Path and directory

I have ever my problem with a connection to a .txt file. Do you know if they exist a small app to know the path of the app on the support. I think that my problem could be a path problem.

Have a look at the link in my earlier post:
http://forums.ni.com/ni/board/message?board.id=170&message.id=267439#M267439
LabVIEW Champion . Do more with less code and in less time .

Similar Messages

The file path and directory about the .exe file created in LabView, who knows?

I have a project, in the project, VIs and documents(.doc,.txt,.tdms,etc.) in different directories, and when the project run in the labView, it can find the directories and files, but when I created .exe file, I found the directories had changed and I didn't know the directory structure in the file, anyone know it?
Thanks for any reply!
YangAfreet

Have a look at the link in my earlier post:
http://forums.ni.com/ni/board/message?board.id=170&message.id=267439#M267439
LabVIEW Champion . Do more with less code and in less time .

Both file and directory path selection

Hi,
Is it possible to select both file path and directory path from one parameter using F4?If yes how?
Thanks.
Edited by: Ginger on Oct 5, 2009 7:51 AM

>
Ginger wrote:
> No...file can be anywhere....file path selection means will have file name(c:\abc.txt) and directory selection means only select 'C:\'
> from which files can be selected.
>
> Now I want to enable both the options in one parameter path.Is it possible?
AFAIK, it is not possible.
But you can try a workaround though :-P
Make two radio-buttons:
1. Select File (rb_file)
2. Select Directory (rb_dir)
Then based on the radio-button call the respective method
IF RB_FILE = 'X'.
Call Method CL_GUI_FRONTEND_SERVICES=>FILE_OPEN_DIALOG
ELSEIF RB_DIR = 'X'.
Call Method CL_GUI_FRONTEND_SERVICES=>DIRECTORY_BROWSE
ENDIF.
BR,
Suhas

HOw to create a text file in the given path and delete it after the use?

Hi all,
I am trying to create a text file at the given path and delete the created file after the use.
I am using following code.:
import java.io.*;
// write binary data as characters
public class RanIO {
                                        public static void main(String f[])
                                                  // First illustrate append
                                                  String lineSep = "\n";
                                                  try {
                                                                 File temp= new File("C:/Ash","cute.txt");
                                                  boolean ch=temp.createNewFile();
                                                  if(ch)
                                                       System.out.println("file created");
                                                  else
                                                  System.out.println("file Not created");
                                                  //writing to file
                                             /*     PrintWriter p = new PrintWriter(new BufferedWriter(new FileWriter("cute.txt",true)));
                                                  p.print("Emp NO");
                                                  p.close();*/
                                                            // Open fileWriter in append mode
                                                                           FileWriter fos = new FileWriter(temp, true);
                                                                           BufferedWriter bw = new BufferedWriter(fos);
                                                                           PrintWriter pw = new PrintWriter(fos);
                                                                           double d=550;
                                                                      // lineSep = System.getProperty("line.separator");
                                                                      pw.print("Hello");
                                                                      //pw.print( lineSep );
                                                                      pw.print( d );
                                                                      pw.close();
                              boolean det=temp.delete();
                                             if(det)
                                                  System.out.println("File deleted");
                                             else
                                                  System.out.println("File not deleted");
                                             } catch (IOException ioe)
                                                            System.out.println( "Append IO error:" + ioe );
My problem:
1)
I am not able to write to the file. I want to know, where i am going wrong.
It is giving error message like
"Canot resolve Symbol: temp,"
But, FileWriter Constructor should accept a File type parameter.
here temp is a file parameter.
If i am not using file=new file();
i can't delete the file after the use. i.e if i use
PrintWriter p = new PrintWriter(new BufferedWriter(new FileWriter("cute.txt",true)));
how can i delete cute.txt after the use?
2)
I am not able to write to the text file. file is created but, a blank file.
"Hello" is not written into the text file.
can anyone help me in this regard
Thanks in advance
Ashvini

