File Name from File Adpter in IDOC

Hi all,
We have scenario of file->xi->zidoc.
Now in one field for segement i also want to store the file name which has been processed by xi , is there anyway to get it ?
Thanks & Regards,
Tuhin

Hi Holger,
Thanks but can u guide me how to include customize code in module for communication channel and does it need to be EJB only or jar file can also be included.
If i get filename thorough this customize code but how to get this name in graphical mapping ??
Please send me any code or docs on [email protected] .
Thanks & Regards,
Tuhin

Similar Messages

File name from file path

Hi Friends,
i am using METHOD CL_GUI_FRONTEND_SERVICES=>FILE_OPEN_DIALOG
i want to know the file name selected from the path.
example : c:\folder\file.txt.
here i want to know which file i have selected. if it is a small path then we can use SPLIT command but if the path selected is long with special chareters then?
urget requirement.
regards
Kumar M.

u can use this function...
CALL FUNCTION 'WS_FILENAME_GET'
   EXPORTING
     DEF_PATH         = 'C:\'
     MASK             = '*.xls'
     MODE             = 'O'
     TITLE            = SY-TITLE
   IMPORTING
     FILENAME         = FILENAME
   EXCEPTIONS
     INV_WINSYS       = 1
     NO_BATCH         = 2
     SELECTION_CANCEL = 3
     SELECTION_ERROR = 4
     OTHERS           = 5.
have a nice day...

How to get file name from file adapter (Read Operation)

Hi All,
I am reading files from local location, in jca file i gave filename property =*.* ( which reads all files), and now I want to read the incoming file name inside bpel and I want to copy the file name to some variable. Is this possible?

Yes, its possible.
In your receive activity,set below property in its properties tab:
<bpelx:property name="jca.file.FileName" variable="FileName"/>
Before that make sure to create a variable "FileName" of string type.
Whatever file your file adapter will read,its name will be stored in this variable.
Hope this helps.
Regards,
Karan
http://learn-oraclesoa.blogspot.com/

Using the array of file names from File.list()

Hi
I'm having trouble with my code, similar to this:
File temp=new File(path); //where 'path' is a String containing the directory path
String [ ] arr = temp.list();
Then when trying to use arr, I come up with NullPointerExceptions.
The Java API says that the File.list() method returns a Stirng array.
But then how do I use/get this array in a nice usable form?
(eg. to be able to loop through the list of {returned} Files)

Yes I have referred closely to the documentation.
The method list() for my File object is definitely not returning null;
it seems that as soon as the code inside the loop tries to access the next element of the returned array [the one after the last existing element] - which is actually non-existent (ie overstepped the bounds of the array's index) - the program comes up with an Exception*. So the method list() does not return null...
I have managed a workaround similar to this:
        File temp = new File (path);
        String[] list = new String [temp.list ().length + 1];
        try
            for (int i = 0 ; temp.list () != null ; i++)
list [i] = temp.list () [i];
catch (ArrayIndexOutOfBoundsException iobe)
* The program is not coming up with NullPointerExceptions anymore; only an ArrayIndexOutOfBoundsException if I comment out the try and catch coding.... (?)
Thanks for your suggestion about posting this at the Java Programming Forum.
My workaround works, but is there a way to do this without catching an exception?

File name from an attribute of an element of the payload

Dear Experts,
Could you suggest the syntax for specifying the file name from a variable in the file receiver communication channel? I need to grab the file name from an ATTRIBUTE of an element of the payload.
Thanks a lot.
Example: Need to pass the attribute "descReceipt" of the element "promotionEvent" to the variable in the following payload:
<?xml version="1.0" encoding="UTF-8"?>
<promotionExport>
<promotionEvent eventIdentifier="30" status="A" descReceipt="FILE1">

The scenario is IDOC to FILE. I need to get the value of the receiving partner.
The message header has the node "DynamicConfiguration". Please see the xml below. I need to grab the key RCVPRN (0000002104) from this and pass as the file name in the receiver file adapter.
How do I setup my receiver file adapter for this? I tried changing the values for the variable, and also enabling the adapter specific message attributes. No solution yet.
<SAP:DynamicConfiguration xmlns:SAP="http://sap.com/xi/XI/Message/30" xmlns:SOAP="http://schemas.xmlsoap.org/soap/envelope/" SOAP:mustUnderstand="1">
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPOR">LXDCLNT250</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPRT">LS</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPRN">0000002104</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPOR">SAPLCD</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPRN">LCDCLNT220</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="CIMTYP" />
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="IDOCTYP">WPDBBY01</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="MESTYP">WPDBBY</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPRT">KU</SAP:Record>
</SAP:DynamicConfiguration>

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?

