How to get the absolute path of a DTD referenced in an XML?

Hello!
My OS is Windows.
I use "org.apache.xerces.parsers.SAXParser" which implements LexicalHandler to parse XML files. When method "startDTD" starts and "systemId" is got, how can I get the absolute path? Is there any simple way?
Regards!

Hi,
http://forum.java.sun.com/thread.jsp?forum=34&thread=36
612The methods in the above topic seem to be available when use a DOM parser..

Similar Messages

How to get the absolute path of logicalhost server domain on Windows Sun

i am reading a file from Xsql Folder, that is located in the logicalhost Sun\AppServer\domains\domain1\applications\j2ee-apps.(IN Sun Application Server)
I am pretty sure that using the absolute path will solve this issue, so my first question is: How to get the absolute path of logicalhost server domain on Windows?
i tried with System.getProperty("com.sun.aas.instanceRoot").
but i am able to retrive Sun\AppServer\domains\domain1 upto this .i am unable to retrive Sun\AppServer\domains\domain1\applications\j2ee-apps.
please suggest me how u can get absolute path in sun application server

Take a look here

How to get the absolute path of logicalhost server domain on Windows?

My logicalhost server domain behaves strangely. I am reading a file from collaboration definiton, that is located in the logicalhost/is/domains/domain1/config folder. I thought, that this folder is used as an application root folder so I can read files like ./file from there. And it worked.
But then I've installed the domain as a Windows service, restarted the PC. When the domain1 gets started, I get exceptions saying that the file can't be found. Then I restart the domain (in domainmgr.bat) and it works again.
I am pretty sure that using the absolute path will solve this issue, so my first question is: How to get the absolute path of logicalhost server domain on Windows?
And my second question is: Why does this happen?

The default folder for a Windows Service is the system32 folder contrary to the instance root folder when starting it as a normal process.
That's the why, haven't tried the how, but you should be able to get the value by calling System.getProperty("com.sun.aas.instanceRoot").
Hope this helps
Paul

How to get the absolute path of a file from the local disk given the file n

how to get the absolute path of a file from the local disk given the file name

// will look for the file at the current working directory
// this is wherever you start the Java application (cound be C: and your
// application is located in C:/myapp, but the working dir is C:/
File file = new File("README.txt");
if (file != null && file.exists())
String absolutePath = file.getAbsolutePath();

Getting the absolute path of the current executing file

I have a file which will be placed in window and linux environment. It is not good to change the source code in that way:
String path = "D:\\java\file1.txt"; (Window)
String path = "/java/file1.txt"; (Linux)
So I would like to ask how to get the absolute path based on that executing file?
I referred to the reference in jsp, but I don't know what class and the coding syntax in console environment.
Thx for any help.

If you are looking to "get" a file that is located in a directory with (or somewhere under) the class file than use
myClass.getClass().getResource(<relativePathWithFileFromClass>); //URL
myClass.getClass().getResourceAsStream(<relativePathWithFileFromClass>); //InputStreamAs far as determining which Operating System you are on, there are a number of environment variables that will tell you that, and you can then format your file path accordingly.
You can use "/" in your path regardless of which OS you are using. You do not have to use "\\" on Windows (excpet maybe inside of a Runtime.exec command string).
Here is a very small program you can compile and run if you wish to see the list of System Properties available:
public class ShowProperties {
public static void main(String[] args) {
System.getProperties().list(System.out);
}Just save that to a file (named ShowProperties.java of course) and compile and run it, and you will get a list of your available System Properties and their current values

How to get the ablolute path of the web application in WebSphere?

How to get the ablolute path of the web application in WebSphere?
For example:
I have installed IBM WebSphere on D:\WebSphere\Appserver, and I created a new appliction named "myapp" on D:\myapp,. How can I get the absolute path of application "myapp"? In other words,how can I get the absolute path of the application's
root directory?

In the WebSphere(default), what directory is the Java Bean's root directory ?

