Android how to write a file to internal storage

Question

I am developing a simple android application and I need to write a text file in internal storage device. I know there are a lot of questions (and answers) about this matter but I really cannot understand what I am doing in the wrong way.

This is the piece of code I use in my activity in order to write the file:

public void writeAFile(){
    String fileName = "myFile.txt";
    String textToWrite = "This is some text!";
    FileOutputStream outputStream;

   try {
      outputStream = openFileOutput(fileName , Context.MODE_PRIVATE);
      outputStream.write(textToWrite.getBytes());
      outputStream.close();
   } catch (Exception e) {
     e.printStackTrace();
   }
}

I really cannot understand which mistake I am doing. In addition, I have tried this project on my emulator in Android Studio and my phone in order to understand where I am doing something wrong but even with that project no file is written neither on the phone or on the emulator.

EDIT: I know that no file is written to my internal storage because I try to read the content of the file, after I have written to it, with this code:

public void ReadBtn(View v) {
    //reading text from file
    try {
        FileInputStream fileIn=openFileInput("myFile.txt");
        InputStreamReader InputRead= new InputStreamReader(fileIn);

        char[] inputBuffer= new char[READ_BLOCK_SIZE];
        String s="";
        int charRead;

        while ((charRead=InputRead.read(inputBuffer))>0) {
            String readstring=String.copyValueOf(inputBuffer,0,charRead);
            s +=readstring;
        }
        InputRead.close();
        textmsg.setText(s);
    } catch (Exception e) {
        e.printStackTrace();
    }
}

Nothing is shown at all.

Answer 1

Android offers openFileInput and openFileOutput from the Java I/O classes to modify reading and writing streams from and to local files. openFileOutput(): This method is used to create and save a file. Its syntax is given below: FileOutputStream fOut = openFileOutput("file name",Context.