Thank you Ram,
But, i want to create a text file in Append mode.
for that i used
FileWriter fos = new FileWriter(temp,true); But, it is not accepting FileWriter constructor in
this format. if i use
FileWriter fos = new
FileWriter("c:/ash/cute.txt",true); it works fine. !!!!!Here's the javadoc
public FileWriter(File file,
boolean append)
throws IOExceptionConstructs a FileWriter object given a File object. If the second argument is true, then bytes will be written to the end of the file rather than the beginning.
Parameters:
file - a File object to write to
append - if true, then bytes will be written to the end of the file rather than the beginning
Throws:
IOException - if the file exists but is a directory rather than a regular file, does not exist but cannot be created, or cannot be opened for any other reason
Since:
1.4
Are you using jdk.13 or lower ?
>
ONe more doubt, Does flush method deletes a file?
if not, then i need to use
File temp=new File("c:/ash/cute.txt");
FileWriter fos = new FileWriter(temp,true); //which
is again a problem
if(temp.delete())
out.println("File is deleted");
}I don't know whether i am taking it wrong ! or
anything wrong with my coding ! but, after creating
and writing data into a text file. I must delete it
as it contains confidential informations.
Regards,
Thanks.'flush' writes to a file immediately. Else you should explicitly call 'flush' to write contents from buffer to underlying source.
javadoc again
PrintWriter
public PrintWriter(Writer out,
boolean autoFlush)Create a new PrintWriter.
Parameters:
out - A character-output stream
autoFlush - A boolean; if true, the println, printf, or format methods will flush the output buffer
cheers,
ram.
Question; What do you gain by opening a file, writing to it and deleting it in the same program ?

I have a question on paths and templates

I have a question on how I should structure my site
At my root directory on my server I have this set up.
index.html
/pages
     page1.html (all of these were created from my own template)
     page2.html
     page3.html
     and so on....
/webReadyGraphics
/css
/library
/templates
     genericTemplate.dwt
Each folder has their own respective files in it.
The problem I'm bumping into is relative and absolute paths in my DW templates/assest. I will do my best to explain.
I've create a menu with text that links to various pages. An example would be a home button which always returns you to the index.html. I turned that home button into an asset. I've placed that asset into my genericTemplate.dwt. The link attached to the home button is a relative path and looks something like ../index.html.
I have used my genericTemplate to create all my pages, even index.html. So on my index page is my home button which tries to link to a parent directory ../index.html. I decided to use absolute paths instead which is great on my local machine but when I go to upload to my server won't this be a problem? As browsers will be looking for a local absolute path?
How should I structure my website? I've been thinking about putting all my pages at the root directory but that just gets mess looking.
I hope I explained that well. My understanding of web theory is a bit novice.
Thanks!
-Dweinin

How do I do that? The pages which have the asset (button) have different paths to get to the index.html... some paths would be ../index others would be ../../index.html
I don't know how to explain this. I have 1 button. that button is on 15 webpages. Those pages are located at various directories from within my website. So all their paths would be different but I only have 1 button. So how do I put a relative path in there?
I can put all the pages in 1 location so the button path would be the same. That's not a problem BUT! I also have that button on my index.html page which is at the root of my site.
All of my pages would be located at root/pages/______.html. The button would have a relative path of ../index.html
The index is located at root/index.html and that SAME button would have an incorrect path of ../index.html and return an error saying page not found.
I want to keep the button an asset since I will likely be changing it's destination in the future or may want it to link a different page. So I made the button an asset so that I when I do want to change it I only have to do it once and it updates across all my pages.
Does that make sense? Should I just put all my pages at the root? Is that normal to do? My root would start to look pretty mess with 15-20+ html pages.
-Drew

Library paths and environment variables

Hello,
I am using 10g AS 9.0.4 on a Unix system.
I am trying to convert an old JSP web application to work with 10g. It is not an EAR/WAR file so I am manually trying to set it up under the "default-web-app" folder. It uses a number of java class files which currently reside in $J2EE_HOME/default-web-app/WEB-INF/classes/mystuff.
These class files require access to a number of libraries. I foolishly believed that they would be accessible if I were to place them in the $J2EE_HOME/default-web-app/WEB-INF/lib folder, but that did not work.
I read that OC4J uses a "catch all" lib directory at $J2EE_HOME/applib. I copied the library files there and they still did not work.
Frustrated, I tried to print out the environment variables for the server - only to find that this function is not supported.
I managed to print out the library path by creating a Java class files which returns the following as a string:
System.getProperty("java.library.path");
This (when called from a JSP) prints a number of different library paths, none of which match those found listed in the $J2EE_HOME/config/application.xml file. I placed the library files in one of the paths listed by the "java.library.path" and it worked!
3 questions:
1. How do you print the server's environment variables using a JSP? This strikes me as a very useful function for sorting out library and path issues. Do a search on Google for "print environment variables OC4J" or something similar and there are no useful pages suggesting how this can be done.
2. Why are the default library paths different on the server than to those specified in application.xml?
3. How do I change the LD_LIBRARY_PATH (or any environment variable for that matter)? I have tried adding the following line to $ORACLE_HOME/Apache/Jserv/etc/jserv.properties without success:
wrapper.env=LD_LIBRARY_PATH=/my/custom/library/path
Any help you can offer would be most appreciated.

