How to get the source file name and transform into database

Hi,
We need to load data from source file abc.csv into oracle table. during loading, we need to keep the file name abc.csv and write into the database table as one field. Is there anyone who did the similar task? and how to make it? Pls help!

Hi,
If you can write the filename to a text file, you may do the following:
(1) create a text-file containing the name of your source-file.
(2) create an external table using that text-file as input.
Now, you can SELECT the filename from the external table.
Of course, you need a way to : create the flat file, handle the flat file, etc.
Grtz.

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Getting the Source File name Info into Target Message

Hi all,
I want to get the Source file name Info into Target message of one of the fields.
i followed Michal BLOG /people/michal.krawczyk2/blog/2005/11/10/xi-the-same-filename-from-a-sender-to-a-receiver-file-adapter--sp14
Requirement :
1) I am able to get the Target file name as same as the source file name when i check the ASMA in Sender & Receiver Adapter , with out any UDF...............this thing is OK
2) I took One field extra in the target structure Like "FileName" & I mapped it Like
Constant(" " )--UDF-----FileName
I Checked the Option ASMA in Both Sender & Receiver Adapters
Here iam getting the Target File name as same as Source file name + Source File name Info in the Target Field " FileName".
I Dont want to get the Target File name as same as Source file name. I want like Out.xml as Target file name.
If i de-select the Option ASMA in Adapters means it is showing " null" value in the target field "FileName".
Please Provide the Solution for this
Regards
Bopanna

Hi All,
Iam able to do this by checking the Option ASMA in only sender adapter itself
Regards
Bopanna

How to print the report file name and path and the last mod date

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How to get the SQL file name in SQL*plus

hi all,
     I have created two sql file at C drive as "c:\Createtable.sql" and "c:\Deletetable.sql"
afterwards i open
C:\>sqlplus
SQL*Plus: Release 10.2.0.1.0 - Production on Wed Jan 30 11:37:10 2008
Copyright (c) 1982, 2005, Oracle. All rights reserved.
Enter user-name: scott/tiger
Connected to:
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Singh

Dear Damorgan,
     >>your version number to three decimal places
     My Oracle DB Version i have already stated in my previous post is 10.2.0.1.0
Actually my problem is to get the sql files name we run in sqlplus enviornment with @ symbol. like
i have created one sql file in c drive as
"C:\Createtable.sql"
afterwords i have connected to sqlplus as
sql> conn scott/tiger
sql>@c:\createtable.sql
Now i want some query to get the name of the file which is run.
In actual my problem is as
i have suppose 10 or more SQL files in some folder ( sql1.sql, sql2.sql, sql3.sql ....).
i created one file to call all the 10 sql files (main.sql)
i have also one track_table which will keep track that which sql file is runned.
I want some automated script which will insert the record in that track_table....... for that i need the name of sql file which is runned.
Hope this will help you.
Thanks & Regards
Singh

How to get the Output File Name as One of the Field Value From Payload

Hi All,
I want to get the Output file name as one of the Field value from payload.
Example:
Source XML
<?xml version="1.0" encoding="UTF-8" ?>
- <ns0:MT_TEST xmlns:ns0="http://sample.com">
- <Header>
<NAME>Bopanna</NAME>
</Header>
</ns0:MT_TEST>
I want to get the Output file name as " Bopanna.xml"
Please suggest me on this.
Regards
Bopanna

Hi,
There are couple of links already available for this. Just for info see the below details,
The Output file name could be used from the field value of payload. For this you need to use the UDF DynamicFile name with below code,
// Description: Function to create dynamic Filename
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File" , "FileName");
conf.put(key,a);
return "";
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File" , "FileName");
conf.put(key,a);
return "";
With this udf map it with the MessageType as
(File Name field from Payload) > DynamicFileConfiguration>MTReceiver
Thanks
Swarup

How to get the document type, name and revision from the search page cv04n

Hello,
After performing cv04n and getting a list of documents satisfying the search criteria, how can I then get the document type, name and revision of the selected document using ABAP?
Thanks

