie" not "/movie.mp4".
Refer to this link:
Raw folder url path?

We can use this Alternative method for finding the path of Files in Raw Folder.
InputStream inputStreamdat=getResources().openRawResource(R.raw.face_landmark_model);
File model=getDir("model", Context.MODE_PRIVATE);
File modelFile=new
File(model,"face_landmark_model.dat");
FileOutputStream os1 = new
FileOutputStream(modelFile);
byte[] buffer1 = new byte[4096];
int bytesRead1;
while ((bytesRead1=inputStreamdat.read(buffer1))!=-1) {
os1.write(buffer1, 0, bytesRead1);
}
inputStreamdat.close();
os1.close();
You Can Use
modelFile.getAbsolutePath()); For getting the Path

How can I retrieve an Android resource from its name (in Kotlin)? [duplicate]

I want to open a file from the folder res/raw/.
I am absolutely sure that the file exists.
To open the file I have tried
File ddd = new File("res/raw/example.png");
The command
ddd.exists();
yields FALSE. So this method does not work.
Trying
MyContext.getAssets().open("example.png");
ends up in an exception with getMessage() "null".
Simply using
R.raw.example
is not possible because the filename is only known during runtime as a string.
Why is it so difficult to access a file in the folder /res/raw/ ?

With the help of the given links I was able to solve the problem myself. The correct way is to get the resource ID with
getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
"raw", getPackageName());
To get it as a InputStream
InputStream ins = getResources().openRawResource(
getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
"raw", getPackageName()));

Here is example of taking XML file from raw folder:
InputStream XmlFileInputStream = getResources().openRawResource(R.raw.taskslists5items); // getting XML
Then you can:
String sxml = readTextFile(XmlFileInputStream);
when:
public String readTextFile(InputStream inputStream) {
ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
byte buf[] = new byte[1024];
int len;
try {
while ((len = inputStream.read(buf)) != -1) {
outputStream.write(buf, 0, len);
}
outputStream.close();
inputStream.close();
} catch (IOException e) {
}
return outputStream.toString();
}

You can read files in raw/res using getResources().openRawResource(R.raw.myfilename).
BUT there is an IDE limitation that the file name you use can only contain lower case alphanumeric characters and dot. So file names like XYZ.txt or my_data.bin will not be listed in R.

Here are two approaches you can read raw resources using Kotlin.
You can get it by getting the resource id. Or, you can use string identifier in which you can programmatically change the filename with incrementation.
Cheers mate 🎉
// R.raw.data_post
this.context.resources.openRawResource(R.raw.data_post)
this.context.resources.getIdentifier("data_post", "raw", this.context.packageName)

FileReader file path in Android

I created A Buffered Reader, File Reader in my Android app and placed two text files inside raw folder. Now I have to pass a file path into File Reader, but what is the path of my file now?
BufferedReader reader = new BufferedReader(new FileReader("path"));
image link

but what is the path of my file now?
You don't have to pass file path into FileReader for reading file, here you can check following code snippet.
InputStream inputStream = null;
try {
inputStream = getResources().openRawResource(R.raw.hello_world);
byte[] reader = new byte[inputStream.available()];
while (inputStream.read(reader) != -1) {}
editField.setText(new String(reader));
editField.setSelection(editField.getText().length());
} catch(IOException e) {
Log.e(LOG_APP_TAG, e.getMessage());
} finally {
if (inputStream != null) {
try {
inputStream.close();
} catch (IOException e) {
Log.e(LOG_APP_TAG, e.getMessage());
}
}
}
}
Raw folder is for what in ANDROID
Arbitrary files to save in their raw form. To open these resources with a raw InputStream, call Resources.openRawResource() with the resource ID, which is R.raw.filename.In above example R.raw.hello_world.
However, if you need access to original file names and file hierarchy, you might consider saving some resources in the assets/ directory (instead of res/raw/). Files in assets/ are not given a resource ID, so you can read them only using AssetManager.
Raw mostly used with media files.

By default, you will be working in your present class path. So, you have to go back to the folder where the raw files is present and go inside it and access the files you want to.
file:///
will go back to your project directory, the one that contains all the folders. From there , you can access the raw folder.

Why does AssetManager list() not show my assets folder?

In my Android app I have packaged a file in the /assets folder that needs to be copied to the SDCARD. When I list the contents of the assets folder to a String[] I get "images", "sounds", "webkit" and "kioskmode" but not my file manually added to the assets folder.
My code is here:
private void copyAsset () {
AssetManager am = getApplicationContext().getAssets();
String[] files = null;
try {
files = am.list("");
} catch (IOException e) {
}
for (String filename : files) {
if (filename.equals("images") || filename.equals("kioskmode") ||
filename.equals("sounds") || filename.equals("webkit")) {
Log.i(TAG, "Skipping folder " + filename);
continue;
}
InputStream in = null;
OutputStream out = null;
try {
in = am.open(filename);
File outFile = new File(Environment.getExternalStorageDirectory().getPath(), filename);
out = new FileOutputStream(outFile);
copyFile(in, out);
in.close();
in = null;
out.flush();
out.close();
out = null;
} catch (IOException e) {
e.printStackTrace();
Log.e(TAG, "Error copying asset ", e);
}
}
}
Does it make a difference that this is a second class in my app and is called in my MainActivity using Intent showHelper = new Intent(this, HelperManager.class);
startActivity(showHelper); ?
I have tried the 2nd line (AssetManager am = ...) with and without the getApplicationContext() bit, tried moving the file into a subfolder of /assets and tried files = am.list("") with leading and trailing slashes. If I use a subfolder the files array is empty when the code runs (set a breakpoint on the files = am.list(""); line and inspected it at run time.
The strange thing is that it worked once - when I first wrote the code, but for further testing, I deleted the file from the /sdcard folder on the phone, and it never worked since even though the file is still in the assets folder.
I am using Android Studio if that matters.
Thanks

Managed to get a solution using Load a simple text file in Android Studio as a fix. It still puts the 4 folders in the files array but I Can skip them using code as given above, although I should rather check for the file I want rather than the 4 I don't!

Android - How do I open varying files from res folder?

Ok, so far all the answers I've found seem to be answered to by people who do not know the answer...
This should be a simple one (free rep for you yay):
I have a file in res/raw/ called overworld_a.tmx
I need to load it using the path as a string then the filename added to the string, since it varies.
Like so:
String mapName = "overworld_a.tmx";
try {
TMXMapReader mapReader = new TMXMapReader();
map = mapReader.readMap("raw/"+mapName);
} catch (Exception e) {
System.out.println("Error while reading the map:\n" + e.getMessage());
return;
}
Problem is, this obviously doesn't work.... Simple Question is, what is the relative path to that file?
And if this is not a possibility:
Complex question is, how do I open varying files from res folder and its children?

You can read overworld_a.tmx from res/raw dir as :
1. use openRawResource for reading overworld_a.tmx as InputStream from raw folder :
String mapName = "overworld_a"; //<< just pass name without file extension
resID = getResources().getIdentifier(mapName, "raw", getPackageName());
InputStream inputStream = getResources().openRawResource(resID);
2. Get ByteArrayOutputStream from inputStream :
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
//... your code for reading byteArray from inputStream
3. Pass byteArrayOutputStream to mapReader.readMap :
TMXMapReader mapReader = new TMXMapReader();
map = mapReader.readMap(byteArrayOutputStream);
//.....