>
Thanks very much for a such a comprehensive response
- it has helped answer a lot of questions, especially
being able to print out the system properties. That
will be really useful!
No worries at all Rob.
The main hurdle is that the libraries are indeed
native (*.so) libraries. I hadn't considered that
until you pointed it out in your message. Can I
assume, then, that $J2EE_HOME/applib and
$J2EE_HOME/default-web-app/WEB-INF/lib are only
designed to handle Java's .JAR librares? That would
explain why placing the native libraries in there
didn't work, but is there a way of getting such
native libraries recognised and bundled up in a
WAR/EAR file?
Yep, that's it.
Our proprietary applib and the servlet specs web-inf/lib are only dealing with Java archives.
The J2EE specs don't cover how to deal with native libs like this.
Now I've NEVER EVER tried it, but it's worth a shot of at least trying -- for simplified packaging purposes, what you may be able to do is to distribute the .so's within the web-inf/lib -- and then just set that directory (where's its realized on the server after the app is deployed) to be referenced witihn the LD_LIBRARY_PATH. I'd expect our deployment manager will just ignore files it doesn't know about, so arguably you should be able to at least distribute them with the standard archive.
You'll need to specifically set the LD_LIBRARY_PATH as a separate task outside of the deployment operation though.
>
This is where things get a bit weird, and is partly
my fault during the installation of 10g AS.
We already had an existing installation of Oracle 10g
Database (10.1.0) at $ORACLE_HOME under the user
"oracle". I logged in as the user "oracle" to install
the Application Server into a different location (In
this case, "$ORACLE_HOME/../ias10g_9.0.4" - let's
call this $IAS_HOME). So there is no separate user
account for Oracle Database and Application Server
(and hence no separate $LD_LIBRARY_PATH).
If I log into UNIX as the user "oracle" and echo the
LD_LIBRARY_PATH, I get the following directories:
$ORACLE_HOME/lib
/usr/openv/netbackup/bin
/usr/dt/lib
/usr/lib
I use this user to do "opmnctl stopall" and "opmnctl
startall". If I then print out the library path from
a JSP using
"System.getProperty("java.library.path");", I get the
following paths:
$IAS_HOME/jdk/jre/lib/sparc/server
$IAS_HOME/jdk/jre/lib/sparc
$IAS_HOME/jdk/jre/../lib/sparc
$IAS_HOME/opmn/lib
$IAS_HOME/lib
$IAS_HOME/usr/lib
Which is totally different to those listed by
$LD_LIBRARY_PATH at the command line, which means
they must be coming from somewhere else.
It gets worse - my application is trying to call a
native library that is only present in the
$ORACLE_HOME/lib32 directory (libclntsh.so.10.1), so
unless I can include this path in the
"java.library.path" on the Applicaiton Server, then
my program will not work.
So, how do I change the "java.library.path" when the
values aren't coming from $LD_LIBRARY_PATH in the
first place?
The utility $IAS_HOME/opmn/bin/opmnctl is actually a shell script.
It has this section which sets the LD_LIBRARY_PATH:
if [ -z "$LD_LIBRARY_PATH" ]
then
LD_LIBRARY_PATH=$ORACLE_HOME/opmn/lib:$ORACLE_HOME/lib ; export LD_LIBRARY_PATH
else LD_LIBRARY_PATH=$ORACLE_HOME/opmn/lib:$ORACLE_HOME/lib:${LD_LIBRARY_PATH} ; export LD_LIBRARY_PATH
fi
So if you have an existing LD_LIBRARY_PATH env var set, it should just be putting its paths onto the front of it.
What you could try as a quick test is to edit the opmnctl script (after taking a backup of course) and either appending the path you need to it, or just add some debug statements to output the LD_LIBRARY_PATH value it ends up setting so you can try and work out what's causing it.
I expect these other entries:
$IAS_HOME/jdk/jre/lib/sparc/server
$IAS_HOME/jdk/jre/lib/sparc
$IAS_HOME/jdk/jre/../lib/sparc
are coming from the JRE when it is launched, in probably a similar manner to our opmn entries.
I really appreciate your help, and I bet if anyone
can answer these questions, you can! ;-)
You're very welcome.
cheers
-steve-