How to get the current path of my application in java ?

how to get the current path of my application in java ?
thanks

To get the path where your application has been installed you have to do the following:
have a class called "what_ever" in the folder.
then you do a litte:
String path=
what_ever.class.getRessource("what_ever.class").toString()
That get you a string like:
file:/C:/Program Files/Cool_program/what_ever.class
Then you process the result a little to remove anything you don't want:
path=path.substring(path.indexOf('/')+1),path.lastIndexOf('/'))
//Might be a little error here but you should find out //quickly if it's the case
And here you go, you have a nice
C:/Program Files/Cool_program
which is the path to your application.
Hooray

How to get the complete path of the file that is selected using FormFile

i m working on struts..
i hv used FormFile like
<html:file property="xsdpath" value="Browse" />
need to get the whole path that i will select using browse button
for example d:\foldername\filename.java
but FormFile Api has a method getFileName(); which returns the filename, for getting the absolute path wat has to be done.
please reply bak soon
thanks in advance

here i use formfile <html:file> just to allow the
user to select a xml file .
so i need to get the whole path of the selectedfile
to parse the xml file.No you dont.
You would definitely benefit from further reading on
file upload.
<html:file> tag renders an HTML <input> element of
type file.
When a user uploads a file, this file is sent as a
stream of data, which a program (jsp/servlet) on the
server, reads and stores the data back in the form
of a file on the server.
Any server program that needs to parse the file,
should do so on the file stored on the server.
There's no point in knowing the absolute path of the
file on the client machine. If a server program can
parse a file on the client machine, why upload the
file in first case ? Get the drift ?
i also want to show my user the path he hadselected.
If you have such a requirement, then yes.
But it sounds weird to me. If you see my response
above, you will realize that the server has a copy of
the client's file uploaded and then parsed. What if
the client has changed his file after upload ?
cheers,
ram.I also have a requirement to get the whole filepath of the file selected and place this information into a table. From FormFile I can only retreive the absolute filename
Any suggestions would be helpful.
Thanks, dam

How to get the full path instead of just the file name, in �FileChooser� ?

In the FileChooserDemo example :
In the statement : log.append("Saving: " + file.getName() + "." + newline);
�file.getName()� returns the �file name�.
My question is : How to get the full path instead of just the file name,
e.g. C:/xdirectory/ydirectory/abc.gif instead of just abc.gif
import java.io.*;
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
import javax.swing.filechooser.*;
public class FileChooserDemo extends JFrame {
static private final String newline = "\n";
public FileChooserDemo() {
super("FileChooserDemo");
//Create the log first, because the action listeners
//need to refer to it.
final JTextArea log = new JTextArea(5,20);
log.setMargin(new Insets(5,5,5,5));
log.setEditable(false);
JScrollPane logScrollPane = new JScrollPane(log);
//Create a file chooser
final JFileChooser fc = new JFileChooser();
//Create the open button
ImageIcon openIcon = new ImageIcon("images/open.gif");
JButton openButton = new JButton("Open a File...", openIcon);
openButton.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent e) {
int returnVal = fc.showOpenDialog(FileChooserDemo.this);
if (returnVal == JFileChooser.APPROVE_OPTION) {
File file = fc.getSelectedFile();
//this is where a real application would open the file.
log.append("Opening: " + file.getName() + "." + newline);
} else {
log.append("Open command cancelled by user." + newline);
//Create the save button
ImageIcon saveIcon = new ImageIcon("images/save.gif");
JButton saveButton = new JButton("Save a File...", saveIcon);
saveButton.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent e) {
int returnVal = fc.showSaveDialog(FileChooserDemo.this);
if (returnVal == JFileChooser.APPROVE_OPTION) {
File file = fc.getSelectedFile();
//this is where a real application would save the file.
log.append("Saving: " + file.getName() + "." + newline);
} else {
log.append("Save command cancelled by user." + newline);
//For layout purposes, put the buttons in a separate panel
JPanel buttonPanel = new JPanel();
buttonPanel.add(openButton);
buttonPanel.add(saveButton);
//Explicitly set the focus sequence.
openButton.setNextFocusableComponent(saveButton);
saveButton.setNextFocusableComponent(openButton);
//Add the buttons and the log to the frame
Container contentPane = getContentPane();
contentPane.add(buttonPanel, BorderLayout.NORTH);
contentPane.add(logScrollPane, BorderLayout.CENTER);
public static void main(String[] args) {
JFrame frame = new FileChooserDemo();
frame.addWindowListener(new WindowAdapter() {
public void windowClosing(WindowEvent e) {
System.exit(0);
frame.pack();
frame.setVisible(true);

simply use file.getPath()
That should do it!Thank you !
It takes care of the problem !!

