How to customize the XML file name from user entered field value when submitting

How can I customize the name of an HTTP submitted XML file from data entered by the user in a text field?
The field in question contains user identification information and would ease the sorting of files received for easy identification.
Thank you,

Similar Messages

How to load the .xml file list from 'C:\Drag_list\Modified' into xmltype tb

Hi all experts,
I am in Oracle Enterprise Manager 11g 11.2.0.1.0.
SQL*Plus: Release 11.2.0.1.0 Production on Tue Feb 22 11:40:23 2011
is there anyone know why
SQL> create or replace directory XMLDIR as '/xdb/faq/testdata'
2 /
SQL> set long 10000 pages 200 lines 150
SQL> --
SQL> select xmltype(bfilename('XMLDIR','2003.xml'),nls_charset_id('AL32UTF8'))
2 from dual
worked. but
SQL> create or replace directory XMLDIR as 'C:\Drag_list\Modified';
SQL> select xmltype(bfilename('XMLDIR','2011.xml'),nls_charset_id('AL32UTF8'))
from dual
did not work?
IS there any way I can load the .xml file list from 'C:\Drag_list\Modified' into xmltype table?
Thanks.

did not work?Generally we'll need a bit more info than "it didn't work", was there an ora-X error message? What was the error?
Assuming you're on *nix a 'C:\<folder name>\...' directory spec just plain won't work, the directory has to point to a valid storage location on the database server host.
Try a host command (at the database server) and make sure the .xml file name and location is valid, in sqlplus a "bang" (exclamation) runs a host command, i.e.:
SQL> !ls -l /xdb/faq/testdata
... usr grp ... 2003.xml
...But if you are on windows, its the `host` command:
SQL> host dir c:\drag_list\modified
mm/dd/yyyy ... 2011.xml
... The create directory ... as ... must point to a valid storage location for it to work, at least that is step one.

How to load the .xml file list from 'C:\Drag_list\Modif into xmltype table?

Hi all experts,
I am in Oracle Enterprise Manager 11g 11.2.0.1.0.
SQL*Plus: Release 11.2.0.1.0 Production on Tue Feb 22 11:40:23 2011
is there anyone know why
SQL> create or replace directory XMLDIR as '/xdb/faq/testdata'
2 /
SQL> set long 10000 pages 200 lines 150
SQL> --
SQL> select xmltype(bfilename('XMLDIR','2003.xml'),nls_charset_id('AL32UTF8'))
2 from dual
worked. but
SQL> create or replace directory XMLDIR as 'C:\Drag_list\Modified';
SQL> select xmltype(bfilename('XMLDIR','2011.xml'),nls_charset_id('AL32UTF8'))
from dual
did not work?
IS there any way I can load the .xml file list from 'C:\Drag_list\Modified' into xmltype table?
Thanks.
Edited by: Cow on Apr 13, 2011 12:58 AM

This is a question better suited for the sql and pl/sql support forum, since it is NOT an APEX based question..: PL/SQL
Thank you,
Tony Miller
Webster, TX
On the road of life...There are 'windshields', and there are 'bugs'
(splat!)
"Squeegees Wanted"
If this question is answered, please mark the thread as closed and assign points where earned..

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;
     }

Mail to File - how to read the attachment file name from the subject.

I need to use the SHeaderSUBJECT's value in the receiver file adapter's variable substitution.
This is a Mail to File scenario without design part where the attachment file name comes in the subject of the mail.
I see the below in the dynamicconfiguration section. How can i retrive the value from dynamicconfiguration section to the filename.
<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/Mail" name="SHeaderSUBJECT">PlainAttachment.txt</SAP:Record>
</SAP:DynamicConfiguration>
Points will be rewarded.