Create logical file path and read it in main program

Hello Experts,
My requirement is, I have to read files from application server.
File path/ Directory is different in different system (systid). i have to read all files starting with (xyz*) from mentioned directory.
This path i must define in logical path.
Following are the steps i followed.
1) In FILE (TCODE) i have created a logical path and assigned a physical path to it.
2) I have created logical file ( I am not sure about this, as i dont have to create logical file, i just need to read file from above path starting with xyz*, but FM FILE_GET_NAME requires logical file name so i have created it).
3) I will pass logical file name to FM FILE_GET_NAME and recieve physical file path(with file name which i can ignore).
these all steps i found on SDN. but its not working.
FM FILE_GET_NAME is always returning same file path ( Not even close to what i have defined ).
I can not read data from Table path as it will return physical path as it is, but i need it replaced with system id.
Any suggestion will be really helpful.
Thanks in Advance.
Regards,
Amar

Hi Amar,
one difficulty defining physical path name is to know the right Syntax group.
For example if you habe OS Linux, you need Syntax Group UNIX here. In Table OPSYSTEM you can see, which OS (OPSYS) needs which FILESYS (Syntax Group).
If you always get a wrong physical path, this may be caused ba a wrong choice of the Syntax Group. Only the correct one will lead to the correct path.
Regards,
Klaus

Regarding the Logical Path and file name

Hi all ,
I am using LSMW for creating the BOM of a material .
I am getting problem at step 7 , saying the logical path is pointing to physical directory .
Then i did some R & D . I created the logical path and logical file name using transaction FILE . Then i got the physical path using FM file_get_name.
But when i give those logical path and logical file names , still i am getting prob.... at step 7 of lsmw.
Pls let me know . I did my best unti l now ....
Regards

Hi Raj,
Can you send the Error which you got at Step 7??
It would be of great help if you can send the File Path and the file names you have used.
Regards,
-Syed.

Set path and classpath in ANT

by using ANT, how can set the JDK path and classpath..
jdk version exists in the : C:\Program Files\Java\jdk1.6.0
This is the build.xml that i need to modify!
-Do i need to have a serperate JAVA_HOME variable?
<?xml version="1.0"?>

<project name="Chat" default="jar" basedir=".">
   
   
   
   <target name="init">
      <property name="appname"        value="chat"   />
      <property name="src.dir"        value="src"               />
      <property name="lib.dir"        value="lib"              />
      <property name="build.dir"      value="Chat"            />
      <property name="build.compiler" value="classic"          />
   </target>
   
   
   
   <target name="prepare" depends="init">
      <mkdir dir="${lib.dir}" />
      <mkdir dir="${lib.dir}/META-INF" />
      <mkdir dir="${build.dir}" />
   </target>
   
   
   
   <target name="classes" depends="prepare">
      <javac srcdir="${src.dir}/oracle/otnsamples/oc4jjms"
             destdir="${lib.dir}"
             includes="**" />
   </target>
   
   
   
   <target name="j2ee-meta-inf" depends="classes, prepare">
      <copy file="${src.dir}/META-INF/application-client.xml"
            tofile="${lib.dir}/META-INF/application-client.xml" />
      <copy file="${src.dir}/images/oralogo.gif"
            tofile="${lib.dir}/oralogo.gif" />
   </target>
   <target name="jar" depends="j2ee-meta-inf">
   <jar jarfile="${build.dir}/${appname}.jar"
       basedir="${lib.dir}" manifest="${src.dir}/META-INF/MANIFEST.MF"/>
      <delete dir="${lib.dir}" />
   </target>
</project>Message was edited by:
jugp

[url ='http://www.jguru.com/forums/home.jsp?topic=Ant']answer here

Setting PATH and CLASSPATH in Java Applet

Hi,
I have been wondering if there was a way to set the PATH and the CLASSPATH in the HTML file within the <APPLET> tag...
What I would like to do is to read some xml files outside of the directory where my html file is located to run my test application with the configuration on my computer.
(The applet and the html file from which it is called is on my local host.)
Therefore, I have seen that in order to access local files, I have to sign my jar files, which I did.
But the tricky part is that for my test application to work, I need to set the PATH and the CLASSPATH... and I have no idea if this is possible or not...
I have seen that:
*CODEBASE: I specify here the main directory where my jar files are located
*ARCHIVE: I specify here the list of my jar files
In my case:
CODEBASE=../../ (this is the root directory of my whole application)
ARCHIVE=dir/lib/jar1.jar,dir2/jar2.jar,configXML/,etc...I don't know if I can specify the filepath of my xml files (here "configXML") in ARCHIVE (I suspect we can only specify jar files there... right ?)
Maybe I can use ARCHIVE this way to specify my CLASSPATH (and well... not completely the way I would like I suppose), but what about the PATH ?
I have been searching this for soooo long now...
Any help, hint or suggestions (I might doing this all the wrong way...) would be greatly appreciated !
Thanks in advance !

