File names within a directory

Hello,
I am trying to get a list of all the files in a specific directory for a statistics program.
It would be good if I could get the list back as a string[] or something similar.
I'm using windows and the files are contained in \logs\
This is to allow me to scan all of the files without any user input.
ps. sorry for my lack of english skills

Look at the java.io.File class and at the method list().

Similar Messages

  • Error while reading file name in a directory

    Hi,
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    Run-time error '5':
    Invalid procedure call or argument
    The actual path is "Q:\Budget\Historical Budgets\FY15\*.xls*"
    and ThisWorkbook.Sheets(1).Range("A1").Value = FY15 in my excel sheet.
    "Below is the code I am using"
    Dim file As Variant
    file = Dir("Q:\Budget\Historical Budgets\" & ThisWorkbook.Sheets(1).Range("A1").Value & "\*.xls*")
    If file = "" Then
            MsgBox "no files"
            Exit Sub
          Else
            ' ... else, count the files
            x = 0
            Do While file <> ""
                x = x + 1
                file = Dir         
    <----  I get the error at this line.
            Loop
    End If
    Could you please help me to solve this problem
    Regards, Hitesh

    Do you want to generate a list of all files in a folder, in your spreadsheet?  If so, please try this sample code?
    Option Explicit
    Private cnt As Long
    Private arfiles
    Private level As Long
    Sub Folders()
    Dim i As Long
    Dim sFolder As String
    Dim iStart As Long
    Dim iEnd As Long
    Dim fOutline As Boolean
    arfiles = Array()
    cnt = -1
    level = 1
    sFolder = "C:\Users\Excel\Desktop\Coding\Microsoft Excel\Work Samples\"
    ReDim arfiles(2, 0)
    If sFolder <> "" Then
    SelectFiles sFolder
    Application.DisplayAlerts = False
    On Error Resume Next
    Worksheets("Files").Delete
    On Error GoTo 0
    Application.DisplayAlerts = True
    Worksheets.Add.Name = "Files"
    With ActiveSheet
    For i = LBound(arfiles, 2) To UBound(arfiles, 2)
    If arfiles(0, i) = "" Then
    If fOutline Then
    Rows(iStart + 1 & ":" & iEnd).Rows.Group
    End If
    With .Cells(i + 1, arfiles(2, i))
    .Value = arfiles(1, i)
    .Font.Bold = True
    End With
    iStart = i + 1
    iEnd = iStart
    fOutline = False
    Else
    .Hyperlinks.Add Anchor:=.Cells(i + 1, arfiles(2, i)), _
    Address:=arfiles(0, i), _
    TextToDisplay:=arfiles(1, i)
    iEnd = iEnd + 1
    fOutline = True
    End If
    Next
    .Columns("A:Z").ColumnWidth = 5
    End With
    End If
    'just in case there is another set to group
    If fOutline Then
    Rows(iStart + 1 & ":" & iEnd).Rows.Group
    End If
    Columns("A:Z").ColumnWidth = 5
    ActiveSheet.Outline.ShowLevels RowLevels:=1
    ActiveWindow.DisplayGridlines = False
    End Sub
    Sub SelectFiles(Optional sPath As String)
    Static FSO As Object
    Dim oSubFolder As Object
    Dim oFolder As Object
    Dim oFile As Object
    Dim oFiles As Object
    Dim arPath
    If FSO Is Nothing Then
    Set FSO = CreateObject("Scripting.FileSystemObject")
    End If
    If sPath = "" Then
    sPath = CurDir
    End If
    arPath = Split(sPath, "\")
    cnt = cnt + 1
    ReDim Preserve arfiles(2, cnt)
    arfiles(0, cnt) = ""
    arfiles(1, cnt) = arPath(level - 1)
    arfiles(2, cnt) = level
    Set oFolder = FSO.GetFolder(sPath)
    Set oFiles = oFolder.Files
    For Each oFile In oFiles
    cnt = cnt + 1
    ReDim Preserve arfiles(2, cnt)
    arfiles(0, cnt) = oFolder.Path & "\" & oFile.Name
    arfiles(1, cnt) = oFile.Name
    arfiles(2, cnt) = level + 1
    Next oFile
    level = level + 1
    For Each oSubFolder In oFolder.Subfolders
    SelectFiles oSubFolder.Path
    Next
    level = level - 1
    End Sub
    Knowledge is the only thing that I can give you, and still retain, and we are both better off for it.

