Scanning multiple files and printing them alphabetically

Question

I'm reading multiple text files using Java and printing their directories and I wonder why the output is not alphabetically arranged?

Code Snippet (got it from the internet also)

File dir = new File("/home/dilapitan/Desktop/xml-parsing/files/");
File[] listOfFiles = dir.listFiles();
for (File path : listOfFiles) {
    System.out.println(path);
}

Output:

dilapitan@NT071855:~/Desktop/xml-parsing$ java Multiple 
/home/dilapitan/Desktop/xml-parsing/files/c.txt
/home/dilapitan/Desktop/xml-parsing/files/b.txt
/home/dilapitan/Desktop/xml-parsing/files/a.txt
dilapitan@NT071855:~/Desktop/xml-parsing$

Can I do it with the output being:

a.txt
b.txt
c.txt

Thank you in advance!

Possible duplicate of [How to sort file names in ascending order?](https://stackoverflow.com/questions/16898029/how-to-sort-file-names-in-ascending-order) — Ori Marko, Jul 04 '17 at 08:16
see https://stackoverflow.com/questions/16898029/how-to-sort-file-names-in-ascending-order — Ori Marko, Jul 04 '17 at 08:16
"why the output is not alphabetically arranged?" see https://docs.oracle.com/javase/8/docs/api/java/io/File.html#listFiles-- -> "***There is no guarantee that the name strings in the resulting array will appear in any specific order***; they are not, in particular, guaranteed to appear in alphabetical order.". You can use `File#list()` to get files as String[]. Since String is Comparable it provides default order so you can sort this array with `Arrays.sort`. — Pshemo, Jul 04 '17 at 08:17

score 0 · Answer 1 · answered Jul 04 '17 at 08:21

To quote File#listDir's documentation:

There is no guarantee that the name strings in the resulting array will appear in any specific order; they are not, in particular, guaranteed to appear in alphabetical order.

If you want to print them in a specific order, you'll have to do so yourself. Java 8 streams git you an elegant way of combining the extraction of the file's name from the patch and sorting them in a single statement:

Arrays.stream(listOfFiles)
      .map(File::getName)
      .sorted()
      .forEach(System.out::println);

ΦXocę 웃 Пepeúpa ツ · Answer 2 · 2017-07-04T10:30:11.593

Just another way to do it, but using the power of java8

List<Path> x = Files.list(Paths.get("C:\\myPath\\Tools"))
            .filter(p -> Files.exists(p))
            .map(s -> s.getFileName())
            .sorted()
            .collect(Collectors.toList());

x.forEach(System.out::println);

or even better

List<Path> x = Files.list(Paths.get("C:\\Users\\myPath\\Tools"))
                .filter(Files::exists)
                .map(Path::getFileName)
                .sorted()
                .collect(Collectors.toList());

x.forEach(System.out::println);

score 0 · Answer 3 · answered Jul 05 '17 at 08:13

Thank you to those who answered. I thought of a very simple solution:

File dir = new File("/home/dilapitan/Desktop/xml-parsing/files/");
File[] listOfFiles = dir.listFiles();
Arrays.sort(listOfFiles);

for (File path : listOfFiles) {
    System.out.println(path);
}

hahahaha my apologies everyone.

Scanning multiple files and printing them alphabetically

3 Answers3