File name from a path

hi ,
there is a path "c:\abc\abc\abc\abc\abc.doc", i have to get file name from it.
but i had to write a dynamic code which can get the file name from any given path.

Smoke me a kipper, I'll be back before breakfast.
SQL> var extended_filepath varchar2(128)
SQL> exec :extended_filepath := '/this/is/a/path/to/a/file.txt'
PL/SQL procedure successfully completed.
SQL> select instr(:extended_filepath, '/', -1)
2 from dual
3 /
INSTR(:EXTENDED_FILEPATH,'/',-1)
21
SQL> select substr(:extended_filepath, instr(:extended_filepath, '/', -1) +1)
2 from dual
3 /
SUBSTR(:EXTENDED_FILEPATH,INSTR(:EXTENDED_FILEPATH,'/',-1)+1)
file.txt
SQL> Cheers, APC

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Seperating file name from the path

Hi,
im trying to validate the file path and trying to seperate file name from the path.
for eg.
if the path is f:\sapfilepath\doc\ext.txt
need to get f:\sapfilepath\doc im looking for a dynamic way of doing this.
your help will be appreciated.
Regards,
ravi.

Hi,
Use the function module
SO_SPLIT_FILE_AND_PATH
Thanks,
Naren

Retrieve file name from full file path

I am trying to retrieve a file name from a path (ex. c:\temp\tes.txt) using the following code, but it is given me an error of "The method split(String) in the type String is not applicable for the arguments (char).
{code}
String filepath = item.getName();
String[] buf = filepath.split('/');
String filename = buf[buf.length-1];
{code}
I tried to use double quote instead of single quote, but it is not returning anything.
{code}
String[] buf = filepath.split("/");
{code}
Anyone knows why? Thanks.

How is this related to JDBC?
Anyway, is the path separator actually a forward slash? Isn't it a backward slash?
If you're using Apache Commons FileUpload (which I guess, the 'item.getName()' is recognizeable), then just read their FAQ: [http://commons.apache.org/fileupload/faq.html#whole-path-from-IE].

Need to extract only file name from path.........

Hi All,
I have a parameter.This calls the function
"CALL FUNCTION 'F4_FILENAME' to get the file from C drive.
After selecting the file the path is displayed in the Parameter field.
My problem is I need to extract only file name from the path.Please advice.
Example : Prameter id    C:\folder\file.xls
I shd extract only file.xls from the path.Please advice.

Hi,
Use the below logic:
data: begin of itab,
           val    type char20,
        end of itab.
SPLIT l_f_path AT '\' INTO TABLE itab.
The last record of the internal table holds the file name.
describe table itab lines l_f_lines.
read itab index l_f_lines.
l_f_filaname = itab-val.
Hope this helps u.

Obtaining file name from the file path given

hi,
how to obtain file name from the file path given

Hi bharath,
1. PC_SPLIT_COMPLETE_FILENAME
2.
DATA : path LIKE pcfile-path.
DATA : extension(5) TYPE c.
path = filename.
CALL FUNCTION 'PC_SPLIT_COMPLETE_FILENAME'
EXPORTING
complete_filename = path
* CHECK_DOS_FORMAT =
IMPORTING
* DRIVE =
extension = extension
name = name
* NAME_WITH_EXT =
* PATH =
EXCEPTIONS
invalid_drive = 1
invalid_extension = 2
invalid_name = 3
invalid_path = 4
OTHERS = 5
regards,
amit m.

Flat file connection: The file name "\server\share\path\file.txt" specified in the connection was not valid

I'm trying to execute a SSIS package via SQL agent with a flat file source - however it fails with Code: 0xC001401E The file name "\server\share\path\file.txt" specified in the connection was not valid.
It appears that the problem is with the rights of the user that's running the package (it's a proxy account). If I use a higher-privelege account (domain admin) to run the package it completes successfully. But this is not a long-term solution, and I can't
see a reason why the user doesn't have rights to the file. The effective permissions of the file and parent folder both give the user full control. The user has full control over the share as well. The user can access the file (copy, etc) outside the SSIS
package.
Running the package manually via DTExec gives me the same error - I've tried 32 and 64bit versions with the same result. But running as a domain admin works correctly every time.
I feel like I've been beating my head against a brick wall on this one... Is there some sort of magic permissions, file or otherwise, that are required to use a flat file target in an SSIS package?