HI,
IN table DRAW... u have Document type, Version and document number.
In Table DRAT also u get Document Type, Version, Number and Description of Doucment.
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How to get the incoming file name using JMS adapter and SOAP adapter

Hi Everybody,
   In one of my interface i need to get the file name of incoming flat file using JMS adapter at sender side. and then i am using xslt to convert it to IDOC and then posting to SAP IDOC.
my incoming filname are in this form price<DateTimestamp>.txt. when i do the tranformation this incoming file name should be part of one element in the IDOC which i am posting.
EX:
<IDOC
<REF>price<DateTimestamp>.txt</REF>
</IDOC>
Hope it is clear to everybody. I need your suggestion how i can capture this incoming file name and send it as part of IDOC.
Thanks
raj

If they are passing it in message id or correlation id,
you can access it using
<xsl:variable name="dynamic-conf"
        select="map:get($inputparam, 'DynamicConfiguration')" />
    <xsl:variable name="dynamic-key"
        select="key:create('http://sap.com/xi/XI/System/JMS', 'DCJMSMessageID/ DCJMSCorrelationID')" />
    <xsl:variable name="dynamic-value"
        select="dyn:get($dynamic-conf, $dynamic-key)" />
Check this:
http://help.sap.com/saphelp_nw70/helpdata/en/f4/2d6189f0e27a4894ad517961762db7/content.htm
Thanks,
Beena.

How to get full source file name in Exporter SDK

Hi,
I'm making an Exporter Plug-in using SDK 5.5.
I want to get the path and extension of the source file (when export file) and path of the sequence (when export sequence).
Especially the extension of the source file, which is always be replaced by the output extension.
I can't find it in the SDK guide and sample. Sorry if I'm being a noob.
Thank you.

Sequences are generally made up of multiple clips edited together. It's rare that a user would take a single clip and reexport it. So there isn't usually a one-to-one relationship between source file and exported file. Because of that, PPro provides the sequence name as the default file name.
If you did need to get at the source files, you could do it only under specific circumstances - only when exporting directly from PPro, or when loading a clip directly into AME, but not when exporting a sequence from PPro through AME. You would need to use the Video Segment Suite to parse the timeline segments, as demonstrated in the player sample code in videoSequenceParser->ParseSequence(). This is quite a complicated process and not something worth pursuing unless absolutely necessary.
Regards,
Zac

How to get the right file name in an attachment

I have design a program to get the attach file from an attachment. But when I click the link to save the file, the default name of the file is 'attach' whenever I get different files. Who know what is the problem, please tell me the truth.
Here are some codes of the program.
download.jsp
<%="test.doc"%>
web.xml
<servlet-mapping>
<servlet-name>AttachmentServlet</servlet-name>
<url-pattern>attachment</url-pattern>
</servlet-mapping>
AttachmentServlet.java
public void doGet
Message msg = folder.getMessage(msgNum);
// the message I get is right and the name of the file in
// the attach in this message is right too.(I have printed)
Multipart multipart = (Multipart)msg.getContent();
Part part = multipart.getBodyPart(partNum);
ContentType ct = new ContentType(sct);
response.setContentType(ct.getBaseType());
InputStream is = part.getInputStream();
int i;
while ((i = is.read()) != -1) {
out.write(i);
out.flush();
out.close();

The trick that I used is to append the URL with the name of the file. So if your url is "http://domain.com/attach", use "http://domain.com/attach/filename.doc". IE and Netscape parse the URL to determine what the filename is, so by placing this text at the end, you essentially let the browsers know what the filename is. :)
I have design a program to get the attach file from an
attachment. But when I click the link to save the
file, the default name of the file is 'attach'
whenever I get different files. Who know what is the
problem, please tell me the truth.
Here are some codes of the program.<< snip >>

How to Get the Source File Path in the Receiver Side

Hi Experts,
Here We are Trying to Do How to Get the Sender Information ( File Name & Path ) on the Receiver Side .
According to the Michal Blog
/people/michal.krawczyk2/blog/2005/11/10/xi-the-same-filename-from-a-sender-to-a-receiver-file-adapter--sp14
Here we Are Able to Get only the File Name.
But If I Want to get the Total File Path Also means What Should I Do Here ????
Please Let Me Know
Regards
Khanna

Khanna,
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey fkey = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File","FileName");
DynamicConfigurationKey dkey = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File","Directory");
String fname = conf.get(fkey);
Sring path = conf.get(dkey);
String final=fname""path;
return ""final"";
Best regards,
raj.

