Get the file list of a given directory

does anyone know how to get the file list of a given directory? I tried "list"/"filelist" but the class not found.
THanks!

What made you assume that there exist classes called list or filelist etc.
Do you read API documentation etc ?
Anyways, here is a hint java.io.File.listFiles()

Similar Messages

  • How can I get file list for a given directory?

    Hi,
    How can I get file list for a given directory? Is there a function?
    Thanks.

    Hi friend,
    Try this sample report. It displays all files in a directory.
    While executing give some directory name in input. ex:  C:\
    Mention file type in filter parameter. ex: *.DOC
    REPORT ztests.
    DATA : file_table LIKE TABLE OF sdokpath WITH HEADER LINE .
    DATA : dir_table LIKE TABLE OF sdokpath WITH HEADER LINE .
    PARAMETERS:p_dir(50) TYPE c.
    CALL FUNCTION 'TMP_GUI_DIRECTORY_LIST_FILES'
      EXPORTING
        directory  = p_dir
        filter     = '*.DOC'
      TABLES
        file_table = file_table
        dir_table  = dir_table
      EXCEPTIONS
        cntl_error = 1
        OTHERS     = 2.
    LOOP AT file_table .
      WRITE:/ file_table-pathname.
    ENDLOOP.
    Might be helpful...
    Thanks.....
    Edited by: Guest77 on Feb 11, 2009 5:30 AM

  • Functional module to get the File from a given Directory

    Hi all,
    I am using a FM name 'subst_get_file_list' to get the file from a given directory but it is accepting only 40 Character length file only my requirement is to accept file name other than 40 char,
    give me good sugestion
    regards
    paul

    Hi Paul,
    Check the Function Module Gayathri has given. ie. 'SO_SPLIT_FILE_AND_PATH'.
    In the exporting parameter FULL_NAME , give the path name and in the importing parameter stripped_name , you will get the filename.
    Check this code.
    REPORT ZSHAIL_SPLITFILE.
    data: it_tab type filetable with header line,
          gd_subrc type i.
    tables: rlgrap.
    data: path type string,
          file_name type string.
    parameters file_nam type rlgrap-filename .
    data: user_act type i.
    at selection-screen on value-request for file_nam.
    CALL METHOD cl_gui_frontend_services=>file_open_dialog
      EXPORTING
        WINDOW_TITLE            = 'select a file'
       DEFAULT_EXTENSION       = '*.txt
        DEFAULT_FILENAME        = ''
        FILE_FILTER             = '*.txt'
        INITIAL_DIRECTORY       = ''
        MULTISELECTION          = abap_false
       WITH_ENCODING           =
      CHANGING
        file_table              = it_tab[]
        rc                      = gd_subrc
        USER_ACTION             = user_act
       FILE_ENCODING           =
      EXCEPTIONS
        FILE_OPEN_DIALOG_FAILED = 1
        CNTL_ERROR              = 2
        ERROR_NO_GUI            = 3
        NOT_SUPPORTED_BY_GUI    = 4
        others                  = 5
    IF sy-subrc <> 0.
    MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
               WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
    ENDIF.
    if user_act = '0'.
    loop at it_tab.
    file_nam = it_tab-filename.
    endloop.
    endif.
    path = file_nam.
    CALL FUNCTION 'SO_SPLIT_FILE_AND_PATH'
      EXPORTING
        full_name           = path
    IMPORTING
       STRIPPED_NAME       = file_name
      FILE_PATH           =
    EXCEPTIONS
      X_ERROR             = 1
      OTHERS              = 2
    IF sy-subrc <> 0.
    MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
            WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
    ENDIF.
    at selection-screen.
    message i001(zmess) with file_name.
    Regards,
    SP.