Hi Rossco150,
I have tried to reproduce the issue in my test environment (Windows Server 2012 R2 + SQL Server 2008 R2), however, everything goes well with the permission settings as you mentioned. In my test, the permissions of the folders are set as follows:
\\ServerName\Temp --- Read
\\ServerName\Temp\Source --- No access
\\ServerName\Temp\Source\Flat Files --- Full control
I suspect that your permission settings on the folders are not absolutely as you said above. Could you double check the permission settings on each level of the folder hierarchy? In addition, check the “Execute as user” information from job history to make
sure the job was running in the proxy security context indeed. Which version of SSIS are you using? If possible, I suggest that you install the latest Service Pack for you SQL Server or even install the latest CU patch.
Regards,
Mike Yin
If you have any feedback on our support, please click
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Mike Yin
TechNet Community Support

Logical file name or logical path name incorrectly defined

Dear All,
We are doing archival in our IDES for test purpose before we do it to our Production.
Steps Performed
Copied AM_ASSET archive object to ZAM_ASSET
Logical Path
Logical path    ZAM_ASSET
Name            Asset
Syntax group    UNIX       Unix compatible
Physical path   /archive/test/<FILENAME>
Logical File Name
   Logical file    ZAMASSET
   Name            Asset
   Physical file   FI_<MONTH>_<DAY>.txt
   Data format     ASC
   Applicat.area   AM
   Logical path    ZAM_ASSET
But when we run the WRITE though SARA , in the job log we get the following
Logical file name or logical path name incorrectly defined
When generating a file name for an archive file that is to be created, the system determined that the logical file name FIAA_ARCHIVE_DATA_FILE or the logical path name ARCHIVE_GLOBAL_PATH was defined incorrectly.
But we have maintained a Logical name ZAMASSET , so we are unable to change the location of archived file and as well as the format.
So is there any setting we need to maintain apart from the logical file name and logical file path.
Suggestions are highly appreciated.
Thanks in anticipation

hi,
follow this steps :
- transaction SARA
- enter authorization objects, eg SD_VBAK
- hit button CUSTOMIZING
- Archiving Object-Specific Customizing: execute Technical Setting
- field Logical File Name enter or select ARCHIVE_DATA_FILE
- leave CONTENT REPOSITORY as blank if you are not using 3rd party for storing (eg. IBM Tivoli)
- back to customizing
- from Basis Customizing, execute : Cross-Client File Names/Paths
- on Logical FIle Path Definition, highlight (select) ARCHIVE_GLOBAL_PATH on the right pane
- double click on the Assignment of Physical Paths to Logical Path on the left pane
- double click on OS used, eg. UNIX, define Physical Path where archive file (on WRITE process) will be stored
- save changes made
- double click Logical File Definition, Cross Client on the left pane
- double click ARCHIVE_DATA_FILE on the right pane
- make sure that logical path is already set to ARCHIVE_GLOBAL_PATH
- save changes made
this setting also can be done using transaction FILE
we have experienced on this case using SAP standard archiving (SARA, SARI) and everything is fine with this setting above.
hope it help you.
rgds,
Alfonsus Guritno

How to get the target file name from an URL?

Hi there,
I am trying to download data from an URL and save the content in a file that have the same name as the file on the server. In some way, what I want to do is pretty similar to what you can do when you do a right click on a link in Internet Explorer (or any other web browser) and choose "save target as".
If the URL is a direct link to the file (for example: http://java.sun.com/images/e8_java_logo_red.jpg ), I do not have any problem:
URL url = new URL("http://java.sun.com/images/e8_java_logo_red.jpg");
System.out.println("Opening connection to " + url + "...");
// Copy resource to local file
InputStream is = url.openStream();
FileOutputStream fos=null;
String fileName = null;
StringTokenizer st=new StringTokenizer(url.getFile(), "/");
while (st.hasMoreTokens())
                fileName=st.nextToken();
System.out.println("The file name will be: " + fileName);
File localFile= new File(System.getProperty("user.dir"), fileName);
fos = new FileOutputStream(localFile);
try {
    byte[] buf = new byte[1024];
    int i = 0;
    while ((i = is.read(buf)) != -1) {
        fos.write(buf, 0, i);
} catch (Throwable e) {
    e.printStackTrace();
} finally {
    if (is != null)
        is.close();
    if (fos != null)
        fos.close();
}Everything is fine, the file name I get is "e8_java_logo_red.jpg", which is what I expect to get.
However, if the URL is an indirect link to the file (for example: http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719 , which link to a file named JavaSetup6u18-rv.exe ), the similar code return AutoDL?BundleId=37719 as file name, when I would like to have JavaSetup6u18-rv.exe .
URL url = new URL("http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719");
System.out.println("Opening connection to " + url + "...");
// Copy resource to local file
InputStream is = url.openStream();
FileOutputStream fos=null;
String fileName = null;
StringTokenizer st=new StringTokenizer(url.getFile(), "/");
while (st.hasMoreTokens())
                fileName=st.nextToken();
System.out.println("The file name will be: " + fileName);
File localFile= new File(System.getProperty("user.dir"), fileName);
fos = new FileOutputStream(localFile);
try {
    byte[] buf = new byte[1024];
    int i = 0;
    while ((i = is.read(buf)) != -1) {
        fos.write(buf, 0, i);
} catch (Throwable e) {
    e.printStackTrace();
} finally {
    if (is != null)
        is.close();
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}Do you know how I can do that.
Thanks for your help
// JB
Edited by: jb-from-sydney on Feb 9, 2010 10:37 PM