You need to use the code attribute to specify the class that contains the main method to start the application.
<applet code="myPackage.myClass.class" codeBase="http://www.someServer/someFolder/" archive="myApplet.jar">
I hope this solves part of your problem at least.
Andrew

Sort by Path and Filename - What does a guy have to do?

Hello all,
First post here and such, but as the title reads. I have been using iTunes since early versions of 7 when I aquired my used 2gen Mini.
In a past incident where people were attaching viruses to the ID3 Tags (see: http://en.wikipedia.org/wiki/Talk:ID3#Virusesin_ID3tags) I took the liberty (stupidity) of stripping all my ID3 tags from my music directory due to that security fault at the time, but also because all of them (the tags) were horridly inputted to the files.
As iTunes users we can see the bad mistake I made, so now when I add a folder/directory to iTunes they just get sent to the bottom of the iTunes 'music list'.
'Sort by path and Filename' would do wonders in organization of these untagged files. When adding an entire genre or folder, all **** breaks loose and there are the typical 40 tracks of '01 - songname' '02 - songname' and so on.
This is where the rant part comes in. Here we are on Version 7.4.3.1 and Apple still hasnt added this minute feature to iTunes.
I have to use Yahoo! Musicmatch (yuck) to sort and rename my files, only then to re-import them back into iTunes. I dont really care for that software (yahoo) for anything except sorting. Because it adds the actual folder to the viewable list and allows for batch renaming of directories and ID3 Tag renaming of entire albums/directories. I am not so concerned over the actual filename itself, but at least sorting by Path in iTunes would be nice so that I could give the files their proper ID3 Tags.
Cmon Cupertino, this one isnt that hard.
Thank you for your time.
6

You have several options -- two of which I'll outline below (neither built-in to iTunes, sorry)
1) Use an external program to write ID3 tags based on the path/filename. One such program is "MP3Tag" (http://mp3tag.de/en)
2) I wrote a program that with put the path in the Description field. You select the playlist -- then for every track in the playlist it changes the Description to the path. NOTE: I do NOT recommend that you select the "Library" playlist because it will overwrite existing descriptions for TV Shows, podcasts, etc. You can limit it to music by selecting the "Music" playlist.
EXE version:
http://home.comcast.net/~teridon73/itunesscripts/itunespathdescription.zip
source code:
http://home.comcast.net/~teridon73/itunesscripts/itunespathdescription.pl
If you use my program, and it works for you, please consider a Paypal donation at my home page (http://home.comcast.net/~teridon73/itunesscripts/)

Path and file name to long

Hi
I have a schema with a deep hierarchy that I am running xmlbeans on. Initially
it failed to build because it couldn't find some of the class files it had generated.
It turned out that the path and filename was too long for Windows XP to handle
and moving the directory I was running the buiild to the root directory allowed
the build to complete. Unfortunately I have another schema which includes the
first one. It also has quite a long namespace. Is there any way of changing the
filename convention of the class files so that it doesn't use the XML node hierarchy
for its names?
If not it puts a serious limit on its use with real world schemas.
Andy

On Fri, 14 Nov 2003 01:24:39 -0700, Andy Brown wrote:
>
Hi
I have a schema with a deep hierarchy that I am running xmlbeans on. Initially
it failed to build because it couldn't find some of the class files it had generated.
It turned out that the path and filename was too long for Windows XP to handle
and moving the directory I was running the buiild to the root directory allowed
the build to complete. Unfortunately I have another schema which includes the
first one. It also has quite a long namespace. Is there any way of changing the
filename convention of the class files so that it doesn't use the XML node hierarchy
for its names?
If not it puts a serious limit on its use with real world schemas.
AndyAndy,
This topic in the workshop help demonstrates how to do some simple mapping
from namespaces -> package names and from xsd names -> java names.
http://e-docs.bea.com/workshop/docs81/doc/en/workshop/guide/howdoi/howGuideXMLBeansTypeNaming.html?skipReload=true
hope that helps!
-- kevin krouse

