Specifying servlet classpath in xml file

          Hi there...
          Am trying to achieve dynamic class reloading for my web application, and have
          specified the servlet classpath in my web.xml file. However my server still loads
          classes that exist on the classpath first, and ignores any 'later' versions I
          have placed in the servlet classpath.
          My question would be do I really have to split up my jar files and place the individual
          classes on the servlet classpath, or can I specify two or more parameters under
          the "context-params" in the web.xml.
          Ideally I would like to be able to keep my jars whole, and upload classes to override
          the contents of the jar, and at the moment I can't see a way to achieve this !!
          Many thanks for any help anyone might be able to give !!
          George
          

I'm no expert with WAR files and 5.1 but....
          Have you tried putting your jar files and any non-weblogic specific jars in
          to your WEB-INF/lib directory?
          Then you dont have to specify them in the classpath when starting weblogic.
          Following this - i think you should be a step closer to dynamic class
          reloading.
          "George Cover" <[email protected]> wrote in message
          news:[email protected]...
          >
          > Thanks for the prompt response...unfortunately we are still at version 5,
          with
          > a service pack level of 10 (which has just gone live). However we do have
          our
          > environments set up as much as possible in a J2EE fashion.
          >
          > We still specify a classpath, where the jars are kept, however we also
          have WEB-INF
          > directories, where compiled classes (jsp, and shock horror, jhtml) are
          kept !!
          >
          > I think that placing classes in here, puts me in a similar situation as
          placing
          > them on the servlet classpath, the jars still override them !!
          >
          > Also the halfway house our server environments are in, plus the fact they
          are
          > all constantly changing, makes it difficult to predict behaviour, or even
          come
          > close to understanding what might happen if I change configurations here
          and there
          > !!
          >
          > Many thanks
          > George
          >
          > "Matt Krevs" <[email protected]> wrote:
          > >What version of weblogic are you using?
          > >
          > >If you are using 6.0 or greater - my understanding is that you dont
          specify
          > >your classes in the classpath when starting weblogic. Your classes and
          > >jars
          > >should be in web-inf/classes and web-inf/lib and should be loaded from
          > >there.
          > >
          > >"George Cover" <[email protected]> wrote in message
          > >news:[email protected]...
          > >>
          > >> Hi there...
          > >>
          > >> Am trying to achieve dynamic class reloading for my web application,
          > >and
          > >have
          > >> specified the servlet classpath in my web.xml file. However my server
          > >still loads
          > >> classes that exist on the classpath first, and ignores any 'later'
          > >versions I
          > >> have placed in the servlet classpath.
          > >>
          > >> My question would be do I really have to split up my jar files and
          > >place
          > >the individual
          > >> classes on the servlet classpath, or can I specify two or more
          parameters
          > >under
          > >> the "context-params" in the web.xml.
          > >>
          > >> Ideally I would like to be able to keep my jars whole, and upload
          classes
          > >to override
          > >> the contents of the jar, and at the moment I can't see a way to achieve
          > >this !!
          > >>
          > >> Many thanks for any help anyone might be able to give !!
          > >> George
          > >
          > >
          >
          

Similar Messages

  • Specifying Schema Location in XML file and not in the Java code

    I have a repository of schema xsd files. When I receive my xml file, I need to validate it against the specified schema. The xml file would declare the schema location, using the following syntax:
    <CERD:CERD xsi:schemaLocation = "..\CERD.xsd" xmlns:CERD = "CERD.xsd" xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance">
    Is it really necessary to define the schema location again in the JAVA code? Why set the schemaLocation in the xml file at all then?
    Does anybody have any examples where the schema location is not set in the JAVA code? I am using Java 1.6, and at this point in time I only need to validate. Any help would be appreciated.

    Thank you very much for your quick reply. I have made some progress but I am still stumped.
    In my code I am doing this:
    SchemaFactory factory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
    Schema schema = factory.newSchema();
    Validator validator = schema.newValidator();
    validator.validate(XML_SOURCE);
    I find this works if my schema does not have a target namespace. I have downloaded the following simple example from the internet that uses a target namespace and it fails:
    library1.xsd:
    <xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"
    targetNamespace="http://example.org/prod"
    xmlns:prod="http://example.org/prod">
    <xsd:element name="product" type="prod:ProductType"/>
    <xsd:complexType name="ProductType">
    <xsd:sequence>
    <xsd:element name="number" type="xsd:integer"/>
    <xsd:element name="size" type="prod:SizeType"/>
    </xsd:sequence>
    <xsd:attribute name="effDate" type="xsd:date"/>
    </xsd:complexType>
    <xsd:simpleType name="SizeType">
    <xsd:restriction base="xsd:integer">
    <xsd:minInclusive value="2"/>
    <xsd:maxInclusive value="18"/>
    </xsd:restriction>
    </xsd:simpleType>
    </xsd:schema>
    <prod:product xmlns:prod="http://example.org/prod"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
              xsi:schemaLocation="library1.xsd"
    effDate="2001-04-02">
    <number>557</number>
    <size>10</size>
    </prod:product>
    I get the following SAXParseException when I validate:
    [line 4, col 36|
    cvc-elt.1: Cannot find the declaration of element 'prod:product'.
    Am I doing something wrong with the namespace declaration?
    Edited by: alfredamorrissey on Oct 31, 2007 6:34 AM                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       