Thanks for your answer.
By following your idea, I found out that one of the header ( content-disposition ) can contain the name to be used if the file is downloaded. Here is the full code that allow you to download locally a file on the Internet:
      * Download locally a file from a given URL.
      * @param url - the url.
      * @param destinationFolder - The destination folder.
      * @return the file
      * @throws IOException Signals that an I/O exception has occurred.
     public static final File downloadFile(URL url, File destinationFolder) throws IOException {
          URLConnection urlC = url.openConnection();
          InputStream is = urlC.getInputStream();
          FileOutputStream fos = null;
          String fileName = getFileName(urlC);
          destinationFolder.mkdirs();
          File localFile = new File(destinationFolder, fileName);
          fos = new FileOutputStream(localFile);
          try {
               byte[] buf = new byte[1024];
               int i = 0;
               while ((i = is.read(buf)) != -1) {
                    fos.write(buf, 0, i);
          } finally {
               if (is != null)
                    is.close();
               if (fos != null)
                    fos.close();
          return localFile;
      * Returns the file name associated to an url connection.<br />
      * The result is not a path but just a file name.
      * @param urlC - the url connection
      * @return the file name
      * @throws IOException Signals that an I/O exception has occurred.
     private static final String getFileName(URLConnection urlC) throws IOException {
          String fileName = null;
          String contentDisposition = urlC.getHeaderField("content-disposition");
          if (contentDisposition != null) {
               fileName = extractFileNameFromContentDisposition(contentDisposition);
          // if the file name cannot be extracted from the content-disposition
          // header, using the url.getFilename() method
          if (fileName == null) {
               StringTokenizer st = new StringTokenizer(urlC.getURL().getFile(), "/");
               while (st.hasMoreTokens())
                    fileName = st.nextToken();
          return fileName;
      * Extract the file name from the content disposition header.
      * <p>
      * See <a
      * href="http://www.w3.org/Protocols/rfc2616/rfc2616-sec19.html">http:
      * //www.w3.org/Protocols/rfc2616/rfc2616-sec19.html</a> for detailled
      * information regarding the headers in HTML.
      * @param contentDisposition - the content-disposition header. Cannot be
      *            <code>null>/code>.
      * @return the file name, or <code>null</code> if the content-disposition
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     private static final String extractFileNameFromContentDisposition(
               String contentDisposition) {
          String[] attributes = contentDisposition.split(";");
          for (String a : attributes) {
               if (a.toLowerCase().contains("filename")) {
                    // The attribute is the file name. The filename is between
                    // quotes.
                    return a.substring(a.indexOf('\"') + 1, a.lastIndexOf('\"'));
          // not found
          return null;
     }

How to dynamically define a file name and its path in a web application

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By an external configurationfile? That can be a propertiesfile, a xml file, an ini file, a plain vanilla txt file, etcetera.

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How can I delete file names from the Adobe Reader list? I have deleted the file & it's empty, but deleted file names remain in my list & it's very annoying. I can't seem to find an option anywhere to delete the file names.

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This seems to be the case, you need to restore each file you want in your list one by one by accessing them elsewhere on your pc, & then you must open them with Adobe Reader to have files reappear in your Adobe Reader list. However, this seems to be a tedious process if you only want to delete one obsolete file from your list by first emptying the list & rebuilding it file by file. It works, but I'm wondering if there is a more sensible & efficient way to do this?

How can I assign image file name from Main() class

I am trying to create library class which will be accessed and used by different applications (with different image files to be assigned). So, what image file to call should be determined by and in the Main class.
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And here is the library class
package org.me.lib;
public class MyJNIWindowClass {
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Your problem is one of timing. Consider this simple example.
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        ex.s = "world";
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1. new Example() is executed, causing an object to constructed. In particular:
2. field s is given value null (since no value is explicitly assigned.
3. field message is given value "Hello, null"
4. Back in method main, field s is now given value "world", but that
doesn't change message.
5. Finally, "Hello, null" is output.
The following fixes the above example:
public class Example {
    private String message;
    public Example(String name) {
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        return message;
    public static void main(String[] args) {
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}

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Hi,
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Actually my file is located a directory inside one more directory and one more directory and inside the files are located.
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I need to read the latest file name in the directory.
So that i can do some manipulation after getting the file name.
Is there option like OPEN DATASET , READ DATASET and CLOSE DATASET?
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Regards
Ram

Hi Ram,
Following thread can be helpful for you, were it shows in the tables structure rsfillst a field RSFILLST-TYPE whether its a directory or file..........
http://scn.sap.com/thread/865272
thanks and regards,
narayan

Flat file name from Sender Side Dynamically into subject of Receiver Mail

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File name from a path

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