Taw web xml file ADMINREADCC8 context parameter

when I run the scipt for CCA the Useme.sql I only see the ADMINCC8 user I don't see the ADMINREADCC8 so I'm not sure what to put on the following context for web.xml
<!-- Needs to be removed for MSSQL-->
<!--<context-param>-->
<!--<param-name>databaseSchemaReadOnly</param-name>-->
<!--<param-value>ADMINCCREAD813</param-value>-->
<!--</context-param>-->

This is the eclipse error
The content of element type "web-app" must match "(icon?,display-
name?,description?,distributable?,context-param*,filter*,filter-mapping*,listener*,servlet*,servlet-
mapping*,session-config?,mime-mapping*,welcome-file-list?,error-page*,taglib*,resource-env-
ref*,resource-ref*,security-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-local-
ref*)".
Web.xml file
<?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>tapestryHiber Tapestry 5 Application</display-name>
     <context-param>
          <!-- The only significant configuration for Tapestry 5, this informs Tapestry
               of where to look for pages, components and mixins. -->
          <param-name>tapestry.app-package</param-name>
          <param-value>com.kids.crm</param-value>
     </context-param>
     <filter>
          <filter-name>app</filter-name>
          <filter-class>org.apache.tapestry5.TapestryFilter</filter-class>
     </filter>
     <filter-mapping>
          <filter-name>app</filter-name>
          <url-pattern>/*</url-pattern>
     </filter-mapping>
     <context-param>
          <param-name>contextConfigLocation</param-name>
          <param-value>/WEB-INF/spring/spring-servlet.xml</param-value>
     </context-param>
     <listener>
          <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
     </listener>
</web-app>
      Edited by: csckid on Aug 29, 2011 5:51 AM

Similar Messages

  • 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 >");

  • 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

  • Web.xml file

    Hi
    I have a servlet that is working perfectly through JBuilder. When I try and run it through Tomcat I get an error saying it cannot find the file. In JBuilder I have a html form. The value entered in the form is passed to the servlet. The servlet calls Java classes from a different package. The servlet uses the value from the form as the paramater needed for the classes in the other package and outputs the results.
    I cannot get this working outside JBuilder.
    It could be something to do with the web.xml file. I tried copying the webapp folder from the JBuilder servlet project straight across to Tomcat but it didnt work. This project has its own WEB-INF folder and web.xml file. Should this work or do I need to change the Tomcat web.xml file.
    Any help
    Damul

    Hai,
    First you are taking the parameter value from the HTML page, pass it to the package class and show the out put.
    Fine, to run in Tomcat 4.1.24,first you have to create a context of yours.
    Else you can place your class files or package in the folder
    'examples' of webapps in Tomcat directory.
    Now place your HTML file in 'examples' folder.
    One more thing you have to include your package class name in the web.xml for the new context,if you create.Else if you place in the 'examples' folder , no need to edit web.xml
    to run this, for your package
    http://localhost:8080/examples/servlet/<packagename>.<classname>
    Hope this works out...

  • How to configure a JNDI ressource in the web.xml file

    Hi,
    does someone know if there is a way to configure a JNDI ressource with its parameters only in the web.xml file ?
    In my case, the JNDI ressouce is an oracle database an its conenction paramters: driverClassName, username, password, etc...
    Actually, I put everything in the configuration file of Tomcat but I would like my war file to be 100% independent of the application server.
    Thanks in advance.
    Fred.

