Using Servlet Parameters defined in web.xml

Hi all,
would like to use some of the properties of the UIX-Servlet within my UIX application (they may be configured in web.xml).
But how to get to the Servlet? Typically, one just gets a BajaContext which does not provide access to the servlet itself....
any ideas?
Thanx, Ernst.

Try these:
Servlet servlet = bajaContext.getService(Servlet.class, null);
ServletConfig config = bajaContext.getService(ServletConfig.class, null);

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Try these:
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Hi,
I have a query related to the init parameters defined in web.xml file.
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hai,
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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:
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<param-name>datasource</param-name>
<param-value>jdbc/formacion</param-value>
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</context-param>
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Similar using a simple server can be:
/** Initiates new XServlet */
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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 >");

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Can someone please tell me how the <servlet> entry differs from
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Yes this is the source of my confusiong. It seems to be that WL can be told
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          >
          > --
          >
          > Cameron Purdy
          > http://www.tangosol.com
          >
          >
          > "Robert Nicholson" <[email protected]> wrote in message
          > news:[email protected]...
          > > If what you say is true then why is there the need to also tell WL about
          > the
          > > servlets themselves?
          > >
          > > Presumably this is for when the servlets exist outside of a WAR then?
          > >
          > > "Cameron Purdy" <[email protected]> wrote in message
          > > news:[email protected]...
          > > > The weblogic.properties tells WL to load a WAR or specific servlet.
          The
          > > > <servlet> entry (I assume you refer to
          > > > http://java.sun.com/j2ee/dtds/web-app_2_2.dtd) defines the presence of
          a
          > > > servlet within a WAR (a deployment unit):
          > > >
          > > > >
          > > > The servlet element contains the declarative data of a
          > > > servlet. If a jsp-file is specified and the load-on-startup element is
          > > > present, then the JSP should be precompiled and loaded.
          > > > [end-quote]
          > > >
          > > > In the case of a WAR deployment, you have to tell WL that the WAR
          > exists,
          > > > then it will look in the WAR and figure out what servlets are there
          from
          > > the
          > > > <servlet> tags in your web.xml. You set up a WAR in the
          > > weblogic.properties
          > > > as such:
          > > >
          > > > weblogic.httpd.webApp.<http-path>=<os-path>
          > > >
          > > > For example:
          > > > weblogic.httpd.webApp.website=c:/weblogic/website.war
          > > >
          > > > --
          > > >
          > > > Cameron Purdy
          > > > http://www.tangosol.com
          > > >
          > > >
          > > > "Robert Nicholson" <[email protected]> wrote in message
          > > > news:[email protected]...
          > > > > Can someone please tell me how the <servlet> entry differs from
          > > > >
          > > > > weblogic.httpd.register.snoop=examples.servlets.SnoopServlet
          > > > >
          > > > > in weblogic properties?
          > > > >
          > > > > Do these perform identical functions?
          > > > >
          > > > >
          > > >
          > > >
          > >
          > >
          >
          >

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.

Web-inf\web.xml init parameters

Hi,
I use several servlets/JSPs which are using common init parameter, is there a way to specify that some init parameters are common or must I write all the parameters for all of my servlets/JPSs ?
Thanks

yah that's what I mean, have your parameters in your web.xml file and have a Servlet read them when the application starts up (using ServletConfig). Then you put the values into the application scope where all the JSPs and Servlets can read them. So lets say you specify a parameter called "html-path" with a value of "/www/html/" you could read this in your JSPs like this:
String htmlPath = application.getAttribute("html-path");I can't remember how to do his off-hand but I know it works.

How to use dynamic file as welcome-file-list in web.xml

I have configured my web.xml file as this,
<web-app>
<context-param>
<param-name>javax.faces.DEFAULT_SUFFIX</param-name>
<param-value>.xhtml</param-value>
</context-param>
<servlet>
<servlet-name>Faces Servlet</servlet-name>
<servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>*.jsf</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>login.jsf</welcome-file>
</welcome-file-list>
</web-app>
and my login.xhtml file as this,
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:ui="http://java.sun.com/jsf/facelets"
xmlns:h="http://java.sun.com/jsf/html"
xmlns:c="http://java.sun.com/jstl/core"
xmlns:f="http://java.sun.com/jsf/core">
<body>
<f:view>
<h:form>
UserName:<h:inputText id="userName"
value="#{bean.userName}" rendered="true"
required="true"/>
Password:<h:inputText id="password"
value="#{bean.password}" rendered="true"
required="true"/>
<h:commandButton id="submit" value="Submit" action="#{bean.authenticate}" />
</h:form>
</f:view>
</body>
</html>
but when i deploy this using tomcat and try to put url as this,
http://localhost:8080/project
Its not identifying welcome file from web.xml
I am getting error like this,
The requested resource (/project/) is not available.
How to resolve this,
Thanks,
Vinutha