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.
Here is the Main class
import org.me.lib.MyJNIWindowClass;
public class Main {
public Main() {
public static void main(String[] args) {
MyJNIWindowClass mw = new MyJNIWindowClass();
mw.s = "clock.gif";
And here is the library class
package org.me.lib;
public class MyJNIWindowClass {
public String s;
ImageIcon image = new ImageIcon("C:/Documents and Settings/Administrator/Desktop/" + s);
public MyJNIWindowClass() {
JLabel jl = new JLabel(image);
JFrame jf = new JFrame();
jf.add(jl);
jf.setVisible(true);
jf.pack();
I do understand that when I am making reference from main() method to MyJNIWindowClass() s first initialized to null and that is why clock could not be seen but how can I assign image file name from Main() class for library class without creating reference to Main() from MyJNIWindowClass()? As I said, I want this library class being accessed from different applications (means different Main() classes).
Thank you.

Your problem is one of timing. Consider this simple example.
public class Example {
    public String s;
    private String message = "Hello, " + s;
    public String toString() {
        return message;
    public static void main(String[] args) {
        Example ex = new Example();
        ex.s = "world";
        System.out.println(ex.toString());
}When this code is executed, the following happens in order:
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) {
        message = "Hello, " + name;
    public String toString() {
        return message;
    public static void main(String[] args) {
        Example ex = new Example("world");
        System.out.println(ex.toString());
}

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

Retreiving the file names from directory inside another directory from application server

Hi,
I had a problem in retreiving the file names from a directory inside another directory.
I tried using the FM's SUBST_GET_FILE_LIST, RZL_READ_DIR_LOCAL and EPS_GET_DIRECTORY_LISTING
But here I am getting only one directory details.
Actually my file is located a directory inside one more directory and one more directory and inside the files are located.
i.e total 3 directories inside the 3rd one my files are there.
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?
Can anyone please let me know How can i acheive this one.
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

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;
     }

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

Hi All,
         I am Using a file to Mail Scenario, My requirement is to get dynamically the file name from flat file of the Sender Side into Subject of Receiver Mail Adapter and attachment of file from the Sender Side. Can anyone help me out.
Thanks in Advance

Hi....
          You can write udf in the message mapping to get the Dynamic file name and map that to the subject of the receiver mail.
DynamicConfiguration conf = (DynamicConfiguration) param.get(
          StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey KEY_FILENAME = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File","FileName");
      // read value
      String Filename = conf.get(KEY_FILENAME);
Regards,
Leela

How to store the File Name from a Unix Directory into a ODI variable?

Hi,
I have built a ODI package with the following steps:
1. ODI is polling for a flat file in a Unix directory. I have used OdiFileWait tool for this purpose. Here the file name is not fixed, so I am using wild character (*) to poll for file. Example: DF*ABC1*.DAT where the first wild character denotes 1 letter and the second wild character denotes 2 digits.
2. In the second step, if the file is found, I am moving the file to ODI file server path (../oracledi/demo/file). Here I have used ODIFileMove tool.
3. Then I am using an ODI Interface to Load the file data into a Oracle database table.
4. I am using a Process log table to keep the log for each step I am executing in ODI that is ODIFileWait, ODIFileMove, Interface etc. for each file. In this table a row needs to be inserted after ODIFileWait tool gets the file, with the File Name and File Date. Later on this row will be updated as the consequent steps are executed.
Here is my concern, I need to get the exact File Name of the file after ODIFileWait gets the file and I need to insert that in the Process Log table after the ODIFileWait step gets a file. So if I can store the File Name in a ODI Variable, I can insert it into the Process log table at this point. This I am not able to do.
The files are coming in a different directory (not in ODI file server path), So after getting the file it is moved to ODI file server path (../oracledi/demo/file)
The Files which I am processing are fixed length format. Also, the File name and File Date is stored in the 1st Record(Header record) of the files.
Kindly provide me suggestions to implement this in my code.
Thanks and Regards,
Anik
Edited by: 809820 on Nov 10, 2010 11:36 PM

Look at this link -http://odiexperts.com/getting-one-or-several-unknown-files-from-a-directory
change the command to fetch the ls command and write into File and then using either java or jython break the data and fetch the file name and date and insert into log table.
(or)
you can use the os.system command and get the complete ls command into string and then process and insert it.
Let us know if you need any other help.