How to get the document path of the pictures uploaded for products?

Hi Gurus,
How to get the document path of the pictures uploaded for products uploaded through tcode COMMPR01?
Many Thanks,
Neeraj

client path.
I need to get the client path in order to download files form server to client.
Best regards,
Huy.

How to get the project path ?

In my servlet, how do I get the project path ?
I have the following dir structure :
Web_App
+ build
    lib
+ nbproject
+ src
    test
+ web ( index.jsp , my.jsp , my.html )
    + Dir_Docs ( my file : ABC.txt ) My Servlet is :
public class My_Servlet extends HttpServlet
public void init(ServletConfig config) throws ServletException
    super.init(config);
public void destroy() { }
protected void processRequest(HttpServletRequest request,HttpServletResponse response) throws ServletException,IOException
    response.setContentType("text/html");
    PrintWriter out=response.getWriter();
    out.println(System.getProperty("user.dir"));
}What I got was : C:/apache-tomcat-6.0.14/bin/
How to get the project path ? Which is : C:/Web_App/ ?

Yes, config.getServletContext().getRealPath("index.jsp"); got the job done !
Thanks.

How to get the application path?

Does some body know how to get the application path in Java? Here is an example about what I'm thinking: Let's imagin the application is on "c:\try" or "c:\ProgramFiles\MyApp". My question is how can I find the path where the application is. Please give an example if you know the answer. Thank you so much.

Those two replies give you some useful directories, it's true, but maybe not what the OP asked. However the OP asked for something that doesn't have a meaning (applications don't have to be "on" any directory, whatever that might mean).
Would the OP like to describe what the actual problem is here?

To get the absolute path of a file in JSP

Hi,
I want to include a HTML page in a jsp.The path of the jsp should not be hot coded.So that i can get the path from the jsp file i am running and then concat the correspondinghtml file name to the absolute path of the jsp file i am running.What can i do to get the absolute path of the jsp file.

I dont know if there is a way to get the path of the current jsp - but you sure can get the path to your web application using the ServletContext's getRealPath() method and then if you have stored the html (say x.html) in a folder (called, say 'static') immediately under your application root,
String contextPath = application.getRealPath("/"); //path to your application ctxt
String htmlFilePath = contextPath.append("/static/x.html");ram.

How to get the Real Path of a file which is accessed by URL?

iam using tomcat6.0.
I have a file xyz.xml at the top of the webapplication HFUSE which i can able to access by URL
http://localhost:8080/HFUSE/xyz.xml
My problem is how to get the realpath of the file "xyz.xml" for reading and writing purposes.
I tried various things but i could not able to successfully solved the problem?
1) File f = new File("/xyz.xml");
print(f.getAbsolutePath()) ============== it is not fetching the file @ http://localhost:8080/HFUSE/xyz.xml rather it is creating a file
at the root of the drive where eclipse is running.
2) File f = new File("xyz.xml");============> this is also not working , it is creating the file xyz.xml in the eclipse directory ..................
Can anyone please guide on this problem?

RevertInIslam wrote:
If you want your context root(i.e HFUSE)
use this:
request.getContextPath() //where request is HttpServletRequest object to get the needful path.
e.g:
File f = new File(request.getContextPath()+"/xyz.xml");//it will create the file inside HFUSE.
Hope this helps.
Regards
BWrong. The File constructor expects an absolute filesystem path. The HttpServletRequest#getContextPath() doesn't return the absolute filesystem path, it only returns the relative path from the current context root. Use ServletContext#getRealPath() instead, it returns the absolute filesystem path for the given relative path from the current context root.
File file = new File(servletContext.getRealPath("/"), "xyz.xml");

How to get the current path/directory

Hey,
I was wondering how to figure out the current path or directory where my Application is running (stored) in! I know there is a statement like getCodebase() for Applets to get the actual path, but I couldn't find anything similar for regular Applications. I think I can not just use a absolute path, since it might run on different os with different file seperator (like /root or c:\). Any help is welcome!
Thank you very much,
Marc

Try this ...
System.out.println("current directory = " + System.getProperty("user.dir"));
[/code}

How to get the absolute path of a DTD referenced in an XML?

Similar Messages

Maybe you are looking for