    I wrestled with this for a long time. I may have done something wrong, but to get my mySQL JDNI reference to work, I had to add it to the server.xml file. You can try with web.xml, but I eventually just got frustrated and put it in server.xml.
    The following is between two <context> tags:
    <Resource name="jdbc/instance_name_goes_here"
                   auth="Container"
                   type="javax.sql.DataSource"/>
              <ResourceParams name="jdbc/same_instance_name_here">
                   <parameter>
                        <name>username</name>
                        <value>user_name_goes_here</value>
                   </parameter>
                   <parameter>
                        <name>password</name>
                        <value>password_goes_here</value>
                   </parameter>
                   <parameter>
                        <name>driverClassName</name>
                        <value>jdbc_fully_qualified_class_name</value>
                   </parameter>
                   <parameter>
                        <name>factory</name>
                        <value>org.apache.commons.dbcp.BasicDataSourceFactory</value>
                   </parameter>
                   <parameter>
                        <name>url</name>
                        <value>jdbc_connect_url_database_vendor_specific</value>
                   </parameter>
                   <parameter>
                        <name>maxActive</name>
                        <value>8</value>
                   </parameter>
                   <parameter>
                        <name>maxIdle</name>
                        <value>8</value>
                   </parameter>
              </ResourceParams>
    - Saish
    "My karma ran over your dogma." - Anon

  • 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.

  • 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 roles referenced by this web application -->
    <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/

  • How to setup web.xml files in Tomcat 4.0.6

    Hi, I'm to new to Java Servlets. i am trying to run my first servlet but it's not showing up..can anyone help me out please!!!
    I created a folders like this:
    webapps/sumitapps/WEB-INF\classes\Forms\login.class
    under sumitapps: i'm puttin all of my .html files(Logins.html)
    under WEB-INF: i put web.xml file and classes folder
    <?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>
    <servlet>
    <servlet-name>login</servlet-name>
    <servlet-class>Forms.login</servlet-class>
    </servlet>
    <servlet-mapping>
    <servlet-name>login</servlet-name>
    <url-pattern>/Forms.login</url-pattern>
    </servlet-mapping>
    </web-app>
    <Context path="/sumitapps" docBase="webapps/sumitapps" crossContext="true" debug="0" reloadable="true" trusted="false" >
    </Context>
    under my classes folder i put folder Forms where i'm keeping all of my .class files. my class file name is login.class
    can anyone tell me why i cant run my servlets
    <FORM method="POST" action="Forms.login"
    name=userinfo onsubmit="return checkData()">
    please help me out here
    i can access my html page like this:
    http://lcoalhost:8080/sumitapps/Logins.html
    but when i enter userid and password it doesnt show anything..it takes me to http://localhost:8080/sumitapps/Forms.login
    and shows cannot be displayed

    have a look at www.moreservlets.com. There you can find a free pdf (part of the book, a whole chapter) that explains in deep the use of web.xml.

  • Doubt in locating a file in context parameter

    I have got a file named mylibrary.txt in WEB-INF/data directory.
    I wish to map the file in context Parameter.
    While trying this,
    <context-param>
         <param-name>library-file</param-name>
         <param-value>/WEB-INF/data/my-library.txt</param-value>
    </context-param>     
    I got a null pointerException while reading the file in my servlet. I am using Eclipse.

