Get File name of the inbound file during mapping

Scenario: to read the file name of the inbound file (which has date required for the mapping) during runtime.
The requirement is to read the date of the inbound file (passed to the XI pipline by the file adapter) and populate the same in the outbound mapping structure.
Any idea about how to do this?
(I went through few options of using java.util.map. not successful yet)

Hi Anand,
I posted the same question a time ago, without any help....
Can I find out the full filename of input file in message mapping?
Posted: Nov 23, 2004 1:00 PM
I have in XI 2.0 the following scenario :
In the inbound fileadapter I read my input file. The filename of the input file is part fixed, part variable (Like INDATA01.txt, INDATA03.txt, etc).
So in my Adapterconfiguration, I specify the filename with a wildcard (INDATA*.txt).
What I now want to do, is in my Message Mapping use the full filename , so I can do something different for every filenumber. Is there a way where I can get the full filename available in my message mapping (I did not find the filename in the XML in the message trace).

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How write the Outbound file name in the Inbound file

Friends,
I have a scenario like this.
Outbout filename - sample.txt
In sample.txt, i have having data like 20
Inbound filename - vivek.txt
i need data the like sample, 20
(NOTE: sample is the outbound file)
How to get this scenario.
Thanks in advance.
regards,
Vivek.

Hi Vivek,
If your PI is of SP14 and more you have an option called Adapter Specific Message Attributes.(ASMA) in communicatoin channel configurtaion, Which will help you in capturing the Sender file name during runtime.
So you will have to enable that option (ASMA) in Commchannel and capture file name in UDF using Java Code with the help of sap related API and map it to the target field which u would like to get populated.
follow the thread it has the discussion about acessing the Source file name.
Re: Getting file directory using dynamic configuration:Code needed
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File-2-Proxy: Get file name in the inbound proxy?

Hi,
I need to know the file name in the inbound proxy. After a search I found that's possible mapping the file name to the target structure as it's explained in that link: [http://www.sdn.sap.com/irj/scn/weblogs;jsessionid=(J2EE3417200)ID1731350750DB00274596951164745615End?blog=/pub/wlg/2664]
But I need to know if there is another way to access the file name in the inbound proxy without modifying the mapping.
Thanks.

this is possible by a dynamic configuration (there is no other easy way)
some references -
http://www.sdn.sap.com/irj/scn/weblogs;jsessionid=(J2EE3417500)ID1159980350DB00296046510040545862End?blog=/pub/wlg/13704
the idea is to get the file name in runtime using the dynamic conf code and poulating it to any field that will be part of the proxy message

When someone other than myself downloads an image from my web album, a dynamically generated file name replaces the original file name. How I can prevent the original file name being replaced during the downloading process?

When someone other than myself downloads an image from my web album, a dynamically generated file name replaces the original file name. How I can prevent the file name being changed during this downloading process?

Hi Glenyse,
Here are my steps.
1. I upload multiple image (jpg) files onto my photo album.
2. I select the "sharing" by email option for this album.
3. I enter the recipient's email address.
4. The recipient receives my message and clicks on the link.
5. The recipient accesses my photo album and clicks on one of the images.
6. The image opens up to its own screen.
7. The recipient selects the "download" and then save file option.
Here is the part I do not understand. For some reason, during this "download" process, the original name which I have given to the file is replaced by different name. So I was hoping that someone knows how to prevent the file name from being changed during the "download and save" process.
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Assign Logical file name for the physical file path through Program

Hi all,
I am having a physical file which is getting generated dynamically. It is having the date and time stamp in its name which is added at runtime.
Now I need to assign a logical file name for the physical file path through the program.
Is there any FM or any other method to assign the same.
Gaurav

I think it is not possible. becuase the date & time added at runtime. so if you check the table PATH you can find filename and their definitions
a®

