Servlet parameters in web.xml file

Hi Guys
What is ctx:dynamo: in web.xml why we have to use this one in servlet parameters
Please give me a clear picture on this
<init-param>
    <param-name>jumpServlet</param-name>
    <param-value>
      ctx:dynamo:/atg/dynamo/servlet/dafpipeline/JumpServlet
    </param-value>
</init-param>

Similar Messages

Init parameters in web.xml file

Hi,
I have a query related to the init parameters defined in web.xml file.
When does a servlet read these parameters?
Also, how does servlet use these parameters while processing the client's request?
Thanks in advance......

hai,
In order to make servlet flexible we must avoid hard-coding(writing-directly).so there are two ways in that one one way is init param's.
create ServletConfig object as show below
ServletConfig config=getServletConfig();then retrieve the init param value in to your servlet us getInitParameter(..) method as show below
String something=config.getInitParameter("init param name");

Can i run a servlet without a web.xml file for servlet mapping?

Hello everyone.
The code i want to run compiles and everythink looks ok.
It produces the .class file.
I run Tomcat 4.1 and i build my Web site with Dreamweaver.
I have a form in a page and i want to send the data upon form completion to a database i already have build with MySql.
The database is up and running and the server is set-up ok.
I have changed the port in Tomcat to run on port 80.
The directory i have my site is
Tomcat41\webapps\ROOT\se
and the directory where i keep the servlet class is
Tomcat41\webapps\ROOT\se\WEB-INF\servlet
I have a web.xml file to map the servlet and placed it in
Tomcat41\webapps\ROOT\se\WEB-INF
In the Form action i write action:"/servlets/Classes/GroupRegistration"
and I RECEIVE AN 404 ERROR FROM APACHE.
Somethink is wrong .
The following is the code from the GroupRegistration.java file
and follws the web.xml file.
Please Help.
import java.sql.*;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class GroupRegistration extends HttpServlet
Connection con;
public void doPost (HttpServletRequest req, HttpServletResponse res)
                         throws ServletException, java.io.IOException
     handleForm(req, res);
public void init() throws ServletException {
     try{
     /* Loading the driver for the database */
     Class.forName("com.mysql.jdbc.Driver");
     Connection Con = DriverManager.getConnection("jdbc:mysql://localhost/se?user=luser&password=");
     catch (ClassNotFoundException e) {
     throw new UnavailableException("Couldn't load JdbcOdbcDriver");
     catch (SQLException e) {
     throw new UnavailableException("Couldn't get db connection");
     private void handleForm(HttpServletRequest req, HttpServletResponse res)
     throws ServletException {
     //ServletOutputStream out = res.OutputStream();
     //res.setContentType("text/html");
     //Extract the form Data Here
     String group = req.getParameter("GroupNo");
     String Name1 = req.getParameter("Name1");
     String LoginID1 = req.getParameter("LoginID1");
     String Name2 = req.getParameter("Name2");
     String LoginID2 = req.getParameter("LoginID2");
     String Name3 = req.getParameter("Name3");
     String LoginID3 = req.getParameter("LoginID3");
     String Name4 = req.getParameter("Name4");
     String LoginID4 = req.getParameter("LoginID4");
     String URL = req.getParameter("URL");
     String Title2 = req.getParameter("Title2");
     String date = req.getParameter("date");
     String INSERT = "INSERT INTO registration (groupno, name1, loginid1, name2, loginid2, name3, loginid3, name4, loginid4, url, topic, date) VALUES (" + group + "," + Name1 + "," + LoginID2 + "," + Name2 + "," + LoginID2 + "," + Name3 + "," + LoginID3 + "," + Name4 + "," + LoginID4 + "," + URL + "," + Title2 + "," + date + ")";
     PreparedStatement pstmt = null;
     try{
     pstmt = con.prepareStatement(INSERT);
     pstmt.executeUpdate();
     catch (SQLException e) {
     throw new ServletException(e);
     finally {
     try {
     if (pstmt != null)pstmt.close();
     catch (SQLException ignored){
The web.xml file
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com.xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd" version="2.4">
<servlet>
<servlet-name>GroupRegistration</servlet-name>
<servlet-class>GroupRegistration</servlet-class>
</servlet>
<servlet-maping>
<servlet-name>GroupRegistration</servlet-name>
<url-pattern>/myGroupRegistration</url-patern>
</servlet-mapping>
</web-app>
I apreciate your time.
Thanks for any help.

and the directory where i keep the servlet class is
Tomcat41\webapps\ROOT\se\WEB-INF\servletOthers have pointed out that "servlet" should be "classes", but there is another mistake that hasn't been spotted. If you want your servlet to appear in the root context, you should use:
Tomcat41\webapps\ROOT\WEB-INF\classes
If you want your servlet to appear under the /se context, then you should use:
Tomcat41\webapps\se\WEB-INF\classes
and also, in the latter case, your form action should be /se/myGroupRegistration.