    The code as follows
    public class ListLibraryServlet extends javax.servlet.http.HttpServlet implements javax.servlet.Servlet {
        /* (non-Java-doc)
          * @see javax.servlet.http.HttpServlet#HttpServlet()
         public ListLibraryServlet() {
              super();
         /* (non-Java-doc)
          * @see javax.servlet.http.HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
         protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
              // TODO Auto-generated method stub
              response.setContentType("text/html");
              PrintWriter out=response.getWriter();
              ServletContext context=getServletContext();
              List dvd=(List)context.getAttribute("dvdlist");
              out.println("<head><title>ListLibraryServlet</title></head>");
              out.println("<body bgcolor='D2B48C'>");
              out.println("You currently have <b>3</b> DVDs in your collection<br>");
              out.println("<table>");
              out.println("<tr>");
              out.println("<th>Title</th>");
              out.println("<th>Year</th>");
              out.println("<th>Genre</th>");
              out.println("</tr>");
            Iterator it=dvd.iterator();
            while(it.hasNext())
              DVDItem item=(DVDItem)it.next();
              out.println("<tr>");
              out.println("<td>");
            out.println(item.getTitle());
            out.println("</td>");
              out.println("<td>"+item.getYear()+"</td>");
              out.println("<td>"+item.getGenre()+"</td>");
              out.println("</tr>");
              out.println("<tr>");
            out.println("</table>");
            out.println("</body>");
            out.println("</html>");
         /* (non-Java-doc)
          * @see javax.servlet.http.HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
         protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
              // TODO Auto-generated method stub
         }                   I am retrieving the file as
    public class InitializeLibrary extends javax.servlet.http.HttpServlet implements javax.servlet.ServletContextListener {
        /* (non-Java-doc)
          * @see javax.servlet.http.HttpServlet#HttpServlet()
         public InitializeLibrary() {
              super();
         public void contextInitialized(ServletContextEvent event)
         {   InputStream is=null;
             BufferedReader reader=null;
              List dvdlist=new ArrayList();
              ServletContext context=event.getServletContext();
              String libraryfile=context.getInitParameter("library-file");
              try
              is=context.getResourceAsStream("libraryfile");
              reader=new BufferedReader(new InputStreamReader(is));
              String record;
              while((record=reader.readLine())!=null)
                 String[] fields=record.split("\\|");
                 String title=fields[0];
                 String year=fields[1];
                 String genre=fields[2];
                 DVDItem item=new DVDItem(title,year,genre);
                 dvdlist.add(item);
              context.setAttribute("dvdlist",dvdlist);
              catch(IOException e)
                   e.printStackTrace();
              catch(Exception e)
                   e.printStackTrace();
         public void contextDestroyed(ServletContextEvent event)
    }

  • I  have some more problem in the  web.xml file and url-pattern in my  syste

    HI ALL!
    How can I create and place the web.xml and URL-parrten.Because I run
    the tomcat in my computer with ME system.
    Hoc

    what is the problem? please be more specific and detailed? check a sample web.xml file and make your own one and place it in your <application-context>\WEB-INF folder..

  • How to config the web.xml file, when I use Richfaces + RI 1.2?

    Hi there:
    I want to use Richfaces + RI 1.2 to build a project. I don`t know how to config the web.xml file.
    By the way, my web server is Tomcat 6.0, my JDK's version is 6u6. I don`t want to use the facelets.
    thanks.
    lxm

    just add this before *</web-app>*
    <context-param>
           <param-name>org.richfaces.SKIN</param-name>
           <param-value>blueSky</param-value>
      </context-param>
      <filter>
           <display-name>RichFaces Filter</display-name>
           <filter-name>richfaces</filter-name>
           <filter-class>org.ajax4jsf.Filter</filter-class>
      </filter>
      <filter-mapping>
           <filter-name>richfaces</filter-name>
           <servlet-name>Faces Servlet</servlet-name>
           <dispatcher>REQUEST</dispatcher>
           <dispatcher>FORWARD</dispatcher>
           <dispatcher>INCLUDE</dispatcher>
      </filter-mapping>

  • 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.

  • (newbie) Question about replacing .class files and web.xml file

    I'm new to servlets and I have two quick questions...
    Do I absolutely need a web.xml file to define all my servlets, or can I simply place .class files into the WEB-INF directory and expect them to run?
    If my application server (for example Tomcat) is running and I replace a servlet .class file, do I need to restart the server for the new .class file to take effect?
    ...or are both of these questions specific to the application server I'm using?

