如何让Maven从我的资源文件夹中读取文件

Question

I have been happily using JUnit to run my tests and everything has been fine.

However, I now need to use Maven but for some reason it cannot find any of my resource files.

The files are in the expected place: src/main/resources

I am using the following code to try to read a file:

public Map<String, String> readCsv(String filename) {

    Map<String, String> headersAsMap;
    CSVDataManipulator csvDataManipulator = new CSVDataManipulator();
    ClassLoader classLoader = getClass().getClassLoader();
    String wrkbook = new File(classLoader.getResource(filename).getFile()).toString();
    headersAsMap = csvDataManipulator.getAllRecordsAsMap(wrkbook);
    return headersAsMap;
}

However, try as I might it cannot find the file. I've tried lots of different code and tried moving the files to different locations but I cannot get Maven to find my resource files.

Any help would be greatly appreciated! Thanks

Answer 1

To my understanding classLoader.getResource(..) expects the file to be in a folder structure matching the package of the class. So if the package of your class is com.matt.stuff, then you'll have to put the csv file in src/main/resources/com/matt/stuff.

Or you could just use this to grab your csv file:

private static String readFile(String fileName) throws IOException {
    //filename can be src/main/resources/my-csv.csv
    return new String(Files.readAllBytes(Paths.get(fileName))); 
}

Answer 2

Here's my file structure of a Maven project built from the quick-start-archetype, with comons-io added as a dependency:

src
src/main
src/main/java
src/main/java/com
src/main/java/com/essexboy
src/main/java/com/essexboy/App.java
src/main/resources
src/main/resources/dir1
src/main/resources/dir1/test.txt
src/main/resources/dir2
src/main/resources/dir2/test.txt
src/main/resources/test.txt
src/test
src/test/java
src/test/java/com
src/test/java/com/essexboy
src/test/java/com/essexboy/AppTest.java

Here's my test

public class AppTest {
    @Test
    public void shouldAnswerWithTrue() throws Exception {

        StringWriter writer = new StringWriter();
        IOUtils.copy(getClass().getResourceAsStream("/test.txt"), writer, Charset.defaultCharset());
        assertEquals("from root", writer.toString().trim());

        writer = new StringWriter();
        IOUtils.copy(getClass().getResourceAsStream("/dir1/test.txt"), writer, Charset.defaultCharset());
        assertEquals("from dir1", writer.toString().trim());

        writer = new StringWriter();
        IOUtils.copy(getClass().getResourceAsStream("/dir2/test.txt"), writer, Charset.defaultCharset());
        assertEquals("from dir2", writer.toString().trim());
    }
}

Answer 3

A File is indeed a file on the file system. For a resource which might be a file zipped in a jar, and has a path on the class path, you need something else.

Traditionally one would use a more general InputStream instead of a File.

InputStream in = getClass().getResourceAsStream("/.../x.csv"); // Path on the class path

With the new class Path , more general than File , you can deal with several (virtual) file systems:

URL url = getClass().getResource("/.../x.csv"); // Path on the class path
Path path = Paths.get(url.toURI());

Files.copy(path, Paths.get("..."));

With a bit of luck your CSVManipulator should besides being parametrized with a File, also with an InputStream or Reader ( new InputStreamReader(in, "UTF-8") )