How to read file from res/raw by name

问题

I want to open a file from the folder res/raw/. I am absolutely sure that the file exists. To open the file I have tried

File ddd = new File(\"res/raw/example.png\");

The command

ddd.exists();

yields FALSE. So this method does not work.

Trying

MyContext.getAssets().open(\"example.png\");

ends up in an exception with getMessage() \"null\".

Simply using

R.raw.example

is not possible because the filename is only known during runtime as a string.

Why is it so difficult to access a file in the folder /res/raw/ ?

回答1:

With the help of the given links I was able to solve the problem myself. The correct way is to get the resource ID with

getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
                             "raw", getPackageName());

To get it as a InputStream

InputStream ins = getResources().openRawResource(
            getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
            "raw", getPackageName()));

回答2:

Here is example of taking XML file from raw folder:

 InputStream XmlFileInputStream = getResources().openRawResource(R.raw.taskslists5items); // getting XML

Then you can:

 String sxml = readTextFile(XmlFileInputStream);

when:

 public String readTextFile(InputStream inputStream) {
        ByteArrayOutputStream outputStream = new ByteArrayOutputStream();

        byte buf[] = new byte[1024];
        int len;
        try {
            while ((len = inputStream.read(buf)) != -1) {
                outputStream.write(buf, 0, len);
            }
            outputStream.close();
            inputStream.close();
        } catch (IOException e) {

        }
        return outputStream.toString();
    }

回答3:

You can read files in raw/res using getResources().openRawResource(R.raw.myfilename).

BUT there is an IDE limitation that the file name you use can only contain lower case alphanumeric characters and dot. So file names like XYZ.txt or my_data.bin will not be listed in R.

回答4:

Here are two approaches you can read raw resources using Kotlin.

You can get it by getting the resource id. Or, you can use string identifier in which you can programmatically change the filename with incrementation.

Cheers mate 🎉

// R.raw.data_post

this.context.resources.openRawResource(R.raw.data_post)
this.context.resources.getIdentifier("data_post", "raw", this.context.packageName)

来源：https://stackoverflow.com/questions/15912825/how-to-read-file-from-res-raw-by-name

标签

android

file-io

resources