Have a problem in getting the servlet path

hi
i wanna get the complete path to my servlet folder in TomcatServer
like http://localhost:8084/Test can anyone tell me how can i get it plz.

try ServletContext.getRealPath
Returns a String containing the real path for a given virtual path. For example, the path "/index.html" returns the absolute file path on the server's filesystem would be served by a request for "http://host/contextPath/index.html", where contextPath is the context path of this ServletContext.
The real path returned will be in a form appropriate to the computer and operating system on which the servlet container is running, including the proper path separators. This method returns null if the servlet container cannot translate the virtual path to a real path for any reason (such as when the content is being made available from a .war archive).

Similar Messages

Problem for getting the real path using one servlet and one jsp page

I have one tomcat machine and several virtual domains. Eahc virtual domain has one realpath in the disc.
I am using one servlet and one jsp page for using this servlet.
my purpose is to load, using the servlet , the real path for the domains (eahc domain has its path).
for this i make this:
the servlet code is this:
package utils.ticker;
import java.io.*;
import java.io.File;
import java.io.IOException;
import java.net.*;
import java.util.*;
import java.text.*;
import javax.servlet.*;
import javax.servlet.http.*;
import javax.servlet.http.HttpServlet;
import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.FilterConfig;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
public class Edicion extends HttpServlet{
public Edicion() {
public String cc(){
ServletContext myContext= getServletContext();
String abspath = myContext.getRealPath("/");
//here i want to return the real path
return abspath;
public static void main(String args[]){
and the jsp page is:
<%@ page import="utils.ticker.*" %>
<jsp:useBean id="tick" class="utils.ticker.Edicion" scope="session"/>
<html><head><body>
<%
tick.cc();
%>
</body></head></html>
But this produces one error, NullPointerException and dont shows me the real path.
Can anyone help me?
thanks

i have put this into one sevlet:
package utils.ticker;
import javax.servlet.ServletContext;
public class Edicion{
private ServletContext myContext;
public Edicion(ServletContext myContext) {
this.myContext = myContext;
public String getCC(){
return myContext.getRealPath("/");
and in the jsp page this:
<%@ page contentType="text/html; charset=iso-8859-1" language="java" import="java.sql.*"%>
<%@ page import="utils.ticker.*" %>
<jsp:useBean id="tick" class="utils.ticker.Edicion" scope="session"/>
<html><head><body>
<%=tick.getCC()%>
</body></head></html>
but appear this error in the tomcat.
Can you, please, help me. i am trying to solve this during one week and i am desesperate.
Thanks.
ERROR:
HTTP Status 500 -
type Exception report
message
description The server encountered an internal error () that prevented it from fulfilling this request.
exception
org.apache.jasper.JasperException: class utils.ticker.Edicion : java.lang.InstantiationException: utils.ticker.Edicion
     at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:248)
     at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:295)
     at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:241)
     at javax.servlet.http.HttpServlet.service(HttpServlet.java:853)
     at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:247)
     at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:193)
     at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:260)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardContext.invoke(StandardContext.java:2415)
     at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:180)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.valves.ErrorDispatcherValve.invoke(ErrorDispatcherValve.java:170)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:641)
     at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:172)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:641)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:174)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.coyote.tomcat4.CoyoteAdapter.service(CoyoteAdapter.java:223)
     at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:432)
     at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.processConnection(Http11Protocol.java:386)
     at org.apache.tomcat.util.net.TcpWorkerThread.runIt(PoolTcpEndpoint.java:534)
     at org.apache.tomcat.util.threads.ThreadPool$ControlRunnable.run(ThreadPool.java:530)
     at java.lang.Thread.run(Thread.java:536)
root cause
javax.servlet.ServletException: class utils.ticker.Edicion : java.lang.InstantiationException: utils.ticker.Edicion
     at org.apache.jasper.runtime.PageContextImpl.handlePageException(PageContextImpl.java:533)
     at org.apache.jsp.pruebas_jsp._jspService(pruebas_jsp.java:72)
     at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:137)
     at javax.servlet.http.HttpServlet.service(HttpServlet.java:853)
     at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:204)
     at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:295)
     at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:241)
     at javax.servlet.http.HttpServlet.service(HttpServlet.java:853)
     at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:247)
     at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:193)
     at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:260)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardContext.invoke(StandardContext.java:2415)
     at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:180)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.valves.ErrorDispatcherValve.invoke(ErrorDispatcherValve.java:170)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:641)
     at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:172)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:641)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:174)
     at org.apache.catalina.core.StandardPipeline$StandardPipelineValveContext.invokeNext(StandardPipeline.java:643)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:480)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:995)
     at org.apache.coyote.tomcat4.CoyoteAdapter.service(CoyoteAdapter.java:223)
     at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:432)
     at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.processConnection(Http11Protocol.java:386)
     at org.apache.tomcat.util.net.TcpWorkerThread.runIt(PoolTcpEndpoint.java:534)
     at org.apache.tomcat.util.threads.ThreadPool$ControlRunnable.run(ThreadPool.java:530)
     at java.lang.Thread.run(Thread.java:536)
Apache Tomcat/4.1.18