  • Get file list from URL/web directory

    Hey guys,
    I have created a script for After Effects that can download multiple images from URLs that are given in an array. Now ideally what I want, is to get all the filenames of the files in a certain browsable web directory to be put in an array.
    Unfortunately, I can't seem to figure out how to do this. Does someone have the knowledge here to help me out?
    Thanks in advance,
    Jorge

    This is more difficult to do than it seems. Since the files are on a remote web server, you need to find some way to get the server to divulge its local file structure, which is not something most servers will do. The absolute best way is to have control over the web server and run some server-side code (PHP, node.js, whatever) to generate the file list on the server side and then fetch it with your client/extension/script. Other than that it's almost impossible to get a 3rd party server to divulge its file structure as far as I know, mainly for security reasons. So the #1 question is: do you have control over the server/website that you're trying to fetch from or not?

  • Get the File information(data) from Directory AL11.

    Hi Friends,
    I want to get the file data from the given particular directory(which maintained in AL11) in the selection screen.
    for listing the files from the directory, i used FM 'RZL_READ_DIR_LOCAL'. it displaying only files what ever in the given directory.
    But my requirement is to display the complete data in the file ( log file contents).
    please suggest me with relevant Function module or logic.
    Thank you.
    Regards
    Ramesh M

    HI,
    Try using function module:
    PARAMETERS:     p_fname    LIKE rlgrap-filename               .
    data l_path       TYPE dxlpath       .
    DATA: l_true       TYPE btch0000-char1.
    *-- F4 functionality for filename on Application Server
      CALL FUNCTION '/SAPDMC/LSM_F4_SERVER_FILE'
        EXPORTING
          directory        = '/usr/sap/input'
          filemask         = ''
        IMPORTING
          serverfile       = l_path
        EXCEPTIONS
          canceled_by_user = 1
          OTHERS           = 2.
      IF sy-subrc <> 0.
        MESSAGE ID sy-msgid TYPE sy-msgty NUMBER sy-msgno
                WITH sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4.
      ELSE.
        p_fname = l_path.  ""Here p_fname is the parameter for the user to select the file from the application server
      ENDIF.
    PFL_CHECK_OS_FILE_EXISTENCE
      DATA: l_file       TYPE tpfht-pffile.
      CLEAR l_file.
      l_file = p_fname.    "Here p_fname is the parameter for the user to select the file from the application server
      CALL FUNCTION 'PFL_CHECK_OS_FILE_EXISTENCE'
        EXPORTING
          fully_qualified_filename = l_file
        IMPORTING
          file_exists              = l_true.
      IF l_true = space.
        MESSAGE e001(zmsg).
      ENDIF.
    Hope it helps
    Regards
    Mansi

  • How to get the file size (in bytes) for all files in a directory?

    How to get the file size (in bytes) for all files in a directory?
    The following code does not work. isFile() does NOT recognize files as files but only as directories. Why?
    Furthermore the size is not retrieved correctly.
    How do I have to code it otherwise? Is there a way of not converting f-to-string-to-File again but iterate over all file objects instead?
    Thank you
    Peter
    java.io.File f = new java.io.File("D:/todo/");
    files = f.list();
    for (int i = 0; i < files.length; i++) {
    System.out.println("fn=" + files);
    if (new File(files[i]).isFile())
         System.out.println("file[" + i + "]=" + files[i] + " size=" + (new File(files[i])).length() ); }

    pstein wrote:
    ...The following code does not work. Work?! It does not even compile! Please consider posting code in the form of an SSCCE in future.
    Here is an SSCCE.
    import java.io.File;
    class ListFiles {
        public static void main(String[] args) {
            java.io.File f = new java.io.File("/media/disk");
            // provides only the file names, not the path/name!
            //String[] files = f.list();
            File[] files = f.listFiles();
            for (int i = 0; i < files.length; i++) {
                System.out.println("fn=" + files);
    if (files[i].isFile()) {
    System.out.println(
    "file[" +
    i +
    "]=" +
    files[i] +
    " size=" +
    (files[i]).length() );
    }Edit 1:
    Also, in future, when posting code, code snippets, HTML/XML or input/output, please use the code tags to retain the indentation and formatting.   To do that, select the code and click the CODE button seen on the Plain Text tab of the message posting form.  It took me longer to clean up that code and turn it into an SSCCE, than it took to +solve the problem.+
    Edited by: AndrewThompson64 on Jul 21, 2009 8:47 AM                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