Different log file name in the Control file of SQL Loader

Dear all,
I get every day 3 log files with ftp from a Solaris Server to a Windows 2000 Server machine. In this Windows machine, we have an Oracle Database 9.2. These log files are in the following format: in<date>.log i.e. in20070429.log.
I would like to load this log file's data to an Oracle table every day and I would like to use SQL Loader for this job.
The problem is that the log file name is different every day.
How can I give this variable log file name in the Control file, which is used for the SQL Loader?
file.ctl
LOAD DATA
INFILE 'D:\gbal\in<date>.log'
APPEND INTO TABLE CHAT_SL
FIELDS TERMINATED BY WHITESPACE
TRAILING NULLCOLS
(SL1 DATE "Mon DD, YYYY HH:MI:SS FF3AM",
SL2 char,
SL3 DATE "Mon DD, YYYY HH:MI:SS FF3AM",
SL4 char,
SL5 char,
SL6 char,
SL7 char,
SL8 char,
SL9 char,
SL10 char,
SL11 char,
SL12 char,
SL13 char,
SL14 char,
SL15 char)
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I thought of renaming the log file to an instant name, such as in.log, but how can I distinguish the desired log file, from the other two?
Thank you very much in advance.
Giorgos Baliotis

I don't have a direct solution for your problem.
However if you invoke the SQL loader from an Oracle stored procedure, it is possible to dynamically set control\log file.
# Grant previleges to the user to execute command prompt statements
BEGIN
dbms_java.grant_permission('bc4186ol','java.io.FilePermission','C:\windows\system32\cmd.exe','execute');
END;
* Procedure to execute Operating system commands using PL\SQL(Oracle script making use of Java packages
CREATE OR REPLACE AND COMPILE JAVA SOURCE NAMED "Host" AS
import java.io.*;
public class Host {
public static void executeCommand(String command) {
try {
String[] finalCommand;
finalCommand = new String[4];
finalCommand[0] = "C:\\windows\\system32\\cmd.exe";
finalCommand[1] = "/y";
finalCommand[2] = "/c";
finalCommand[3] = command;
final Process pr = Runtime.getRuntime().exec(finalCommand);
new Thread(new Runnable() {
public void run() {
try {
BufferedReader br_in = new BufferedReader(new InputStreamReader(pr.getInputStream()));
String buff = null;
while ((buff = br_in.readLine()) != null) {
System.out.println("Process out :" + buff);
try {Thread.sleep(100); } catch(Exception e) {}
catch (IOException ioe) {
System.out.println("Exception caught printing process output.");
ioe.printStackTrace();
}).start();
new Thread(new Runnable() {
public void run() {
try {
BufferedReader br_err = new BufferedReader(new InputStreamReader(pr.getErrorStream()));
String buff = null;
while ((buff = br_err.readLine()) != null) {
System.out.println("Process err :" + buff);
try {Thread.sleep(100); } catch(Exception e) {}
catch (IOException ioe) {
System.out.println("Exception caught printing process error.");
ioe.printStackTrace();
}).start();
catch (Exception ex) {
System.out.println(ex.getLocalizedMessage());
public static boolean isWindows() {
if (System.getProperty("os.name").toLowerCase().indexOf("windows") != -1)
return true;
else
return false;
* Oracle wrapper to call the above procedure
CREATE OR REPLACE PROCEDURE Host_Command (p_command IN VARCHAR2)
AS LANGUAGE JAVA
NAME 'Host.executeCommand (java.lang.String)';
* Now invoke the procedure with an operating system command(Execyte SQL-loader)
* The execution of script would ensure the Prod mapping data file is loaded to PROD_5005_710_MAP table
* Change the control\log\discard\bad files as apropriate
BEGIN
Host_Command (p_command => 'sqlldr system/tiburon@orcl control=C:\anupama\emp_join'||1||'.ctl log=C:\anupama\ond_lists.log');
END;Does that help you?
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Error: Enter either the physical file name or the logical file name

