Android open file

Question

I was trying to open a file for reading.

When using: Scanner input = new Scanner(filename); the file could not be found

but when I used:

Answer 1

This drove me crazy couple of minutes ago. I forgot to add this line to manifest:

<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />

I would expect a permission denied message. But just got a file not found...

In your case: openFileInput opens a file in your private app data directory (/data/data/your.package/filename). This never fails. But the scanner tries to open it on the root path. So when you want to read a file from SD card than you would use Environement.getExternalStorageDirectory().getAbsolutePath() + "/" + filename.

Answer 2

Files are stored on the device in a specific, application-dependent location, which is what I suppose openFileInput adds at the beginning of the file name. The final result (location + file name) is constructed as follows:

/data/data/<application-package>/files/<file-name>

Note also that the documentation states that the openFileInput parameter cannot contain path separators.

To avoid hard-coding the location path, which could in principle even be different from device to device, you can obtain a File object pointing to the storage directory by calling getFilesDir, and use it to read whatever file you would like to. For example:

File filesDir = getFilesDir();
Scanner input = new Scanner(new File(filesDir, filename));

Note that constructing a Scanner by passing a String as a parameter would result in the scanner working on the content of the string, i.e. interpreting it as the actual content to scan instead of as the name of a file to open.

Answer 3

Scanner sc = new Scanner(new File(filename));