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HT201412 I have a problem connecting to the server (SSL problem) on my new Apple ipad. I was supplied with a new ID password, but I am unable to get into my settings and email. Could someone please offer a suggestion? Thanks! A.A.

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How to get the pull path name from a file upload window

Hello everyone!
I have encountered the following problem with the following JSP code:
<form method="post" action="filename.jsp">
Upload JAVA program:
<input type=file size=20 name="fname" accept="java">
<input type=submit value="go">
</form>
<%
String s = "";
if (request.getParameter("fname") != null)
s = request.getParameter("fname")
%>
The value of s is alway the filename. However I want to get the full path in addition to the filename, so that I can read the file. Does anyone know how to get the pull name of the file?
thanks a lot in advance,

Dear Sir,
thanks a lot for your reply. Please let me explain what I intended to do: I want to upload a file from the local machine and then read the content of the file. Therefore I need to the fullpath of the filename like /var/local/file.java instead of file.java. The latter is what I got.
The problem I have with your code is that the function like "request.getServerScheme()" is not recognized. Maybe is it because I didn't install servelet package? I only installed javax package btw. Also my application runns on Tomcat server if this could give you some information. The error message I had is as follows:
org.apache.jasper.JasperException: Unable to compile class for JSP
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
[javac] Compiling 1 source file
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:133: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
url = request.getServerScheme()
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:136: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
+ ((("http".equals(request.getServerScheme()) && request.getServerPort() != 80)
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:137: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
||("https".equals(request.getServerScheme()) && request.getServerPort() != 443))
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:139: cannot resolve symbol
symbol : method getServletConfig ()
location: interface javax.servlet.http.HttpServletRequest
+ "/" + request.getServletConfig().getServletName()
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:140: cannot resolve symbol
symbol : variable path
location: class org.apache.jsp.addExercise_jsp
+ "/" + path
^
5 errors
     org.apache.jasper.compiler.DefaultErrorHandler.javacError(DefaultErrorHandler.java:128)
     org.apache.jasper.compiler.ErrorDispatcher.javacError(ErrorDispatcher.java:351)
     org.apache.jasper.compiler.Compiler.generateClass(Compiler.java:413)
     org.apache.jasper.compiler.Compiler.compile(Compiler.java:453)
     org.apache.jasper.compiler.Compiler.compile(Compiler.java:437)
     org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:555)
     org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:291)
     org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:301)
     org.apache.jasper.servlet.JspServlet.service(JspServlet.java:248)
     javax.servlet.http.HttpServlet.service(HttpServlet.java:856)

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

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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 file path from HTML input form in Firefox3

In IE7 (and probably all famous browsers, including old Firefox 2), if we submit a file like 'C:\folder1\folder2\folder3\filename' it works properly and gives the full path to the file and the filename.
In Firefox 3, it returns only 'filename', because of their new 'security feature' to truncate the path, as explained in Firefox bug tracking system (https://bugzilla.mozilla.org/show_bug.cgi?id=143220)
I have no clue how to overcome this 'new feature'.
Can anyone help to find a single solution to get the file path both on Firefox 3 and IE7?
Thanks in advance....

