File name from path

We are converting an application that has a table with a full path text column and want to extract only the file name and wondered how we can accomplish this. The table is named BatchImport and the column is named FilePath varchar(40). An example of a
row would contain G:\Everyone\Billfiles\AATAX.XLS and we want to extract just AATAX.XLS out of it. Thanks.

REVERSE, CHARINDEX and RIGHT will be your friends here:
DECLARE @table TABLE (value NVARCHAR(MAX))
INSERT INTO @table (value) VALUES ('G:\Everyone\Billfiles\AATAX.XLS')
SELECT *, RIGHT(value,CHARINDEX('\',REVERSE(value))-1)
FROM @table
Don't forget to mark helpful posts, and answers. It helps others to find relevant posts to the same question.

Similar Messages

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.

Separating file name from path...

HI,
I have a jsp page in which the user selects a file by using
<INPUT type=file name =fname value="browse">
I am passing this to the javascript in the same page as:
var file = document.f1.fname.value;
What I want to do is strip of the file name and add another filename to it:
i.e if var file = C:\path\to\folder\myfile.txt
then I want another variable to have the value:
var x = C:\path\to\folder\ .
Which means that I want to remove everything until the first " \ " (from the end)
is there any way to do this (and that too in javascript)
Thanks,
-G.

why are you using java script? I think all this can be done without java script. If you take the file name which is selected by the user in a String. Then you can very easily change the file name once you have the name of the file which the user selected.

Truncate file name from path

hi friends..
i have file path
/abc/xyz/file.pqr
i want to truncate file.pqr
/ may be any numbers .
plz help

Hi,
Search the string for '/' in loop
eg:-
lv_last_index type i.
lv_last_index = 0.
do
SEARCH file_path FOR '/' starting at lv_last_index.
if sy-subrc <> 0.
exit.
else.
lv_last_index = SY-FDPOS.
endif.
lv_last_index -
will contain last ocurrence of '/'.....
no ur filepath will be
file_path = file_path+0( lv_last_index ). "..specifies the within the bracket
check if the last char is '/'.. in that case cate lv_last_index = lv_last_index - 1.
Mark helpfull answers & close the thread.
Regards
Manoj

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

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].

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

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.

File name and path should be passed as parameter

hi all,
i am writing a set of statements into the text file. But i have to pass the file name and also path as a user enterable parameter.
please suggest me to proceed further.
for ex: i am storing it as
output:=text_io.fopen(c:abc\def\out.txt','w');
right now i am giving like this and storing the data, but now i have to allow file name and path as user parameter.
please suggest me.
Thanks..

Hi,
for ex: i am storing it as
output:=text_io.fopen(c:abc\def\out.txt','w');
right now i am giving like this and storing the data, but now i have to allow file name and path as user parameter.
please suggest me.Well, how hard can it be?
You have to get those two parameters from your application into two variables, say "v_filename" and "v_pathname", and then create the file just the same way using the variables to construct the full path :
output:=text_io.fopen(v_pathname || '\' || v_filename,'w');

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();
    if (fos != null)
        fos.close();
}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
      *         header does not contain the filename attribute.
     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;
     }

Adding the file name and path to a document

Is there a way to automatically generate the file name and path (i.e. 'P:\ARF530\BACKGROUND\diagram.ai') to a document without having to copy and paste from the explorer window?

Hi,
So to select the directory, in the WHNE-BUTTON-PRESSED trigger write,
     :<block_name>.<path_item_name> := GET_FILE_NAME(' ', NULL, NULL, 'Choose any directory.', OPEN_FILE, FALSE);
     IF SUBSTR(:<block_name>.<path_item_name>, LENGTH(:<block_name>.<path_item_name>)) != '/' AND SUBSTR(:<block_name>.<path_item_name>, LENGTH(:<block_name>.<path_item_name>)) != '\' THEN
          :<block_name>.<path_item_name> := :<block_name>.<path_item_name> || '\';
     END IF;and to open the file, write,
DECLARE
FT_File TEXT_IO.FILE_TYPE;
BEGIN
IF SUBSTR(:<block_name>.<path_item_name>, LENGTH(:<block_name>.<path_item_name>)) = '/' OR SUBSTR(:<block_name>.<path_item_name>, LENGTH(:<block_name>.<path_item_name>)) = '\' OR :<block_name>.<path_item_name> IS NULL THEN
     MESSAGE('Please Enter file name before continue');
     MESSAGE('Please Enter file name before continue');
     RAISE FORM_TRIGGER_FAILURE;
END IF;
FT_File := TEXT_IO.FOPEN(:<block_name>.<path_item_name>, 'w');
END;Regards,
Manu.
If my response or the response of another was helpful or Correct, please mark it accordingly

Separating file name and path

Hi,
I need to separate the file name and path of the file which I am uploading from my BSP appl.I am using SO_SPLIT_FILNAME_PATH ,but it is not working because this fm identifies '\' only from GUI front end.
How do I separate the file name and path in BSP ?
Regards,
Ananya

Assuming that you are using File Upload tag for file upload:
DATA: fileUpload TYPE REF TO CL_HTMLB_FILEUPLOAD.
fileUpload ?= CL_HTMLB_MANAGER=>GET_DATA(
                       request = request
                       id      = 'myUpload'
                       name    = 'fileUpload' ).
file_name      = fileUpload->file_name.
file_mime_type = fileUpload->file_content_type.
file_length    = fileUpload->file_length.
file_content   = fileUpload->file_content.
Data : filepath type string,
       filename type string,
       t1 type string.
filepath = file_name .
while filepath CA ''.
split filepath at '' into t1 filename.
endwhile.
In my sample code above, filename attribute will contain the name of the file uploaded.
Hope this solves the problem.
Regards,
Ravikiran.

Question about pass file name and path to file write adapter

I need to pass file name and path to file adapter for write. I got partial answers from thread Re: Get File name using File Adapter , but seems InboundHeader_msg or outboundHeader_msg only takes file name, how do I pass file directory?
since I still have to specify file format (like xxx_%xx%.txt) in the file adapter wizard. Will this name conflict with what the name defined in InboundHeader_msg ?
Similarly, how can I pass a file name and path to a file synchread adapter?
Thanks,
Message was edited by:
user531689

Just overwrite the filename in the WSDL file that was generated

Target Data File Name and Path

Hi,
I'm trying to deploy a mapping that writes data into a file, but I need to dynamically set the name and the directory where I want to write this file. The only workaround I found was manually change the pl/sql generated code and change the values in the FOPEN call with a parameter, but this is no what I really want to do. Does anyone know how to tell WB not to hardcode the file name and path?
I really appreciate you time,
thanks in advance,
Matias

Carla,
Unfortunately our releases do not go that fast...
What you could do as an intermediate solution (this is what I would do) is create your mapping and load into a staging table (this can happen in set-based mode so would be fast). In a post mapping process you manually write to the flat file by making the calls to FOPEN etc. by selecting from the staging table. The post mapping process can have an input parameter that you dynamically pass (i.e. via a mapping parameter) and set the file name.
Mark.

How to delete file names from Adobe Reader list

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.

Thanks for the answer, I will try this & post my results with this method. I have set the amount to zero, & it seems if I want to put back valid files into my list I first need to find them in another file in my pc, such as documents, open them in Adobe Reader, & then then they will reappear in the Adobe Reader list. Is this correct? Thank you!
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?

File name from path

Similar Messages

Maybe you are looking for