Do I must put ALL my servlets inside the web.xml file?

Hello everybody,
I am trying to migrate from Webshere to TOMCAT a project with severals packets, one of those have 15 servlets and some classes.
Is there any way to use the servlets without have to declare each one inside the web.xlm file ?
Thanks in advance.

All servlet definitions go into web.xml, though not the actual servlets themselves :) . Say if you wanted servlet A to have a different mapping you would define the mapping in web.xml to something other than the standard one.

Automate creation of web.xml file for tomcat 4.1.29

hi , this is with ref to Tomcat 4.1.29, if i am correct, each Servlet in the application has to be mentioned in the web.xml file for the Tomcat to know about it. Is there any way to automate the creation of web.xml file , depending on the contents of the Servlet folder of the application. Any way to escape from writing each Servlet name in web.xml file.
rc

Hello,
Maybe you should check if you can use Ant tool to do
that. I am not sure if it can help u.
Zeph.
http://ant.apache.org/
It will, specially if used in conjunction with XDoclets :
http://xdoclet.sourceforge.net/
XDoclet has Ant tasks to generate web.xml files.

Help on web.xml file, what if the parameters contains key words ?

Hi:
I am just wondering what should I do if I want to include key words suchs
as <param> in web.xml file for a servlet config.
Example:
<servlet>
<servlet-name>testServlet</servlet-name>
<parameter>
<param-name>some name</param-name>
<param-value>some value</param-value>
</parameter>
</servlet>
What should I do if I want to repleace 'some value' with '</param-value>some
value' and still to prevent the engine to terminate parsing the param-value
at the fake ending? Is there a standard way in XML to distanguish that?
(in URL format it can be replaced %xx for some chars).
ie,
<param-value> </param-value>some value</param-value>
where the second </param-value> is the real ending.
Thank you!
Gang

Hi!
You can use "& lt ;" and "& gt ;" xml entities for that. Or wrap text element in <![CDATA[...]]> section.
Regards,
Ignat.

Where to save web.xml file in apache jserv with servlet 2.0

hi,
we are using apache jserv which included servlet 2.0 we are able to run .class files and
jsp files,but we are using web.xml file for connection pooling which we were used in tomcat
but we are unable to find web-inf folder in apache/jserv to save the file.
please help me out in this problem,
please guide us is there any other method of connection pooling apache jserv,we are using backend
as oracle9i
i will be greatfull to u in this regards
laiq

Hi,
Just create WEB-INF folder and copy the web.xml.
good luck !
...san :-)

How can I get the context-parm from a web.xml file using struts?

Hello:
I need get the context-param from the web.xml file of my web project using struts. I want configurate the jdbc datasource connection pooling here. For example:
<context-param>
<param-name>datasource</param-name>
<param-value>jdbc/formacion</param-value>
<description>Jdbc datasource</description>
</context-param>
and then from any Action class get this parameter.
Similar using a simple server can be:
/** Initiates new XServlet */
public void init(ServletConfig config) throws ServletException {
          for (Enumeration e = config.getInitParameterNames(); e.hasMoreElements();) {
               System.out.println(e.nextElement());
          super.init(config);
          String str = config.getInitParameter("datasource");
          System.out.println(str);
     public void doPost(HttpServletRequest req, HttpServletResponse res)
          throws ServletException, IOException {
          // res.setContentType( );
          System.out.println("Got post request in XServlet");
          PrintWriter out = res.getWriter();
          out.println("nada");
          out.flush();
          out.close();
but only this works for init-params, if I use
<servlet>
     <servlet-name>MyServlet</servlet-name>
     <display-name>MyServlet</display-name>
     <servlet-class>myExamples.servlet.MyServlet</servlet-class>
     <init-param>
     <param-name>datasource</param-name>
     <param-value>jdbc/formacion</param-value>
</init-param>
</servlet>
inside my web.xml. I need something similar, but using struts inside the action class for that I can get the context-params and call my database.
Thank you