How to get the file name from Oracle B2B 10g

Hi My requirement is I am getting a CSV file from Trading partner, I am using oracle 10g b2b to translate the data.
In my BPEL 10g I am using AQ adapter to get the message from IP_IN_QUEUE.
Now I want to get the file name Eg: SampleFile.dat of the CSV file in my BPEL process.
I tried using the b2b.filename property in the receive activity and it is not getting the file name.
<sequence name="main">
    <receive name="Receive_Note" partnerLink="GetB2BNote"
             portType="ns1:Dequeue_ptt" operation="Dequeue"
             variable="Receive_Note_Dequeue_InputVariable"
             createInstance="yes">
             <bpelx:property name="b2b.fileName" variable="WriteFileName"/>
    </receive>
</sequence>
Can you help me to get the file name from Oracle b2b 10g ?
Thanks,
b2b user

Hi My requirement is I am getting a CSV file from Trading partner, I am using oracle 10g b2b to translate the data.
In my BPEL 10g I am using AQ adapter to get the message from IP_IN_QUEUE.
Now I want to get the file name Eg: SampleFile.dat of the CSV file in my BPEL process.
I tried using the b2b.filename property in the receive activity and it is not getting the file name.
<sequence name="main">
    <receive name="Receive_Note" partnerLink="GetB2BNote"
             portType="ns1:Dequeue_ptt" operation="Dequeue"
             variable="Receive_Note_Dequeue_InputVariable"
             createInstance="yes">
             <bpelx:property name="b2b.fileName" variable="WriteFileName"/>
    </receive>
</sequence>
Can you help me to get the file name from Oracle b2b 10g ?
Thanks,
b2b user

File Name and File Content not gettinng passed from Proxy to Business Serv

Hi All ,
I have a requirement in OSB , where i need to pick the file from remote Source and FTP the files to Remote Target .Below are the steps which i did to achieve this.
1.Created a FTP Adapter in JDeveloper to Get the files from Source and a FTP Adpater to Put the files to Target.Inboth the adapters i have choose 'Shema is Opaque'
2.Imported the wsdl and jca file to OSB
3.Generated Proxy Service (PS ) and Business service (BS) out of Step 2
4.I edited the Message flow such a way that , the PS is routed to invoke the BS
Aslo i assigned $inbound/ctx:transport/ctx:request/tp:headers/jca:jca.file.FileName to a variable 'FileName' in PS
and in BS service i passed $outbound/ctx:transport/ctx:request/tp:headers/jca:jca.ftp.FileName = 'FileName'
When i tried to activate my session , the file that is getting written to the target has 0 byte.Also , the file name is also not getting passed from PS to BS
Can some one help me with the steps on how to use the Xpath , so as to pass the file name and file body from proxy servive to business service.
Thanks
John

I search multiple shares to find a common file name then create a single output file. I will be doing this search and file creation
for 5-10 different file names. If there is a better way .. certainly open for suggestions. It's working but having issue with
the cmd file for every file and folder I check. It puts this error out for each run of the process.
Error message in LOG file:
Get-ChildItem : Cannot find path 'F:\powershell\-SearchFor' because it does not exist.
Thanks.
I tried your code with little changes and saved in Temp folder.
My CMD file has the below code
PowerShell C:\Temp\Untitled1.ps1
It worked.
Regards Chen V [MCTS SharePoint 2010]

File type from file name

Hi,
Is there any standard method to get a file type from file name?
File name can be C:\test.doc or C:\test(test is file) i.e. with or without file type extension.
...Naddy

You can use the Function module PC_SPLIT_COMPLETE_FILENAME
export paramter extension will tell the type of file
Sample Output
Import     COMPLETE_FILENAME     C:\TEST.TXT
Import     CHECK_DOS_FORMAT
Export     DRIVE     C
Export     EXTENSION     TXT
Export     NAME     TEST
Export     NAME_WITH_EXT     TEST.TXT
Export     PATH     \

File Name from File Adpter in IDOC

Similar Messages

Maybe you are looking for