    Hi,
    From an article I read:
    With Tomcat 3.x, by default servlet container was set up to allow invoking a servet through a common mapping under the /servlet/ directory.
    A servlet could be accessed by simply using an url like this one:
    http://[domain]:[port]/[context]/servlet/[servlet full qualified name].
    The mapping was set inside the web application descriptor (web.xml), located under $TOMCAT_HOME/conf.
    With Tomcat 4.x the Jakarta developers have decided to stop allowing this by default. The <servlet-mapping> tag that sets this mapping up, has been commented inside the default web application descriptor (web.xml), located under $CATALINA_HOME/conf:
    <!-- The mapping for the invoker servlet -->
    <!--
    <servlet-mapping>
    <servlet-name>invoker</servlet-name>
    <url-pattern>/servlet/*</url-pattern>
    </servlet-mapping>
    -->
    A developer can simply map all the servlet inside the web application descriptor of its own web application (that is highly suggested), or simply uncomment that mapping (that is highly discouraged).
    It is important to notice that the /servlet/ isn't part of Servlet 2.3 specifications so there are no guarantee that the container will support that. So, if the developer decides to uncomment that mapping, the application will loose portabiliy.
    And declangallagher, I will use caution in future :-)

  • About Tomact's Web.xml file

    Starting service Tomcat-Standalone
    Apache Tomcat/4.0.1
    PARSE error at line 38 column 11
    org.xml.sax.SAXParseException: The content of element type "web-app" must match "(icon?,display-name?,description?,
    distributable?,context-param*,servlet*,servlet-mapping*,session-config?,mime-mapping*,welcome-file-list?,error-page
    *,taglib*,resource-ref*,security-constraint*,login-config?,security-role*,env-entry*,ejb-ref*)".
    Starting service Tomcat-Apache
    Apache Tomcat/4.0.1
    I am getting this message when i keep web.xml file in web-inf folder.that is after starting tomcat.

    Hi
    The elements in web.xml file must follow a specific order, it the one tomcat prints in the error-message. (It should also tell THIS IS THE REQUESTED ORDER)
    I made the same problem before... ;)
    Made you have a <servlet-mapping> tag before a <servlet> tag? That is not correct! Hope this solves your problems. :)
    /Tobias

  • XCM configuration path in web.xml file

    Hi,
    Here is a requirement that all the XCM configuration files are stored at local desktop(like QAT, PRD..), and need to use them by modifying web.xml.
    Please let me know at
    1)which web.xml tag the path of XCM confiuration files to be specified.
    2) specify the names of the XCM files for configuration to be put in that folder.
    Thanks,
    Devender V

    Hi,
    1)which web.xml tag the path of XCM confiuration files to be specified.
    Try to find below lines in your web.xml
    <context-param>
            <param-name>path.xcm.config.isa.sap.com</param-name>
            <param-value>/WEB-INF/xcm/sap/system/bootstrap-config.xml</param-value>
            <description>Turns the Extended Configuration Management on if a
                path to the configuration file is specified. It is a
                relative path, with context-root of web application as root</description>
        </context-param>
    bootstrap-config.xml file contains XCM configuration related XML file path. You can modify or extend this file and give your user Define path.
    Also you can find below line in web.xml file.
    <!-- Action Servlet Configuration -->
        <servlet>
            <servlet-name>action</servlet-name>
            <servlet-class>com.sap.isa.core.ActionServlet</servlet-class>
            <init-param>
                <param-name>config</param-name>
                <param-value>/WEB-INF/config.xml,/WEB-INF/xcmadmin-config.xml,/WEB-INF/ccmsims-config.xml,/WEB-INF/ipc-config.xml,/ipc/customer/config.xml,/WEB-INF/config_lwc_b2b.xml,/WEB-INF/scheduler-config.xml,/WEB-INF/ipcpricing-config.xml</param-value>
            </init-param>
            <init-param>
                <param-name>config/user</param-name>
                <param-value>/WEB-INF/config_user.xml</param-value>
            </init-param>
            <init-param>
                <param-name>config/dealerlocator</param-name>
                <param-value>/WEB-INF/config_dealerlocator.xml</param-value>
            </init-param>
            <init-param>
                <param-name>initconfig</param-name>
                <param-value>/WEB-INF/xcm/sap/system/init-config.xml</param-value>
            </init-param>
    I hope this will help you.
    eCommerce Developer

Maybe you are looking for