To get context parameters from your web.xml file you can simply get the ActionServlet object from an implementing action object class. In the perform (or execute) method make the following call.
ServletContext context = getServlet().getServletContext();
String tempContextVar =
context.getInitParameter("<your context param >");

Web.xml file not being read

Hello, I did a quick search but could not find an answer to my specific problem. (this is also my first night tackling servlets) i hope this hasnt been answered before. here goes...
i'm using Tomcat 4.1 and this is my file structure where i put my servlets:<intstall dir>/webappps/ROOT/WEB-INF/classes/TestPackage.
this is my code for the servlet
package TestPackage;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
/** Example using servlet initialization. Here the message
* to print and the number of times the message should be
* repeated is taken from the init parameters.
public class ShowMessage extends HttpServlet
     private String message;
     private String defaultMessage = "No message.";
     private int repeats = 1;
     public void init(ServletConfig config) throws ServletException
          //Always call super.init
          super.init(config);
          message = config.getInitParameter("message");
          if (message == null)
               message = defaultMessage;
          try
               String repeatString = config.getInitParameter("repeats");
               repeats = Integer.parseInt(repeatString);
          catch(NumberFormatException nfe)
               //do nothing
     public void doGet(HttpServletRequest request, HttpServletResponse response) throws
                                        ServletException, IOException
          response.setContentType("text/html");
          PrintWriter out = response.getWriter();
          String title = "The ShowMessage Servlet";
          out.println("<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.0 " +
                           "Transitional//EN\">\n" +
                           "<HTML>\n" +
                         "<body bgcolor=\"#FDF5E6\">\n" +
                         "<h1 align=center>" + title + "</h1>");
          for(int i=0; i<repeats; i++)
               out.println(message + "<br>");
          out.println("</body></html>");
}my web.xml file is in <intstall dir>/webappps/ROOT/WEB-INF. it looks like:
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
    PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
    "http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<display-name>Welcome to Tomcat</display-name>
<description>
     Welcome to Tomcat
</description>
<servlet>
      <servlet-name>
        ShowMsg
      </servlet-name>
      <servlet-class>
        TestPackage.ShowMessage
      </servlet-class>
      <init-param>
        <param-name>
          message
        </param-name>
        <param-value>
          Shibboleth
        </param-value>
      </init-param>
      <init-param>
        <param-name>
          repeats
        </param-name>
        <param-value>
          5
        </param-value>
      </init-param>
</servlet>
</web-app>when i access the servlet i get "No message" one time which is the default message. Any ideas as to why its not picking up the init parameters. I've restarted my laptop and started and stopped Tomcat a few times and i always get the default message...i also tried ShowMessage in replace of ShowMsg as the <servlet-name>, to no avail. I also tried removing the package name in the <servlet-class>, still nothing. Is there another web.xml file some where else that i need to update?
Thanks! (sorry for the long code)
BB

The doGet method works if you are sending data from a form using get method. It is better to use service method.

How to get Context paramters out of Tomcat web.xml file

The Documentation at Tomcat seems to suggest:
Context initialization parameters that define shared
String constants used within your application, which
can be customized by the system administrator who is
installing your application. The values actually
assigned to these parameters can be retrieved in a
servlet or JSP page by calling:
String value =
getServletContext().getInitParameter("name");
Now I want to write a jsp, where the user enters his username and pwd, and this contacts a jdbc connection bean. I want to store the driver name in the web.xml file. Now can some1 please suggest a method as to how I can extract this (It ain't that straightforward as suggested in tomcat doumntation. You cannnot call the above method ithout Servlet Initialization (actually to say ithout HTTP Connection). I succeeded when I had written a servlet, but can some1 please suggest an alternative (bcoz othrwise just to extract that I am writing a servlet).
Hope I am clear.
Thanks in advance.

