CODEWITHSUNDEEP
androidfile
Bee
Bee

Reputation: 104

Why Failed to open file error is happening in this line

Error : E/FileSource: Failed to open file 'android.resource:/com.android.grafika/2130968577'. (No such file or directory)

String vidAddress = "android.resource://"+getPackageName()+"/"+R.raw.b_right;
Uri vidUri = Uri.parse(vidAddress);
File sliders = new File(vidUri.toString());
Log.d("File",sliders.toString());

How to fix it? and what is the root cause? I mean what is wrong with this code snippet?

Upvotes: 3

Views: 5393

Answers (3)

Rohit Rathore
Rohit Rathore

Reputation: 402

final String uriPath="android.resource://"+getPackageName()+"/raw/texivedio";

Upvotes: 0

dsh
dsh

Reputation: 12214

The problem is the file path you constructed does not exist.

To open raw resources, use openRawResource(). For example:

InputStream stream = getContext().getResources().openRawResource( R.raw.b_right );

Upvotes: 1

CommonsWare
CommonsWare

Reputation: 1007286

I mean what is wrong with this code snippet?

First, passing the string representation of a Uri to the File constructor will never work. A Uri has a scheme; the File constructor will not know what to do with the scheme.

Second, the android.resource scheme is for an Android resource. While resources are files on your developer machine, they are not files on the Android device and cannot be represented by a File object.

How to fix it?

Do not attempt to access a resource via a File object.

For example, you can call openRawResource() on a Resources object to get an InputStream on your raw resource, given the resource ID.

Upvotes: 2

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