Try to use sthg like this in a UDF :
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().getStreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create(http://sap.com/xi/XI/System/Mail,SHeaderSUBJECT);
String value = conf.get(key);
or in a JAVA mapping :
DynamicConfiguration dynConf = (DynamicConfiguration) param.get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey dynKey = DynamicConfigurationKey.create((http://sap.com/xi/XI/System/Mail,SHeaderSUBJECT);
String keyValue = dynConf.get(dynKey);
param is the map object from the execute() method of your mapping ...
Hope this helps
Chris
Edited by: Christophe PFERTZEL on Apr 23, 2008 11:34 AM

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 XML TAG name itself instead of TAG value

Hi All,
I have a question here
I want to retrieve the XML tag from a XML file instead of its value.
Example:
<item>Colgate</item>
Now I want to retrieve "item" as output from XPath expression, I dont want its value as "colgate"
How to do that...?
Thanks
-Praveen

You can do this with an axes XPatch expression:
child::node()/name()
For more info see: http://www.w3schools.com/xpath/xpath_axes.asp
HTH,
Bas

Look-up name value from XML file based on user entered number value

My form has an "Employee Number" numeric field and an "Employee Name" textfield. I want the user to enter their employee number and have JavaScript that will "look-up" that employee number in the XML data file and return the corresponding Employee Name into the employee name textfield. I need an example XML data file and form with the appropriate script.
Can anyone help me with this?

Hi ,
Re-read your post, find that the col2 in List1 is people and group type column, SharePoint Lookup type column showing additional fileds option does not support people and group type column, so there should be no OOB method to achieve your
goal.
http://techtrainingnotes.blogspot.com/2012/01/sharepoint-not-all-column-types-can-be.html
I think you can try using InfoPath form to manipulate the correspond col2 value when lookup col1 from List1 then show in list view, or may use custom visual web part to achieve you goal.
http://sharepointsolutions.com/sharepoint-help/blog/2011/11/get-infopath-to-display-lookup-column-value-not-id/
Thanks
Daniel Yang
TechNet Community Support

How to get the XML file if we are using the Product short name.

Hi,
Till now I have used Short name of the Concurrent Program for Code while creating a Data Definition. Now saw a seeded template which has given the Code by Product short name. If we have the concurrent program then it is easy to refer the fields by checking the XML file. In this case how to find the XML file or how to refer all the fields if we have given code with Product short name. I saw this for iReceivables(ARI). Anybody please help me.
Thanks.

Hi Siva
Just to clarify, rather than the short name of the conc program there is a shipped data definition that just uses the product short name? What is the data def so I can check it.
Regards, Tim

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());
}

How to call the XML file as PrivateResourcePath in JSP

Hi All,
I want to call the layer-config.xml values in JSP dropdown list. How to call the XML file as PrivateResourcePath in JSP... please advise me if any solution.
locaiton: dist/PORTAL-INF/layer-config.xml
<?xml version="1.0" encoding="utf-8" ?>
- <layer-config>
- <system>
<layer>Test</layer>
<internal>true</internal>
</system>
- <system>
<layer>Test1</layer>
<internal>false</internal>
</system>
- <system>
<layer>Test3</layer>
<internal>false</internal>
</system>
</layer-config>
Thanks in advance.

Hi,
The below link may help you.
[Accessing image from dist/imgaes folder inside JSPDynpage Component;
Regards,
Suresh Bachimanchi

How to find PG.xml file name and path associated with a FUNCTION

Hi,
I am having a function:IRC_VIS_HOME_PAGE with Web HTML value as below:
OA.jsp?akRegionCode=IRC_VIS_HOME_PAGE&akRegionApplicationId=800&OAPB=IRC_BRAND
How to find PG.xml file name and path assoicated with above funtion.
Thanks,
ashok

Ashok,
Function IRC_VIS_HOME_PAGE will have 2 parameter defined for it which are OASF and OAHP where
OASF=<SelectedFunctionName> - this tells the Framework to select this function in the given "Home Page" menu context.
OAHP=<HomePageMenuName> - this is used ONLY with the OASF parameter, and it is used to establish the current menu context. It should point to a "Home Page" menu.
--Shiv

How to get list of file names from a directory?

How to get list of file names from a directory?
Please help

In addition, this:
String filename = files;Should be this:
String filename = files;
That's just because he didn't use the "code" tags, so [ i ] made everything following it become italicized.