Doubleposted and ignored any previous answer: [http://forums.sun.com/thread.jspa?threadID=5342405]. Don't do that. It's rude.
Regarding to your "problem": why do you need the file path?

How to get the full path or name of the Layer to which Filter is applied?

Hi All,
I have created a layer using some background image as source. How to get the name or ful path of the background image?
I checked the documentation and found GetPathNameProc(SPPlatformFileSpecification*) function, but I couldn't get the parameter SPPlatformFileSpecification*.
Is there any other way to get the file path of background layer in my custom filter?
Thanks,
Dheeraj

If this is a client side application then I'd say look into drag-n-drop tutorials.
i.e. drag file to your application, action listener fires, create File object giving you full name and path of users action.
If this is a web application then look into the "multipart/form-data" content type specifications on how to upload files.
i.e. user specifies file from <input type='file' ... /> type, submits, servlet receives data and recreates file locally on application server side.
If you are thinking that all you need to send a file to a program is the full path and name in a textbox its a little bit more complicated then that.
Good luck, hope that helps!

How to get the real path of the xml file

I have a java application
following is the package structure
com>>gts>>xml
having file---------> MyXML.xml
com>>gts>>java
having java program to read the file
Problem is if I use File file = new File("..\\xml\\MyXML.xml");
java.io.FileNotFoundException: E:\LEARNING_WORK_SPACE\JavaXml\..\xml\MyXml.xml (The system cannot find the path specified)
     at java.io.FileInputStream.open(Native Method)
     at java.io.FileInputStream.<init>(Unknown Source)
     at java.io.FileInputStream.<init>(Unknown Source)
     at sun.net.www.protocol.file.FileURLConnection.connect(Unknown Source)
How do I get the real path of the xml file.
Edited by: shashiwagh on Jan 29, 2010 11:46 AM

Hi,
if your XML file is inside a package you can easily get it from the classloader.
Note that your application maybe packaged inside a jar so it is not safe to use java.io.File for this purpose.
You have an xml file in :
com/gts/xml/MyXML.xml
in a class of the same module (that will packaged in the same jar) for example com.gts.java.XmlLoader :
// To get the stream :
InputStream is = this.getClass().getResourceAsStream("/com/gts/xml/MyXML.xml");
// Or the URL :
URL xml = this.getClass().getResource("/com/gts/xml/MyXML.xml");Hope it helps.

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?

Getting the current path from which the form is runnig in 9i OAS

Hi,
We are using the built in Get_Application_property to get the
current path in which the form is running. However this does not
work in 9ioas and it only returns the form name and not the full
path. Does anyone have any solutions to this problem?
Thanks in Advance
Jeet

Place it in the classpath, then you can use ClassLoader#getResource().

How can I get the servlet Context from a WebService Implementation?

I have made a webservice endpoint, using the conventional way (WSDL->wsimport->Java interface->implementation) . I need to get the servlet context below the implementation class. I haven't found any way to get the servletContext though. Any clues? Any help will be greatly appreciated.

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best wishes thank you very much

How to get the file path in adf application

hii all,
i have a txt file that i am using in my adf application,
i am passing this txt file through a File Reader, for which i have to mention the file path.
The file is in web-content and when i am hard coding the complete file path i.e C:/JDeveloper/myApp/ViewController/public_html/log.txt
the application is working fine when run on integrated weblogic server.
My requirement is to access this file without giving the static file path, as in case i have to use this application on any other machine..
for that how to mention the file path-
i tried using FacesContext to get the context path :-
FacesContext.getCurrentInstance().getExternalContext().getRequestContextPath();
which gives me
\myApp-ViewController-context-root
after appending public_html\log.txt
I am using the following path to access the file :-
\myApp-ViewController-context-root\public_html\log.txt
again i am getting the java.io.FileNotFoundException
Does anyone know how to use file from inside the web-content without giving the complete path..???
Thanks

Hi,
If you put your file under public_html folder, you can use this code to access the file:
For example file is : log.txt
FacesContext.getCurrentInstance().getExternalContext().getRealPath('/log.txt').toString().trim();
Thanks.
- LSR

Have a problem in getting the servlet path

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