CODEWITHSUNDEEP
javaandroidfile
Filippo Iacobellis
Filippo Iacobellis

Reputation: 19

Android open file input stream from uri

I'm trying to upload a file by copying it from mobile app to server app running on windows. I used a file chooser to let the user select the file:

public void openFile(View view) {
        Intent intent = new Intent();
        intent.setType("*/*");
        intent.setAction(Intent.ACTION_GET_CONTENT);
        startActivityForResult(Intent.createChooser(intent, "Select file"), LEARN_TREE);
}

Then i got the uri from the intent of onActivityResult. First question here is why it displays a file named "servo.dat" as numbers (in this case it shows "5889")?

After that I put the uri as an extra into another intent and use that intent to start another activity. In the second activity I retrieve the uri. Now I'd like to use FileInputStream to read bytes from my file in order to write them to the ObjectOutputStream created from Socket.getOutputStream(). Here is where it doesn't work. Basically the path provided here

FileInputStream fis = new FileInputStream(uri.getPath());

is incorrect. If I check on my device the file location is Download/servo.dat, the Uri in the app shows Download/5889 and the absolute path that I tried retrieving using a UriUtils library found online shows storage/emulated/0/Download/servo.dat but this one doesn't actually exist on my phone.

I think it's not so hard but I'm getting confused since I'm new to both Android app development and Android itself, please help!

I'm open to any good solution, I saw online there is the ContentResolver class that should be helpful but I didn't manage to understand how to use it :|

Upvotes: 1

Views: 5230

Answers (2)

Muthukumar
Muthukumar

Reputation: 1

Reading or open .xls(office Excel) type of file through uri with InputStram. In this displayXlsData(), i never use String obj(Ex: userName and password) mentioned below. So, hi viewers, Don't confuse. I just checked the result though Log.d() only.

Program Detail : Uri xls_UriData;

public void displayXlsData() throws IOException {

    InputStream myInput;
    myInput  = getContentResolver().openInputStream(xls_UriData);
    // Create a POI File System object
    POIFSFileSystem myFileSystem = new POIFSFileSystem(myInput);

    // Create a workbook using the File System
    HSSFWorkbook myWorkBook = new HSSFWorkbook(myFileSystem);

    // Get the first sheet from workbook
    HSSFSheet mySheet = myWorkBook.getSheetAt(0);

    // We now need something to iterate through the cells.
    Iterator<Row> rowIter = mySheet.rowIterator();
    int rowno =0;

    while (rowIter.hasNext()) {
            rowno++ ;
        HSSFRow myRow = (HSSFRow) rowIter.next();
        if(rowno !=0) {
            Iterator<Cell> cellIter = myRow.cellIterator();
            int colno =0;
            String userName="", password="";
            while (cellIter.hasNext()) {
                HSSFCell myCell = (HSSFCell) cellIter.next();
                if (colno==0){
                    userName = myCell.toString();
                }else if (colno==1){
                    password = myCell.toString();
                }
                Log.e("Excel_2", " Index :" + myCell.getColumnIndex() + " -- " + myCell);
            colno++;
            }
            
        }
    }
    rowno++;

} // Closing displayXlsData()

Upvotes: 0

CommonsWare
CommonsWare

Reputation: 1007276

First question here is why it displays a file named "servo.dat" as numbers (in this case it shows "5889")?

Because it is not a file. It is a piece of content, and you are attempting to treat the path portion of a Uri as a filesystem path, which is is not.

If you want a display name for the content:

  • Call DocumentFile.fromSingleUri(), passing in your Uri, to create a DocumentFile
  • Call getDisplayName on the DocumentFile

Basically the path provided here is incorrect

That is because you are trying to treat the path portion of a Uri as a filesystem path, which it is not.

To get an InputStream, call openInputStream() on a ContentResolver, passing in your Uri. See the documentation. So, for example, from a method in an Activity, you would use InputStream inputStream = getContentResolver().openInputStream(uri);.

I'm getting confused since I'm new to both Android app development and Android itself

You may wish to consider reading a book on Android app development or taking a course in Android app development.

Upvotes: 1

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