Edit your web.xml file and add these entries,
<context-param>
<param-name>dbUrl</param-name>
<param-value>jdbc:oracle:thin:@server5:1521:pl2java</param-value>
</context-param>
<context-param>
<param-name>dbDriver</param-name>
<param-value>oracle.jdbc.driver.OracleDriver</param-value>
</context-param>
Place the above entries in between your already existing <web-app> and </web-app> tags.
Now in your jsp, you can use, <% application.getInitParameter(); %> and in servlet, getServletContext().getInitParameter();
Hope this helps.
Sudha

Console WAR deployment editor tool damage my web.xml file

I put many servlet and servlet-mapping items in web.xml file. When I deployed my
web app into weblogic 7.0 SP1 server, I use the console to open the deployment
description, and change some parameters for the servlet. After I clicked the persist
button, the web.xml file was changed, but some servlet and servlet-mapping items
was removed unexpected.
Why?
When I start the server in Development mode, this error occured rather often.
In the product mode, this error occured a few times.
My operation steps:
1. undeploy the original app from console
2. copy the new war file to the applications directory, overwrite the old one
3. use the console to open the deployment description, make the changes
4. persist the changes
5. re-deploy the web app
Please give the solution as soon as possible

The tag
<target name="deploy" depends="clover-yes, clover-no">...is never closed.
close that tag:
<target name="deploy" depends="clover-yes, clover-no">
     <javac srcdir="${src.dir}" destdir="${classes.dir}" classpath="${libs}" debug="off" optimize="on" deprecation="on" compiler="${compiler}">
          <include name="org/apache/commons/fileupload/**/*.java" />
          <include name="com/portalbook/portlets/**/*.java" />
          <include name="com/portalbook/portlets/content/**/*.java" />
     </javac>
</target>Second error is that the depends in there (clover-yes, clover-no) are not existing as targets in your xml.

How to run java servlet without using Web.xml?

How to run servlet without using Web.xml? From a book, I know that web.xml descriptor is optional, but the book doesn't tell us how to run java servelet without web.xm descriptor. So how to do that? Thanks a lot.

How to run servlet without using Web.xml?But Tomcat now uses a web.xml for its global server-wide configuration.
If you'd like to invoke a servlet with:
http://host/servlet/ServletName
you have to enable the invoker servlet.
[from an HTML]
<FORM METHOD="POST" ACTION="/servlet/HGrepSearchSJ">
[from resin.conf of Resin Web Server 2.1.12]

<servlet-mapping url-pattern='/servlet/*' servlet-name='invoker'/>
[from TOMCAT5.0.19/conf/web.xml, a global server-wide web.xml file]












    <servlet>
        <servlet-name>invoker</servlet-name>
        <servlet-class>
          org.apache.catalina.servlets.InvokerServlet
        </servlet-class>
        <init-param>
            <param-name>debug</param-name>
            <param-value>0</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>
---comment out below----------------------------------------------------------

JDev 10.1.3 compile error in servlet 2.4 web.xml

I built a Servlet 2.4/JSP 2.0 application in JDev 10.1.3. When I try to make/compile/deploy it the compiler gives me errors like the following for every JSP file that has a JSTL 1.1 taglib:
Error(2,9): Element "web-app' used but not declared
Error(2,63): Attribute 'xmlns:xsi' used but not declared
Error(3,103): Attribute 'xsi:schemaLocation' used but not declared
Error(4,16): Attribute 'version' used but not declared
Error(4,56): Attribute 'xmlns' used but not declared
Error(4,57): Can not build schema 'http://java.sun.com/xml/ns/j2ee' located at 'http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd'
Error(5,15): Element 'description used but not declared
It is obviously havng problems with the web.xml file (which is in the servlet 2.4 format) and is probably not finding the web.xml DTD document. How do I fix it?
Installation is on Solaris 9 using J2SDK 1.4.2_04-b05
By the way, this system is on a disconnected network. How do I get JDeveloper updates for a system not connected to the Internet?

Allen,
A JSP with web.xml web-app element
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee">
runs in JDeveloper 10.1.3
thanks,
Deepak

XML Question for web.xml file for LiteWebServer

Hi,
I have no clue of wut tags are required in the web.xml file.
Currently the following code is in my web.xml file and i am trying to run servlets that are located under web-inf/*.class
Pls tell me wuts wrong....
Do i need those mime-mapping tags, session and welcome tags.......... ?
Right now i just want those 2 servlets to work
And wut is this encoding ="UTF-8" in the first line of the xml file. Wut does that do... ? ARe there any others
Any help would b greatly appreciated
Thanks
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app id="WebApp_1">
<display-name>PDMPortal</display-name>
<session-config id="SessionConfig_1">
<session-timeout>60</session-timeout>
</session-config>
<mime-mapping id="MimeMapping_1">
<extension>htc</extension>
<mime-type>text/x-component</mime-type>
</mime-mapping>
<welcome-file-list id="WelcomeFileList_1">
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>HelloWorld</servlet-name>
<servlet-class>HelloWorld</servlet-class>
</servlet>
<servlet>
<servlet-name>HelloWorldExample</servlet-name>
<servlet-class>HelloWorldExample</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern>HelloWorld</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>HelloWorldExample</servlet-name>
<url-pattern>HelloWorldExample</url-pattern>
</servlet-mapping>
</web-app>

Try putting your servlets in the WEB-INF\classes directory.