  • Allowable Characters in the file names within Oracle iFS

    What characters are allowable in file names within Oracle iFS?
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    Be carefull with file and directory names in iFS. You might be able to create folders and add document to it through the web interface that the windows explorer interface might not be able to interpret. Like a directory named '.'.
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  • Can we possible to retrive the file name from the directory...?

    Can we possible to retrive the list of files or file names from the directory...?
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    Yeah, yeah its very good example for this scenario.
    I agree. But, I want to learn about Java based PL/SQL code development for that just I am asking any link for this kind of material.....:-)

  • How to create Dynamic Selection List containg file names of a directory?

    Hi,
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    Basically, I need an Iterator over all file names in a directory, which could be used for Selection List in a JSF form.
    This iterator has nothing to do with a database, but should get the names directly from the file system (java.io.File).
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    I am working with JDeveloper 10.1.3.2, JSF and Business Services.
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    Create a class like this:
    package list;
    import java.io.File;
    public class fileList {
        public fileList() {
        public String[] listFiles(){
        File dir = new File("c:/temp");
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        for (int i = 0; i < files.length; i++)  {
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    Right click to create a data control from it.
    Then you can drag the return value of the listFiles method to a page and drop as a table for example.
    Also note that the Content Repository data control in 10.1.3.2 has the file system as a possible data source.

  • How to get list of file names from a directory?

    How to get list of file names from a directory?
    Please help

    In addition, this:
    String filename = files;Should be this:
    String filename = files;
    That's just because he didn't use the "code" tags, so [ i ] made everything following it become italicized.                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

  • Listing of all file names in a directory

    Hello everyone,
    Is there a way to get the listing of all file names in a directory
    pointed by an entry in dba_directory in oracle into a collection in
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    Tuncay

    this is the same problem I am trying to solve now. I found some ehlp in Tom Kyte articles on java stored procedures.
    create global temporary table DIR_LIST
    ( filename varchar2(255) )
    on commit delete rows;
    create and compile java source named "DirList" as
    import java.io.*;
    import java.sql.*;
    public class DirList
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    throws SQLException
    File path = new File( directory );
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    String element;
    for(int i = 0; i < list.length; i++)
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    #sql { INSERT INTO DIR_LIST (FILENAME)
    VALUES (:element) };
    create procedure get_dir_list( p_directory in varchar2 )
    as language java
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    ---then you fill the collection by fetching the DIR_LIST table
    good luck,
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  • Load all file names in a directory to a SQL Table

    I need to load all the file names in a directory to a SQL table. It must be done via TSQL or SSIS. If using TSQL cannot use xp_cmdshell or any undocumented sprocs. So I am guessing it will be SSIS. But still open to suggestions. I am an SSIS newbie and need
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    Let me add a bit of background. We are sent a group of files. Each file has the ccyymmdd date attached to the name. Possible that we miss processing one day and the next day there are two sets of files in the directory. file1_20140101, file2_20140101,
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  • Unable to  convert the file name in Target directory

    Dear Experts,
    Currently iam facing a problem with some files.my interface is a File to file .As per the interface the sender will pick the flatfiles from source directory which the file name was BNP_NE201.txt once my file reached to target directory the name of the file willrenamed as VENDOR_Payment.NE201 automatically  for this they have done some OS level script in receiver communication channel of the first interface which we have added "un operating system command after message processing" in processing parameters tab.again one more sender comm channel wil pick the file from first target directory which is having naming  convesion of "VENDOR_Payment.NE201 " and post it to target. all the two interfaces are file to file (NFS) . for a parctuclar day files has been posted to 1 st target directory and the receiver communication channel is not yet convert the file names bcz of that the second communication channel can't able to pick the files from target directory. for some times that is working fine and some times it is unable to convert.
    from our side i check every thing .could you please suggest me the way to approch
    Reagrs,
    Kiran tanuku