How to get the long file name from an 8.3 name in Windows.

Because of a legacy string limitation, I have a list of files in 8.3 format. The actual files are saved with long names. How can I get the long name from the short name?
In JavaScript, I would get a file system object and get the file object from there with the short name. Then it's no problem to get the full name. But I don't see anything in Java that matches that. ???
Ideas?
Frank Perry, MSEE

Here is what I did.
String displayName = "somefi~1.txt";
File fO = new File("c:\\"); // I have a more involved path but that's not important here.
String stPath = fO.getPath() + displayName;
File fD = new File(stPath); // get the file using the short name.
File fDc = new File(fD.getCanonicalPath()); // get another file using the cononical name
String FulldisplayName = fDc.getName(); // get the long name from there.
It's roundabout but it works. Since getName() on the file read with the short name only returns the name used to get the file, I open it twice. The alternative is to parse the cononical name for the file name but that's clumbsy too.
Frank

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) {
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          for (String a : attributes) {
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                    // quotes.
                    return a.substring(a.indexOf('\"') + 1, a.lastIndexOf('\"'));
          // not found
          return null;
     }

How to get the original file name in chinese with upload component?

It is inconvenience for user to upload file with name only in english. When upload a file with chinese file name, the file name gotten from "fileUpload1.getUploadedFile().getOriginalName()" is just like "缃戠粶鍥�.jpg". Do somebody have a good idea?

I spoke to the component engineers regarding this issue
First please check if the s page encoding is set correctly, as browsers tend to default to use the encoding that was used for the page when submitting the request.
(Because of how the JSF response writer is written, multibyte characters can be written correctly even if the encoding is not set correctly so the fact that form correctly rendered chinese characters is not proof that the encoding was set correctly).
If you established that the page encoding is correct, then there is a
bug in the FileUpload. The method that gets the name is a straight
shot through to an Apache FileUpload library method. Technically this
ought to be resolved by that method checking the request body
character set before parsing the parameters.
- Winston
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How to get the trace file name for current running application?

Hi, I want to know if it is possible to get the file name directly for current running application instance which is launched by javaws.
There is a property "deployment.user.logdir" tells the log directory, it would be great if a file name property
is available. something like "instance.trace.file".
Our application wants it because we would like our client send use the application log by clicking a "send error"
button, the codes finds the trace file and compress it and send it by using a smtp server.
In 1.5, we can do it by using a shell program.

I found other asked it before, but I tried to set both properties, but neither works. my sun JRE version :java version "1.6.0_04"
<property
name="deployment.javaws.traceFileName"
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<property
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How to get the complete file name?

Good day!
My task is to upload files to the database and also save its path. I am already through with the Blob files but I have a problem with its filename. My boss wants me to get the complete path/location of the file that I am uploading. I mean, she wants to get where it is saved first before I upload it.
Ex.
C:/My Documents/User Files/WebApps/myWebApps/test.txt
currently I am using the following codes:
CODE 1:
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Field = test.txt
which is not complete
CODE 2:
     File tempFileRef = new File(fi.getName());
     System.out.println("Field = "+tempFileRef.getAbsolutePath());which also gives this wrong output:
Field = C:\Tomcat 5.5.23\bin\test.txt
Can someone please help me?
Thanks so much.
God bless.

Thank you for your reply.
I am using Apache Commons File Upload. Here is the code where I got the "fi".
     DiskFileUpload fu = new DiskFileUpload();
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          FileItem fi = (FileItem)itr.next();
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How to get the source file name and transform into database

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