Get the file's path

I am using the file upload component to choose a file. I need to save the file's path in database. The problem is that the attribute "fileName" which is blinded to this component contains the name of the file not the file's path as I need. Could you please guide me how to reach the path of the chosen file?
Thanks ahead,

This may be helpful.
http://nsnotfound.blogspot.com/2007/08/nsimageview-and-image-filenames.html
Eric

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How to get the file path from HTML input form in Firefox3

In IE7 (and probably all famous browsers, including old Firefox 2), if we submit a file like 'C:\folder1\folder2\folder3\filename' it works properly and gives the full path to the file and the filename.
In Firefox 3, it returns only 'filename', because of their new 'security feature' to truncate the path, as explained in Firefox bug tracking system (https://bugzilla.mozilla.org/show_bug.cgi?id=143220)
I have no clue how to overcome this 'new feature'.
Can anyone help to find a single solution to get the file path both on Firefox 3 and IE7?
Thanks in advance....

Doubleposted and ignored any previous answer: [http://forums.sun.com/thread.jspa?threadID=5342405]. Don't do that. It's rude.
Regarding to your "problem": why do you need the file path?

How to get the file path from HTML input form in Firefox 3...

In IE7 (and probably all famous browsers, including old Firefox 2), if we submit a file like 'C:\folder1\folder2\folder3\filename' it works properly and gives the full path to the file and the filename.
In Firefox 3, it returns only 'filename', because of their new 'security feature' to truncate the path, as explained in Firefox bug tracking system (https://bugzilla.mozilla.org/show_bug.cgi?id=143220)
I have no clue how to overcome this 'new feature' because it causes all upload forms in my webapp to stop working on Firefox 3.
Can anyone help to find a single solution to get the file path both on Firefox 3 and IE7?
Thanks in advance....

As you're posting this in the JSF forum and looking at your [previous topic|http://forums.sun.com/thread.jspa?threadID=5342365], I assume that you're using Tomahawk's t:inputFileUpload. In this case, just use UploadedFile#getInputStream(). You may find this article useful: [http://balusc.blogspot.com/2008/02/uploading-files-with-jsf.html].
By the way, are you lazy or just dumb? You wasted one week to this! I've posted the aforementioned article one week ago. How did you stuck?

Getting the local full path name of a file from an item File Browser

Hi all,
I would like to get the local full path name, of a file selected from an item 'File browser'.
I see in wwv_flow_files the column 'name', but actually it's a kind of id like 'F1542335/myFile.gif' for example.
What i want is the local path of this file ("C:\Documents and Settings\All Users\....\myFile.gif").
I hope this is possible??
Anyway, thanks all for reading.
Ludo

Hello, thank you for the answer.
Please understand that I searched before posting, and not ony 15 min.
What i mean is that the value of a File Browser in an Apex applocation is something like 'F1542335/myFile.gif', but what I need is the local path location.
And 'F1542335/' is not the local path location, i can't use it. Or maybe there is a way to convert it in a proper path name, such as "C:\Documents and Settings\....", but I don't think so.
So my question is : how to get the local path name of a file in Apex?
I'll answer if I find some good things.
Thanks
Ludo
Edited by: user12945874 on 06-avr.-2010 3:59

Function module to get the file path in the system for TEMP folder

Hi All,
Is there any function module that I can use to get the file path in the system for TEMP folder.
I mean, i am supposed to give only TEMP as the input for that function module and I need to get the path of that in the system as the output.
I am unsing 4.0 version.
Please advice.
Regards
Ramesh

In Higher versions, we can use the below code:
call method CL_GUI_FRONTEND_SERVICES=>ENVIRONMENT_GET_VARIABLE
    exporting
      VARIABLE   = 'TEMP'
    importing
      VALUE      = LV_TMP
    exceptions
      CNTL_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.
call method CL_GUI_CFW=>FLUSH
    exceptions
      CNTL_SYSTEM_ERROR = 1
      CNTL_ERROR        = 2
      others            = 3.
if SY-SUBRC <> 0.
Error handling
endif.
concatenate lv_tmp '\' into folder_path.
But need to know in the lower versions like 3.1h and 4.0,

To get the File path dynamically

Hi all,
I am using the <b>Gui_upload function, and i want to get the file dynamically from the search help</b>,
So please help me in finding the answer.
Thanks,
Girish