    >> for a parctuclar day files has been posted to 1 st target directory and the receiver communication channel is not yet convert the file names bcz of that the second communication channel can't able to pick the files from target directory. for some times that is working fine and some times it is unable to convert.
    You need to wait until the file is converted. After when it is converted then the second interface sender communication channel will pick the files.
    What do u mean by unable to convert?
    Is it unable to do the content conversion? if so then check the data as you said that it sometimes it works fine.
    > The best approach is to change the name of the file in the receiver communciation channel instead of using a script.
    Thanks,

  • FILE SENDER ADAPTER -  TO GET A DINAMIC FILE NAME IN THE DIRECTORY

    Hi,
         Please help me.
         I have to read by a file sender adapter a specific file in the directory that its name is Dyyyymmdd, where yyyymmdd is variable according with the current day:
         D      - fixed letter
         yymmdd - year,month and day of the current day.
         I cant´t use D* in the File Name Scheme because I must read only the one file generated in the day.
         How can I configure my file sender adapter to read this especific file in the directory?
          Thanks in advance.
          Mider.

    Hi Krishna,
        I must access only one file by time ( of the current day ), but in the directory I have a lot of them of different dates. I mean, in the of the read ( in the adapter configuration ) , I need to configure the name of the file Dyyyymmdd before of the reading in the directory, in other words, I need to construct the name of file (withe the current date) before the communication channel access the file to read it.
        I think that OS command just can be applicated in file receivers, but if possible, at your purpose, how can I access the name of this especific file via OS command for senders?
        Thanks,
             Mider.
    Message was edited by: Midervilson  Andrade
    Message was edited by: Midervilson  Andrade

  • File Name and Target Directory name issue in FTP CC

    Hi All ,
    I am using following code to assing file name and directory name in UDF. I don;t want these parameters to be part of my target structure. Since i have created these files names in UDF, what are the values do i need to mention in FTP CC's "File access Parameter" which are mandatory fiels in CC ?
    DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
    DynamicConfigurationKey key = DynamicConfigurationKey.create(u201Chttp://sap.com/xi/XI/System/Fileu201D,u201CFileNameu201D);
    String myFileName =biz +".dat";
    conf.put(key, myFileName);
    //Similar code for direcotry
    Thanks for your support.
    MK

    Hi,
    in ID in receiver file comm channel, just tick the option of Adapter Specific messge attributes and in it tick FileName and DirectoryName.............
    the filename and directory you give in comm channel will be treated as dummy and the values will be taken from your UDF.......
    Regards,
    Rajeev Gupta

  • File name in different directory

    i have a problem when saving files:
    I copied all the files from one directory to another ---with the same file name and I tried to make chnages in the new directory, But I found that all the changes I made in the new directory also happened in the old directory. So Labview recogonize file names only, not file paths. I don't like this feature!!!!!

    This is an old problem with no easy solution as of now. For a better description of what is going on here, you can read this discussion where I answered a similar question a while ago.
    Ed
    Ed Dickens - Certified LabVIEW Architect - DISTek Integration, Inc. - NI Certified Alliance Partner
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  • Need to reference file name within the XML Output

    Not sure if this is possible, as I'm new to the livecycle/XML world. But I need to be able to reference a file within the XML output from the PDF.
    The Scenario is that a pdf form of a business card will be issued to 50 franchisees for them to type in their names and mobiles, click submit and the XML file to be sent to myself, and be imported in to InDesign populating their data ready for printing. Up to here I have it working perfectly.
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    Hi MadhavaGanji,
    I have post how to validate the file name, header row against definition table which stored the file name and column definition. 
    Take a look and see if this is helpful.
    http://sqlage.blogspot.com/2013/11/ssis-validate-file-header-against.html
    http://sqlage.blogspot.com/

  • How to get JSP file name within a servlet?

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    The Referer is an optional HTTP header, and it only exists when a page is called from an hyperlink (i.e. from another web page). The general meaning of Referer is the "caller page". There at least two situations in which there will be no Referer header:
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  • Method to get all File names in a directory

    Hallo together,
    has Java any method, to get the file names in a specific directory?
    Regards,
    Martin

    ... or simply list() if you're interested in file names only.

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