how to File.listFiles in alphabetical order?

Question

I've got code as below:

class ListPageXMLFiles implements FileFilter {

        @Override
        public boolean accept(File pathname) {
                DebugLog.i("ListPageXMLFiles", "pathname is " + pathname);

                String regex = ".*page_\\d{2}\\.xml";
                if(pathname.getAbsolutePath().matches(regex)) {
                        return true;
                }
                return false;
        }
}

public void loadPageTrees(String xml_dir_path) {
        ListPageXMLFiles filter_xml_files = new ListPageXMLFiles();
        File XMLDirectory = new File(xml_dir_path);
        for(File _xml_file : XMLDirectory.listFiles(filter_xml_files)) {
                loadPageTree(_xml_file);
        }
}

The FileFilter is working nicely, but listFiles() seems to be listing the files in reverse alphabetical order. Is there some quick way of telling listFile() to list the files in alphabetical order?

Answer 1

The listFiles method, with or without a filter does not guarantee any order.

It does, however, return an array, which you can sort with Arrays.sort().

File[] files = XMLDirectory.listFiles(filter_xml_files);
Arrays.sort(files);
for(File _xml_file : files) {
    ...
}

This works because File is a comparable class, which by default sorts pathnames lexicographically. If you want to sort them differently, you can define your own comparator.

If you prefer using Streams:

A more modern approach is the following. To print the names of all files in a given directory, in alphabetical order, do:

Files.list(Paths.get(dirName)).sorted().forEach(System.out::println)

Replace the System.out::println with whatever you want to do with the file names. If you want only filenames that end with "xml" just do:

Files.list(Paths.get(dirName))
    .filter(s -> s.toString().endsWith(".xml"))
    .sorted()
    .forEach(System.out::println)

Again, replace the printing with whichever processing operation you would like.

Answer 2

I think the previous answer is the best way to do it here is another simple way. just to print the sorted results.

 String path="/tmp";
 String[] dirListing = null;
 File dir = new File(path);
 dirListing = dir.list();
 Arrays.sort(dirListing);
 System.out.println(Arrays.deepToString(dirListing));

Answer 3

In Java 8:

Arrays.sort(files, (a, b) -> a.getName().compareTo(b.getName()));

Reverse order:

Arrays.sort(files, (a, b) -> -a.getName().compareTo(b.getName()));

Answer 4

This is my code:

        try {
            String folderPath = "../" + filePath.trim() + "/";
            logger.info("Path: " + folderPath);
            File folder = new File(folderPath);
            File[] listOfFiles = folder.listFiles();
            int length = listOfFiles.length;
            logger.info("So luong files: " + length);
            ArrayList<CdrFileBO> lstFile = new ArrayList< CdrFileBO>();

            if (listOfFiles != null && length > 0) {
                int count = 0;
                for (int i = 0; i < length; i++) {
                    if (listOfFiles[i].isFile()) {
                        lstFile.add(new CdrFileBO(listOfFiles[i]));
                    }
                }
                Collections.sort(lstFile);
                for (CdrFileBO bo : lstFile) {
                    //String newName = START_NAME + "_" + getSeq(SEQ_START) + "_" + DateSTR + ".s";
                    String newName = START_NAME + DateSTR + getSeq(SEQ_START) + ".DAT";
                    SEQ_START = SEQ_START + 1;
                    bo.getFile().renameTo(new File(folderPath + newName));
                    logger.info("newName: " + newName);
                    logger.info("Next file: " + getSeq(SEQ_START));
                }

            }
        } catch (Exception ex) {
            logger.error(ex);
            ex.printStackTrace();
        }