Hai Girish
Check the following Code
DATA: I_FILETABLE TYPE FILETABLE,
      V_RC TYPE I.
PARAMETERS: P_FILE LIKE RLGRAP-FILENAME.     "local file with contracts
AT SELECTION-SCREEN ON VALUE-REQUEST FOR P_FILE.
CALL METHOD CL_GUI_FRONTEND_SERVICES=>FILE_OPEN_DIALOG
   EXPORTING
     WINDOW_TITLE            = 'Find File'
     DEFAULT_EXTENSION       = 'C:\'
     DEFAULT_FILENAME        = ''
     FILE_FILTER             = ',..'
   INITIAL_DIRECTORY       =
   MULTISELECTION          =
   WITH_ENCODING           =
CHANGING
    FILE_TABLE               = I_FILETABLE
    RC                       = V_RC
   USER_ACTION             =
   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.
READ TABLE I_FILETABLE INTO P_FILE INDEX 1.
Thanks & Regards
Sreenivasulu P
Message was edited by: Sreenivasulu Ponnadi

How is it posible to get the File name, size and type from a File out the H

How is it posible to get the File name, size and type from a File out the HttpServletRequest. I want to upload a File from a client and save it on a server with the client name. I want to conrole before saving the name, type and size of the file.How is it posible to get the File name, size and type from a File out the HttpServletRequest.
form JSP
<form name="form" method="post" action="procesuploading.jsp" ENCTYPE="multipart/form-data">
File: <input type="file" name="filename"/
Path: <input type="text" readonly="" name="path" value="c:"/
Saveas: <input type="text" name="saveas"/>
<input name="submit" type="submit" value="Upload" />
</form>
proces JSP
<%@ page language="java" %>
<%@ page import="java.util.*" %>
<%@ page import="FileUploadBean" %>
<jsp:useBean id="TheBean" scope="page" class="FileUploadBean" />
<%
TheBean.doUpload(request);
%>
BEAN
import java.io.*;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.ServletInputStream;
public class FileUploadBean {
public void doUpload(HttpServletRequest request) throws IOException
          String melding = "";
          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;
     }

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.

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 can i get the document full path in S_TieDocument?

I create a new S_TieDocument and override 2 methods:
extendedPreInsert and extendedPostInsert
how can i get the document full path in these 2 methods?
(e.g. if i use the system user to upload a new file named 'test.file' in system root path /home/system, and i want to get the document full path '/home/system/test.file', how to get it?
anybody know it, pls tell me, thanks a lot.
java source is as below
package oracle.ifs.server;
import oracle.ifs.beans.Document;
import oracle.ifs.beans.DocumentDefinition;
import oracle.ifs.beans.LibraryObjectDefinition;
import oracle.ifs.beans.LibrarySession;
import oracle.ifs.beans.PublicObject;
import oracle.ifs.common.AttributeValue;
import oracle.ifs.common.IfsException;
import oracle.ord.itech.cmsdk.ws.view.EmbeddedCMSDKWebServicesWSStub;
public class S_TieDocument extends S_Document
public S_TieDocument(S_LibrarySession session, S_LibraryObjectData data)
throws IfsException
super(session, data);
public S_TieDocument(S_LibrarySession session, java.lang.Long classID)
throws IfsException
super(session, classID);
// Overrides
public void extendedPreInsert(OperationState opState,
S_LibraryObjectDefinition def)
throws IfsException
// Always call super
super.extendedPreInsert(opState, def);
// Get the LibrarySession
S_LibrarySession session = getSession();
def.setUserSetAttribute("CUSTOMATTRIBUTE",
AttributeValue.newAttributeValue(""));
//how can i get the document file full path here?
public void extendedPostInsert(OperationState opState,
S_LibraryObjectDefinition def)
throws IfsException
// Get the LibrarySession
S_LibrarySession session = getSession();
//how can i get the document file full path here?
}

In the new version of Numbers this option is not available. You will have to type the file name into the header yourself. You can post feedback to Apple using the menu item "Numbers > Provide Numbers Feedback"
or use Numbers2.3 which should be in the folder "/Applications/iWork '09" if you previously had it installed

How to get rejected file absolute path location in file rejection handler

HI,
Right now I am working in the below scenario in SOA11G.
1) Create a file adapter service and wire it to a BPEL process.
2) Create a Fault-policies file where we have referred a Java Action as a Fault policy.
3) Refer that fault-policy to the Fault-Binding file and put both the file in the same directory with composite.xml file.
4) So when any rejection message happened with any wrong incoming file, the Java Action is getting called.
But as per my requirement I want to get the absolute path of the incoming file in my java action, so that I can write some other logic to copy that file to some other location.
So is there any API available to achieve this or any logic about how to get the absolute path of a rejected file at runtime.
Regards,
Tripathy.

