zSt zSt - 1 year ago 160
Android Question

Android open file

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:

InputStream in = openFileInput(filename);
Scanner input = new Scanner(in);

It worked. Why was the first line of code wrong?

Answer Source

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:


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.

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