Read file from classpath with Java 7 NIO

Question

I've googled around for quite a while for this, but all the results point to pre-Java 7 NIO solutions. I've used the NIO stuff to read in files from the a specific place on the file system, and it was so much easier than before (Files.readAllBytes(path)). Now, I'm wanting to read in a file that is packaged in my WAR and on the classpath. We currently do that with code similar to the following:

Input inputStream = this.getClass().getClassLoader().getResourceAsStream(fileName);
ByteArrayOutputStream byteStream = new ByteArrayOutputStream();

/* iterate through the input stream to get all the bytes (no way to reliably find the size of the 
 *     file behind the inputStream (see http://docs.oracle.com/javase/6/docs/api/java/io/InputStream.html#available()))
 */
int byteInt = -1;
try
{
    byteInt = inputStream.read();
    while (byteInt != -1)
    {
        byteStream.write(byteInt);
        byteInt = inputStream.read();
    }

    byteArray = byteStream.toByteArray();
    inputStream.close();
    return byteArray;
}
catch (IOException e)
{
    //...
}

While this works, I was hoping there was an easier/better way to do this with the NIO stuff in Java 7. I'm guessing I'll need to get a Path object that represents this path on the classpath, but I'm not sure how to do that.

I apologize if this is some super easy thing to do. I just cannot figure it out. Thanks for the help.

Have you tried YourClass.class.getResourceAsStream(filename);? — Dan W
– Dan W, Commented Feb 4, 2013 at 18:10

shreyas · Accepted Answer · 2015-03-04 16:51:57Z

30

This works for me.

import java.nio.file.Files;
import java.nio.file.Paths;

// fileName: foo.txt which lives under src/main/resources
public String readFileFromClasspath(final String fileName) throws IOException, URISyntaxException {
    return new String(Files.readAllBytes(
                Paths.get(getClass().getClassLoader()
                        .getResource(fileName)
                        .toURI())));
}

answered Mar 4, 2015 at 16:51

shreyas

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2 Comments

macemers Over a year ago

It doesn't work if you execute this peice of code with command line if foo.txt is under src/main/resources, because calling toURI() from jar:file will give you NullPointerException

Praytic Over a year ago

You can use utility method which reads directly into String, instead of byte array: Files.readString(Paths.get(getClass().getClassLoader().getResource(fileName).toURI()));

JB Nizet · Accepted Answer · 2013-02-04 18:24:52Z

8

A Path represents a file on the file system. It doesn't help to read a resource from the classpath. What you're looking after is a helper method that reads everything fro a stream (more efficiently than how you're doing) and writes it to a byte array. Apache commons-io or Guava can help you with that. For example with Guava:

byte[] array = 
    ByteStreams.toByteArray(this.getClass().getClassLoader().getResourceAsStream(resourceName));

If you don't want to add Guava or commons-io to your dependencies just for that, you can always read their source code and duplicate it to your own helper method.

answered Feb 4, 2013 at 18:24

JB Nizet

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2 Comments

Paul Draper Over a year ago

That is not asynchronous which is one of the primary purposes of NIO.

Tilman Hausherr Over a year ago

stream is not closed

Joachim Isaksson · Accepted Answer · 2013-02-04 18:24:43Z

3

As far as I understand, what you want is to open a ReadableByteChannel to your resource, so you can use NIO for reading it.

This should be a good start,

// Opens a resource from the current class' defining class loader
InputStream istream = getClass().getResourceAsStream("/filename.txt");

// Create a NIO ReadableByteChannel from the stream
ReadableByteChannel channel = java.nio.channels.Channels.newChannel(istream);

answered Feb 4, 2013 at 18:24

Joachim Isaksson

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Comments

Will Hartung · Accepted Answer · 2013-02-04 18:22:15Z

1

You should look at ClassLoader.getResource(). This returns a URL which represents the resource. If it's local to the file system, it will be a file:// URL. At that point you can strip off the scheme etc., and then you have the file name with which you can do whatever you want.

However, if it's not a file:// path, then you can fall back to the normal InputStream.

answered Feb 4, 2013 at 18:22

Will Hartung

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