How to call the XML file as Private

Hi All,
I want to call the layer-config.xml values in JSP dropdown list. How to call the XML file as PrivateResourcePath in JSP... please advise me if any solution.
locaiton: dist/PORTAL-INF/layer-config.xml
<?xml version="1.0" encoding="utf-8" ?>
- <layer-config>
- <system>
<layer>Test</layer>
<internal>true</internal>
</system>
- <system>
<layer>Test1</layer>
<internal>false</internal>
</system>
- <system>
<layer>Test3</layer>
<internal>false</internal>
</system>
</layer-config>
Thanks in advance.

Hi,
The below link may help you.
[Accessing image from dist/imgaes folder inside JSPDynpage Component;
Regards,
Suresh Bachimanchi

How to get the Portal Page name from PLSQL?

Can anyone tell me how to get the portal page name from my dynamic page using plsql?
Apparently you can get the page id and work it out from there, but my calls to get the page id are not returning any values anyway.
My code for attempting to get the page id is below.
<oracle>
declare
v_pageid varchar2(30);
begin
v_pageid := wwpro_api_parameters.get_value('_pageid', '/pls/portal30');
htp.print('Page is '|| v_pageid);
end;
</oracle>
Ideally I'd actually just like to get the page name. Is there a straightforward way to do this?
Thanks in advance!
Sarah

Few clarifications -
1. wwpro_api_parameters cannot be used to get default portal
page parameters such as '_pageid', '_dad', '_schema' etc.,
2. Page information can be obtained through any components which
are available in that particular page. For example, in case of
dynamic page, we need to publish it as a portlet and add it to the
page. This process creates necessary packages in the DB, but we
will not have access to the portlet methods.
So, I would prefer creating a simple DB provider & portlet and access
page title from its show method as follows -
//Declare local variable l_page_id, l_page_title as varchar2
select page_id into l_page_id from wwpob_portlet_instance$ where
portlet_id = p_portlet_record.portlet_id and
provider_id = p_portlet_record.provider_id;
select name into l_page_title from wwpob_page$ where id=l_page_id;
More information on DB provider can be found at
http://portalstudio.oracle.com/pls/ops/docs/FOLDER/COMMUNITY/PDK/articles/understanding.database.providers.html
Secondly, usage of wwpro_api_parameters.get_value method is
incorrect. This method expects two arguments -
<ul>
<li><b>p_name : </b> The name of the parameter to be returned.</li>
<li><b>p_reference_path : </b> An unique identifier for a portlet instance on the current page.</li>
</ul>
p_reference_path would be something like 99_SNOOP_PORTLET_76535103 and not some type of path as its name suggests.
The following code fragment fetches all parameters available
for a portlet.
Note : Copy this code into 'show' method of your portlet.
//Declare l_names, l_values as owa.vc_arr
* Retreive all of the names of parameters for this portlet
l_names := wwpro_api_parameters.get_names(
p_reference_path=>p_portlet_record.reference_path);
* Retreive all of the values of parameters for this portlet
l_values := wwpro_api_parameters.get_values(p_names=>l_names,
p_reference_path=>p_portlet_record.reference_path);
//Loop through these arrays to get parameter information
htp.p('<center><table BORDER COLS=2 WIDTH="90%" >');
htp.p('<tr ALIGN=LEFT VALIGN=TOP>');
htp.tableData(wwui_api_portlet.portlet_heading('Name',1));
htp.tableData(wwui_api_portlet.portlet_heading('Value',1));
htp.tableRowClose;
if l_names.count = 0 then
htp.p('<tr ALIGN=LEFT VALIGN=TOP>');
htp.p('<td COLSPAN="2">'
||wwui_api_portlet.portlet_text(
'No portlet parameters were passed on the URL.',1)
||'</td>');
htp.tableRowClose;
else
for i in 1..l_names.count loop
htp.p('<tr ALIGN=LEFT VALIGN=TOP>');
htp.tableData(l_names(i));
htp.tableData(l_values(i));
htp.tableRowClose;
end loop;
end if;
htp.p('</table></center>');
Hope it helps...
-aMJAD.

How to customize the XML file name from user entered field value when submitting

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