  • How to get the path when i select a directory or a file in a JTree

    How to get the path when i select a directory or a file in a JTree

    import java.lang.*;
    import java.io.*;
    import javax.swing.*;
    import javax.swing.tree.*;
    import java.awt.HeadlessException;
    import javax.swing.event.*;
    import java.awt.*;
    import java.awt.event.*;
    import java.util.Iterator;
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    * @version 1.0
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    JTree tree;
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    * @author Frederic FOURGEOT
    * @version 1.0
    private class FSNode extends DefaultMutableTreeNode {
    File file; // contient le fichier li� au noeud
    * Constructeur non visible
    private FSNode() {
    super();
    * Constructeur par initialisation
    * @param userObject Object
    FSNode(Object userObject) {
    super(userObject);
    * Constructeur par initialisation
    * @param userObject Object
    * @param newFile File
    FSNode(Object userObject, File newFile) {
    super(userObject);
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    * Definit le fichier lie au noeud
    * @param newFile File
    public void setFile(File newFile) {
    file = newFile;
    * Renvoi le fichier lie au noeud
    * @return File
    public File getFile() {
    return file;
    public JTree getJTree(){
         return tree ;
    * Constructeur
    * @throws HeadlessException
    public JTreeFolder() throws HeadlessException {
    File[] drive;
    tree = new JTree();
    // cr�ation du noeud sup�rieur
    racine = new DefaultMutableTreeNode("Poste de travail");
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    drive = File.listRoots();
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    //RIEN
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    listFile[posa] = listFile[posb];
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    7/3                                                                            
    PL/SQL: Statement ignored                                                      
    8/54                                                                           
    PL/SQL: ORA-00942: table or view does not exist                                
    8/20                                                                           
    PL/SQL: SQL Statement ignored                                                  
    10/11                                                                          
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  • File Adapter - how to get the file count from a folder

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              String filename = request.getParameter("saveas");
              String path = request.getParameter("path");
              PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("test.java")));
              ServletInputStream in = request.getInputStream();
              int i = in.read();
              System.out.println("filename:"+filename);
              System.out.println("path:"+path);
              while (i != -1)
                   pw.print((char) i);
                   i = in.read();
              pw.close();
    }

    Thanks it works great.
    Here an excample from my code
    import org.apache.commons.fileupload.*;
    public class FileUploadBean extends Object implements java.io.Serializable{
    String foutmelding = "geen";
    String path;
    String filename;
    public boolean doUpload(HttpServletRequest request) throws IOException
         try
         // Create a new file upload handler
         FileUpload upload = new FileUpload();
         // Set upload parameters
         upload.setSizeMax(100000);
         upload.setSizeThreshold(100000000);
         upload.setRepositoryPath("/");
         // Parse the request
         List items = upload.parseRequest(request);
         // Process the uploaded fields
         Iterator iter = items.iterator();
         while (iter.hasNext())
         FileItem item = (FileItem) iter.next();
              if (item.isFormField())
                   String stringitem = item.getString();
         else
              String filename = "";
                   int temp = item.getName().lastIndexOf("\\");
                   filename = item.getName().substring(temp,item.getName().length());
                   File bestand = new File(path+filename);
                   if(item.getSize() > SizeMax && SizeMax != -1){foutmelding = "bestand is te groot.";return false;}
                   if(bestand.exists()){foutmelding ="bestand bestaat al";return false;}
                   FileOutputStream fOut = new FileOutputStream(bestand);     
                   BufferedOutputStream bOut = new BufferedOutputStream(fOut);
                   int bytesRead =0;
                   byte[] data = item.get();
                   bOut.write(data, 0 , data.length);     
                   bOut.close();
         catch(Exception e)
              System.out.println("er is een foutontstaan bij het opslaan de een bestand "+e);
              foutmelding = "Bestand opsturen is fout gegaan";
         return true;
         }

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