Counting the number of files in a directory using Java

Question

How do I count the number of files in a directory using Java   For simplicity  lets assume that the directory doesn t have any sub-directories   I know the standard method of    new File  lt directory path gt   listFiles   length   But this will effectively go through all the files in the directory  which might take long if the number of files is large  Also  I don t care about the actual files in the directory unless their number is greater than some fixed large number  say 5000     I am guessing  but doesn t the directory  or its i-node in case of Unix  store the number of files contained in it  If I could get that number straight away from the file system  it would be much faster  I need to do this check for every HTTP request on a Tomcat server before the back-end starts doing the real processing  Therefore  speed is of paramount importance   I could run a daemon every once in a while to clear the directory  I know that  so please don t give me that solution

User · Answer

Ah    the rationale for not having a straightforward method in Java to do that is file storage abstraction  some filesystems may not have the number of files in a directory readily available    that count may not even have any meaning at all  see for example distributed  P2P filesystems  fs that store file lists as a linked list  or database-backed filesystems      So yes   new File  lt directory path gt   list   length   is probably your best bet

User · Answer

This might not be appropriate for your application  but you could always try a native call  using jni or jna   or exec a platform-specific command and read the output before falling back to list   length   On  nix  you could exec ls -1a   wc -l  note - that s dash-one-a for the first command  and dash-lowercase-L for the second    Not sure what would be right on windows - perhaps just a dir and look for the summary   Before bothering with something like this I d strongly recommend you create a directory with a very large number of files and just see if list   length really does take too long   As this blogger suggests  you may not want to sweat this   I d probably go with Varkhan s answer myself

User · Answer

Unfortunately  as mmyers said  File list   is about as fast as you are going to get using Java  If speed is as important as you say  you may want to consider doing this particular operation using JNI  You can then tailor your code to your particular situation and filesystem

User · Answer

Since Java 8  you can do that in three lines   try  Stream lt Path gt  files   Files list Paths get  your path here           long count   files count        Regarding the 5000 child nodes and inode aspects   This method will iterate over the entries but as Varkhan suggested you probably can t do better besides playing with JNI or direct system commands calls  but even then  you can never be sure these methods don t do the same thing   However  let s dig into this a little   Looking at JDK8 source  Files list exposes a stream that uses an Iterable from Files newDirectoryStream that delegates to FileSystemProvider newDirectoryStream   On UNIX systems  decompiled sun nio fs UnixFileSystemProvider class   it loads an iterator  A sun nio fs UnixSecureDirectoryStream is used  with file locks while iterating through the directory    So  there is an iterator that will loop through the entries here   Now  let s look to the counting mechanism   The actual count is performed by the count sum reducing API exposed by Java 8 streams  In theory  this API can perform parallel operations without much effort  with multihtreading   However the stream is created with parallelism disabled so it s a no go     The good side of this approach is that it won t load the array in memory as the entries will be counted by an iterator as they are read by the underlying  Filesystem  API   Finally  for the information  conceptually in a filesystem  a directory node is not required to hold the number of the files that it contains  it can just contain the list of it s child nodes  list of inodes   I m not an expert on filesystems  but I believe that UNIX filesystems work just like that  So you can t assume there is a way to have this information directly  i e  there can always be some list of child nodes hidden somewhere

User · Answer

Using sigar should help  Sigar has native hooks to get the stats  new Sigar   getDirStat dir  getTotal

User · Answer

Since you don t really need the total number  and in fact want to perform an action after a certain number  in your case 5000   you can use java nio file Files newDirectoryStream  The benefit is that you can exit early instead having to go through the entire directory just to get a count   public boolean isOverMax        Path dir   Paths get  C  foo bar        int i   1       try  DirectoryStream lt Path gt  stream   Files newDirectoryStream dir             for  Path p   stream                  larger than max files  exit             if    i  gt  MAX FILES                    return true                                catch  IOException ex            ex printStackTrace               return false      The interface doc for DirectoryStream also has some good examples

User · Answer

This method works for me very well         Recursive method to recover files and folders and to print the information public static void listFiles String directoryName         File file   new File directoryName       File   fileList   file listFiles       List files inside the main dir     int j      String extension      String fileName       if  fileList    null            for  int i   0  i  lt  fileList length  i                  extension    quot  quot               if  fileList i  isFile                      fileName   fileList i  getName                     if  fileName lastIndexOf  quot   quot      -1  amp  amp  fileName lastIndexOf  quot   quot      0                        extension   fileName substring fileName lastIndexOf  quot   quot     1                       System out println  quot THE  quot    fileName    quot   has the extension      quot    extension                     else                       extension    quot Unknown quot                       System out println  quot extension2       quot    extension                                      filesCount                    allStats add new FilePropBean filesCount  fileList i  getName    fileList i  length    extension                          fileList i  getParent                    else if  fileList i  isDirectory                      filesCount                    extension    quot  quot                   allStats add new FilePropBean filesCount  fileList i  getName    fileList i  length    extension                          fileList i  getParent                      listFiles String valueOf fileList i

User · Answer

If you have directories containing really   100 000  many files  here is a  non-portable  way to go    String directoryPath    a path       -f flag is important  because this way ls does not sort it output     which is way faster String   params       bin sh    -c        ls -f     directoryPath       wc -l     Process process   Runtime getRuntime   exec params   BufferedReader reader   new BufferedReader new InputStreamReader      process getInputStream      String fileCount   reader readLine   trim   - 2     accounting for    and   reader close    System out println fileCount

User · Answer

Unfortunately  I believe that is already the best way  although list   is slightly better than listFiles    since it doesn t construct File objects

User · Answer

In spring batch I did below  private int getFilesCount   throws IOException           ResourcePatternResolver resolver   new PathMatchingResourcePatternResolver            Resource   resources   resolver getResources  file     projectFilesFolder        input splitFolder   csv            return resources length

User · Answer

public void shouldGetTotalFilesCount         Integer reduce   of listRoots    parallel   map this  getFilesCount  reduce 0    a  b  - gt  a   b       private int getFilesCount File directory        File   files   directory listFiles        return Objects isNull files    1   Stream of files               parallel                reduce 0   Integer acc  File p  - gt  acc   getFilesCount p    a  b  - gt  a   b

[java] Counting the number of files in a directory using Java

Examples related to java

Examples related to performance

Examples related to file

Examples related to directory