Hi Expert,
I have error: Enter either the physical file name or the logical file name, when upload BP with external data transfer.
My Step:
1. Define Sender Structure
2. Define Transfer Rules
3. Start Transfer
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Any document/link for upload BP with external data transfer.
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Hi,
See thread: [CO-PA External Data Transfer|CO-PA External Data Transfer]
Regards
Leon

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Hi All,
i have 2 files - File1 from File2 from 2 different folders- single comm channel FTP (FCC). i have to pass a constant A and B respectively to a field in target structure.
my options are -
either create a new field in source str populating different constants and then map to the target str
or a udf to get the source file name or the source folder so that i can map the constant using the value mapping/FixValue mapping.
please suggest. thanks.
rajib

Hi,
please use the UDF to determine the file name. Then based on file name either pass A or B to your target field.
The code is as given below. Just tweak the code to pass either A or B as per your requirement.
public String G_GetFileName(Container container) throws StreamTransformationException{
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File","FileName");
String ourSourceFileName = conf.get(key);
return ourSourceFileName;
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Deepak.

Locating source file NAME from the executable file?

Hi all,
I wrote a vi a couple of year ago (call it file.vi), and created an executable out of it using the Application Builder (call it newfile.exe). I know I renamed the file
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murchak wrote:
Dennis,
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How to obtain the file name of the cached file

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Dynamic file name inside the ZIP file

Hi Experts,
I have done proxy to file scenario, reciever side is the file ZIP file, i have done successfully done Zip file using adapter modules in Communication Channel.
The file name is hard coded like this.
contentType-----Transform.ContentType--
text/plain;charset="UTF-8";name="Billing.txt"
Is there any possible to do dynamic file name instead of hardcoded like add time stamp.
Kind Regards,
Praveen.

hi,
you can always do it using an adapter module like shown in my blog:
/people/michal.krawczyk2/blog/2006/02/23/xi-dynamic-name-in-the-mail-attachment--pseudo-variable-substitution
Regards,
Michal Krawczyk

File name in the target file field

Hi,
In this below particular blog we are creating the UDF and assigning the output of UDF to a filed in the target file structure.
/people/sravya.talanki2/blog/2006/12/20/accessing-system-parameters-in-mapping-runtime
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Please correct if my understanding is wrong.
thanks
kumar

Hi Kumar,
Whatever the filename that we enter in the UDF that will also be displayed as the target file name
Yes
To do this even we are selecting the adapter specific attributes.
"We could be" instead of "we are". If u want to access sender metadata, u should use ASMA.
Check this out
https://www.sdn.sap.com/irj/sdn/go/portal/prtroot/docs/library/uuid/70c89607-e4d9-2910-7280-f6746e964516
[Reward if useful :)]
Regards,
Prateek

How to print .sql file name in the same file using sqlplus

Oracle 10g on Linux platform.
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test.sql
select '&0' from dual;But instead of printing, it is asking for input:
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Enter value for 0: Can any body give me a hint as how can i print the file name.

Use this
SELECT FNAME_KRBMSFT AS name FROM X$KRBMSFT
and see
SQL> create or replace procedure get_dir_list
2 (p_directory varchar2)
3 is
4 l_null varchar2(100);
5 l_directory varchar2(100);
6 BEGIN
7 l_directory := p_directory;
8 sys.dbms_backup_restore.searchfiles(l_directory, l_null);
9 FOR x IN (select fname_krbmsft fname from x$krbmsft) LOOP
10 dbms_output.put_line(x.fname);
11 END LOOP;
12
13 END;
14 /
Edited by: HuaMin Chen on Jun 26, 2012 12:26 PM

Where (drive, folder, file name) are the database files that store catalog information such as tags and changed dates in Elements 13?

I want to backup just the datebae file that holds catalog data. I can't find it in Elements 13 but could in Elements 10. Is it hidden in a secret file with a secret name? Why?

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Get File name of the inbound file during mapping

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