This might help:
http://forum.java.sun.com/thread.jspa?threadID=696586
As well, you might have to change the servlet-mapping in your web.xml. The url-pattern, I think, needs to be .xhtml. Your login.jsf file in the welcome list will need to be renamed to login.xhtml.
CowKing

How to retrieve information from XML using servlets

Hi
I have a scenario like this
File-->XI-->J2ee application(HTTP Receiver adapter)
I want to know how XI sends xml file as a query string file name?
What ever may be the case i just need to retrieve the information from Request object and display the same in browser(JSP).How to do that, I am totally confused, please anyone help me!
Code help is highly rewarded.
Thanx

Hi Rajesh!
I have tried your code. But some where i went wrong. Please correct me: I am using NWDS2.0.9
my web.xml:
<?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>
    <display-name>WEB APP</display-name>
    <description>WEB APP description</description>
    <servlet>
        <servlet-name>DisplayRes</servlet-name>
        <servlet-class>com.quinnox.DisplayRes</servlet-class>
    </servlet>
    <servlet-mapping>
          <servlet-name>DisplayRes</servlet-name>
          <url-pattern>/DisplayRes/*</url-pattern>
     </servlet-mapping>
</web-app>
my application.xml:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE application PUBLIC "-//Sun Microsystems, Inc.//DTD J2EE Application 1.3//EN"
                             "http://java.sun.com/dtd/application_1_3.dtd">
<application>
    <display-name>ReceiverEntpr</display-name>
    <description>EAR description</description>
    <module>
        <web>
            <web-uri>HttpReceiver.war</web-uri>
            <context-root>/HttpReceiver</context-root>
        </web>
    </module>
</application>
and my servlet code:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.Writer;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.Result;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.xml.sax.EntityResolver;
import org.xml.sax.SAXException;
import com.sun.java_cup.internal.parser;
import com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderFactoryImpl;
public class DisplayRes extends HttpServlet {
     public void doGet(HttpServletRequest requset, HttpServletResponse response) throws ServletException, IOException
          doWork(requset, response);
     public void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
          doWork(req, resp);
     private void doWork(HttpServletRequest req, HttpServletResponse resp) throws IOException
          String path = null;
          PrintWriter out = null;
          PrintWriter p2=resp.getWriter();
          try
               resp.setContentType("text/xml");
               out = resp.getWriter();
               out.println("hi");
               path = req.getPathInfo();
               if(req.getContentLength() != -1){
               outputURI(req.getInputStream(), out);
          } catch (IOException ioe) {     return;
     //private void outputURI(InputStream resultStream, Writer out) {
          private void outputURI(InputStream resultStream, PrintWriter out) {
          if (resultStream == null) {
//             no default file
//            logger.error("No File to return");
          return;
          try {
          DocumentBuilder parser = DocumentBuilderFactory.newInstance().newDocumentBuilder();
//            Class clazz = loader.loadClass("org.apache.xerces.jaxp.DocumentBuilderFactoryImpl");
//            DocumentBuilderFactoryImpl factory = (DocumentBuilderFactoryImpl) clazz.newInstance();
//            DocumentBuilder parser = factory.newDocumentBuilder();
          parser.setEntityResolver(new ClassPathEntityResolver());
          printXML(parser.parse(resultStream), out);
          resultStream.close();
          } catch (Exception e) {
//            logger.error("Trying to parse the output " , e);
     //private void printXML(Document document, Writer writer) throws Exception {
          private void printXML(Document document, PrintWriter writer) throws Exception {
          Transformer transformer = TransformerFactory.newInstance().newTransformer();
          Source source = new DOMSource(document);
          Result output = new StreamResult(System.out);
          transformer.transform(source, output);
//             Write as XML so that entity references can be resolved.
          if (writer != null) {
          transformer.transform(source, new StreamResult(writer));
Please tell me how to display in servlet.
If it is better to display in a browser. Please send me the code.
Thanks

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)
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public void init() throws ServletException {
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     Class.forName("com.mysql.jdbc.Driver");
     Connection Con = DriverManager.getConnection("jdbc:mysql://localhost/se?user=luser&password=");
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     throw new UnavailableException("Couldn't load JdbcOdbcDriver");
     catch (SQLException e) {
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     //res.setContentType("text/html");
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     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 + ")";
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     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.

Using Servlet Parameters defined in web.xml

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