Problem in web.xml file with weblogic server 8.1

Hi frnds,
I was deployed one Enterprise Application,it deploys successfully. But in server side thows Exeception in web.xml file.
Here the actual Exception
<HTTP> <BEA-101248> <[Application:
'G:\bea\user_projects\domains\mydomain\myserver\upload\jasmine.ear', Module: 'Ja
smine']: Deployment descriptor "web.xml" is malformed. Check against the DTD: or
g.xml.sax.SAXParseException: The content of element type "web-app" must match "(
icon?,display-name?,description?,distributable?,context-param*,filter*,filter-ma
pping*,listener*,servlet*,servlet-mapping*,session-config?,mime-mapping*,welcome
-file-list?,error-page*,taglib*,resource-env-ref*,resource-ref*,security-constra
int*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-local-ref*)". (line 61
, column 11).>
My web.xml file as follws....
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.4//EN"
"http://java.sun.com/dtd/web-app_2_4.dtd">
<web-app>
<display-name>Jasmine Applications</display-name>
<description>
Jasmine Applications
</description>
<servlet>
<servlet-name>LoginServlet</servlet-name>
<servlet-class>examples.LoginServlet</servlet-class>
<init-param>
<param-name>java.naming.factory.initial</param-name>
<param-value>weblogic.jndi.WLInitialContextFactory</param-value>
</init-param>
<init-param>
<param-name>java.naming.provider.url</param-name>
<param-value>t3://localhost:7001</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>ShowQuoteServlet</servlet-name>
<servlet-class>examples.ShowQuoteServlet</servlet-class>
<init-param>
<param-name>java.naming.factory.initial</param-name>
<param-value>weblogic.jndi.WLInitialContextFactory</param-value>
</init-param>
<init-param>
<param-name>java.naming.provider.url</param-name>
<param-value>t3://localhost:7001</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>CatalogServlet</servlet-name>
<servlet-class>examples.CatalogServlet</servlet-class>
<init-param>
<param-name>java.naming.factory.initial</param-name>
<param-value>weblogic.jndi.WLInitialContextFactory</param-value>
</init-param>
<init-param>
<param-name>java.naming.provider.url</param-name>
<param-value>t3://localhost:7001</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>LoginServlet</servlet-name>
<url-pattern>/login/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>ShowQuoteServlet</servlet-name>
<url-pattern>/showQuote/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>CatalogServlet</servlet-name>
<url-pattern>/catalog/*</url-pattern>
</servlet-mapping>
<security-constraint>
<web-resource-collection>
<web-resource-name>My secure resources</web-resource-name>
<description>Resources to be placed under security control.</description>
<url-pattern>/private/*</url-pattern>
<url-pattern>/registered/*</url-pattern>
</web-resource-collection>
<auth-constraint>
<role-name>guest</role-name>
</auth-constraint>
</security-constraint>
<login-config>
<auth-method>FORM</auth-method>
<realm-name>WebApp</realm-name>
<form-login-config>
<form-login-page>/login.jsp</form-login-page>
<form-error-page>/error.jsp</form-error-page>
</form-login-config>
</login-config>

<security-role>
<description>The role allowed to access our content</description>
<role-name>guest</role-name>
</security-role>
</web-app>
pls give me a good solution this exception.. I tried lot..
Thanks in Advance
Regards
Priya

Your DOCTYPE references 2.4, it should be 2.3. WLS 8.1 supports J2EE 1.3 which was servlet 2.3.
Servlet 2.4 is part of J2EE 1.4 and is supported by WLS 9.0/9.1. Also it uses XML Schema not a DTD.
-- Rob
WLS Blog http://dev2dev.bea.com/blog/rwoollen/

Servlet parameters in web.xml file

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