Hi Omar,
You need to use OPEN_DATASET & READ DATASET statements to get the file from application server.
Steps:
Open dataset in read mode
Read the data set and collect into internal table
Now, convert this into XSTRING by using function module SCMS_STRING_TO_XSTRING
After converting into XSTRING, you can use CL_WD_RUNTIME_SERVICES=>ATTACH_FILE_TO_RESPONSE to download into local system
Please refer the below links
Upload File from SAP Application Server using ABAP
Application Server Program - ABAP Development - SCN Wiki
Hope this helps you.
Regards,
Rama

How do I get the file name?

I'm kind of new to javascript -- working with a colleague who has Adobe Acrobat Professional -- I've got an *old* version of Adobe Acrobat and most of the functions don't work -- but how do we get the file name -- we're trying to develop an application -- so if the user opens c:\something.pdf, we'd like to know the file name (something.pdf).
If we get a fully qualified path, that's fine -- I think I can parse it.
Thank you in advance.

The 'paty' property of the 'doc' object has been available since version 3.01.
Using the following in the JavaScript console when the Acobat JS API Refenece is open:
this.path;
returns:
this.path;
/C/Program Files/Adobe/Acrobat 8.0/Help/ENU/JS_api_reference.pdf

Get the current absolute path name?

How to get the current absolute path and file name in javascript?

If using CS3 or CS4
alert($.fileName);
or this should work on all versions.
function WhoAmI() {
var where;
try {var F = FO;
}catch( err ) {where = File(err.fileName);}
return where;
NB: Neither will work if you are using binary scripts!

File Adapter - how to get the file count from a folder

Hi All,
I have a requirement that have to poll a directory when the file count is reached to number N (ex:number of files avilable in folder is 5) otherwise it should wait and not pick any of the files. Is it possible to get the file count from a folder using file adapter ?? otherwise please suggest me an approach to achieve this requirement.
Thanks,
JJ

Hi Sarath,
Thank you for your reply.
Go with the list files operation of file adapter it will gives you the number of files in the specified folder as you given. . - this step is already done.
When the number of files reaches your count startup your webservice that which can polls the files. . . - how can i acheive this?? Have to poll the directory and process the number files - please let me know, what could be added to the webservice which is being invoked after cheking file count from parent process.
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Thanks,
JJ

FM to get the file from application server to presenatation server automati

Hi!
In my upload program I am egtting the error log file onmy application server directly. Is there a function module or a way that in my abap code I can mention to bring the file or transfer the file to the presentation server . As my program is been run in the back ground so I cannot directly get the file on the presenattion server, so I need to bring it back on teh presentation server back to see teh eror log.
I would like my program to do the transfer in background itself so that I can see the error log on a daily basis from my local file .
Thanks
when 'X'.                      "X is application server
*         TRANSLATE p_err USING '\/'." correct slash for unix
      open dataset p_err2 for output in text mode encoding default.
      if sy-subrc = 0.
        loop at t_err into s_nts.
          transfer s_nts to p_err2.
          if sy-subrc ne 0.
            message i010(ad) with p_err2 'Download Failed'.
          endif.
        endloop.
        close dataset p_err2.
      else.
            message i010(ad) with p_err 'dataset could not be found'.
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      endif.
*        endif.
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        call function 'DOWNLOAD'
          EXPORTING
            filename = p_err
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          sy-subrc = -1.             "Maintain error condition
          exit.
        endif.
    endcase.
ENDIF.
Thanks

SO , after reading all , I assume that there is no way in which when I run the background job and get a errro file on application server can be retrievd on my presenattion serevr during the background job run. I was thinking that theer might be a functionn module which will convert the application server file into presenttion server file during the program run in background and store it in the location provided.
I was trying to use FM CALL FUNCTION 'C13Z_APPL_TO_FRONT_END' and CALL FUNCTION 'ARCHIVFILE_SERVER_TO_SERVER' but using teh first one asks for source file and destination file name during teh program run and as it runs in background it cannot be provided again and again and teh second one dosent seem to work.
Thanks

Get the file's path

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