Zip and directory or package

How on earth do I zip a package like an Omnigraffle file? I have a nice folder-action script that zips a file into an archive when it is dropped into the folder, but when I dropped an omnigraffle file into the folder it broke it into all the files that the original omnigraffle file consisted of, in other words it treats the file like a directory and breaks it open. I guess omnigraffle files are packages. If I use ditto it complains and tells me that the file is a directory. I cannot find any zip parameter that allows me to compress an omnigraffle file and keep it is a single compressed file. If I right-click and select compress on a file OS X has no problem and I get a single compressed file with the name and a .zip appended, but if I execute the zip application in terminal, or from my script, I get all the tiffs and jpegs and bits and bobs that the omnigraffle file is built with and the file it a goner.
Most frustrating.
Thanks for any help.
Lawrence

Well - that's done, works fine. The issue I was having was that I wanted files to be zipped into an archive without any path, so I would include the -j parameter. The -j parameter however would not only NOT store the path of the file, but it would also NOT store directory names.
Okay - I thought. There is the -r parameter, this tells zip to travel the directory structure recursively thus picking up sub-directory content. However the -j conflicts with this and what I found was that zip would take the contents of any sub-directory and stick it at the same level as anything outside the folder.
At first I thought this would be okay as I wanted this script to be used for ONLY files. However when I added an Omnigraffle file I was stuck as these are packages, and zip views them as directories. So any omnigraffle file would essentially be burst into its component parts, all rendered as files, at the root level of the archive folder. A total no-no.
If I removed the -j, so that directories would be read and compressed, as well as the omnigraffle files, everything worked fine. Only now each file had a great long file path rooted at /System/user/me..... If I decompressed the zip archive I got a folder named "System" inside which was a folder named "user" and inside that was "me" and on and on until I finally got a folder with my unzipped files in it.
Horrible.
I realized that the key word in "man zip" was that the file path zip uses is relative to where the zip is running, and I was running it as a folder action which defaulted to the root, hence the vast great file path. So all I needed to do was to modify the script so that the zip command references just the file names without path, and then do a cd to the archive directory immediately prior to the zip command itself. Thus I implemented a "do shell script" which had two commands on the same line, separated by a semi-colon, the first command being the cd with the full path to the archive folder and the second being the zip without any paths at all and using just the -r parameter.
Bingo.
Thanks for everyone's help.
Lawrence

Directory logical path to directory physical path

Hello experts,
I have question related to getting path from defined earlier logcal path in FILE transaction.
I know there are FMs called 'FILE_GET_NAME' and 'FILE_GET_NAME_AND_LOGICAL_PATH' but they are not solving my problem.
Let's say I know only physical file name and logical file directory, but I don't know the full physical path of file.
I now want to build a program where user writes down the physical name of the file and in the code side there is defined logical file directory from which we can get the physical directory path and concatenate it with entered filename. Is there a possibility to do that?
Thank you in advance.
Best regards,
Andrew

function module FILE_GET_NAME_USING_PATH is doing exactly that.

Physical path and Logical name in File adapter

Hi All-
I am using a file adapter, in which i need to specify the physical path or logical path.
In dev environment my file adapter's input path is /c01/interface/dev/input, but in my test environment it is /u01/interface/test/input.
So everytime I have to change the path and deploy it in two different environment.
I am using unix environment, is there any way i can use some environmental variable $INPUT_DIR and use it in logical name so that in logical name i can use as $INPUT_DIR/input, where in dev $INPUT_DIR will be /c01/interface/dev and in test environment it is /u01/interface/test.
I have tried this but there is still some probelm, I want to know is the approach I am using is correct? or is there any other solution for it?
Regards,
Sreejit

Hi,
1. physical name: use a specific directory (e.g. /u01/....
2. logical name: use a LogicalDirectory like "InputFileDir" and rewrite this inside the bpel.xml
http://download-uk.oracle.com/docs/cd/B31017_01/integrate.1013/b28994/adptr_file.htm#CACDEBBH
and
http://download-uk.oracle.com/docs/cd/B31017_01/integrate.1013/b28981/appx_deploydesc.htm#CHDBDIIF
But what about creating a softlink like this on your testsystem:
ln -s /u01 /c01

Path and directory

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