  • Read specified content of a xml file

    How to get specified part of a xml file
    This is the content of the xml file, I just want the content in the <body>...</body> without the <p></p> tag. How can I do that?
    <?xml version="1.0" encoding="UTF-8"?>
    <TEI.2>
    <teiHeader>
    <fileDesc>
    <titleStmt>
    <title>Pastikan kita terus bebas -- PM: Kekalkan perpaduan, jangan terpengaruh dakyah asing di alaf baru
    </title>
    <author>MOHD. NOR AB. SAMAD
    </author>
    </titleStmt>
    <publicationStmt>
    <publisher/>
    </publicationStmt>
    <sourceDesc>
    <!-- ... -->
    </sourceDesc>
    </fileDesc>
    <profileDesc>
    <textClass>
    <catRef target="Muka_Hadapan"/>
    </textClass>
    </profileDesc>
    </teiHeader>
    <text>
    <front>
    <!--
                      front matter ...
                 -->
    </front>
    <body>
    <p>- Datuk Seri Dr. Mahathir Mohamad mengajak rakyat supaya mengekalkan perpaduan dan belajar daripada sejarah negara serta tidak terpedaya dengan dakyah asing bagi memastikan Malaysia terus bebas, merdeka dan maju pada alaf akan datang.</p>
    <p>Sambil mengingatkan mereka usaha menjajah semula negara masih diteruskan oleh bekas penjajah, Perdana Menteri berkata, kemerdekaan negara tidak akan kekal selama-lamanya jika rakyat mengabaikan perkara tersebut.</p>
    <p/>
    <p/>
    <p/>
    </body>
    <back>
    <!--
                      back matter ...
                 -->
    </back>
    </text>
    </TEI.2>

    I can get the first <p> tag only. How to get others <p> tag content?
    How I know how many <p> node that it have if i use while loop?
         public static String ParserWord(String pathname, String tag) {
                Text result = null;
                try{
                  DOMParser parser = new DOMParser();
                    parser.parse(pathname);
                    Document doc = parser.getDocument();
                    NodeList doubles = doc.getElementsByTagName(tag);
                    Node datum = doubles.item(0);
                    result = (Text) datum.getFirstChild();
                 catch (Exception e) {
                  System.err.println(e + "cannot parse");
             return result.getNodeValue();
           }

  • Using servlet to generate XML file

    What I want to do is simple but can't work. Hope you guys can give me a hint.
    I succesfully generate XML file in command line using Oracle XML parser and class generator. Then I was trying to do it using servlet. it compiles fine but generate NullPointer exception at the line:
    Emp e1= new Emp(); //Emp is the root of XML
    I suspect the Emp constructor can't correctly find the globalDTD in its superclass -CGDocument. Please note this only happens in servlet setting, not is command line setting. Any suggestion to get arounf this?
    Another unrelated question is that when I create a XML file using the Oracle XML parser, it seems all the elements a file has to be added once, otherwise the compiler will compalain about the missing element. this will be inconvinient when I constructing a big XML file, which I 'd liek to split into small piece and add them up. Maybe there is a good way but I just don't know it.
    my email: [email protected]

    Hi,
    I'm running into the same problem deploying the classes generated by the class generator. Code works fine from JDeveloper, but had to put my DTD in the directory where my classes are. Deploying the classes with Apache's JServ gives me a NullPointer exception on the first addNode method. I guess it can't find the DTD. I tried to put the DTD in many locations but this didn't fix the problem. Any suggestions?
    Steve,
    Did you fix this problem? Thanx!
    null

  • No table name specified in ejb-jar.xml file

    Hi, I have an entity bean defined as follows:
    <enterprise-beans>
    <entity>
    <display-name>SponsorEJB</display-name>
    <ejb-name>SponsorEJB</ejb-name>
    <home>com.ultradns.j2ee.components.sponsor.ejb.SponsorHome</home>
    <remote>com.ultradns.j2ee.components.sponsor.ejb.Sponsor</remote>
    <ejb-class>com.ultradns.j2ee.components.sponsor.ejb.SponsorEJB</ejb-class>
    <persistence-type>Container</persistence-type>
    <prim-key-class>java.lang.String</prim-key-class>
    <reentrant>False</reentrant>
         <resource-ref>
              <res-ref-name>jdbc/jdbc/OracleDS</res-ref-name>
              <res-type>javax.sql.DataSource</res-type>
              <res-auth>Container</res-auth>
         </resource-ref>     
    <cmp-version>2.x</cmp-version>
    <abstract-schema-name>sponsor</abstract-schema-name>
    <cmp-field>
    <primkey-field>guid</primkey-field>
         <query>
              <query-method>
                   <method-name>findAll</method-name>
                   <method-params></method-params>
              </query-method>
              <ejb-ql>
                   <![CDATA[ SELECT OBJECT(a) FROM sponsor AS a]]>
              </ejb-ql>
         </query>
    </entity>
    </enterprise-beans>
    In the database, I want this entity bean refers to a table named "sponsor", but I could not specify the relation in the ejb-jar.xml file. So when I deployed this bean, it auto created a table called "SponsorEJB". Where is this name come from, from display-name or ejb-name field?
    Basically, how can I specify a bean "SponsorEJB" referes to a table "sponsor"?
    Thanks,

    First thing, to stop OC4J auto-creating tables, edit the file %J2EE_HOME%/config/application.xml. Change the first orion-application tag so that one attribute says autocreate-tables="false"
    The ejb-jar.xml file describes your EJB's to the container. It is not supposed to describe which table each entity bean maps too, as this is an implementation specific property, so would not be portable between databases. Therefore you have to configure the container to point an entity to a table. Sun uses the sun-j2ee-ri.xml file to do this, and this has to be included in the deployment.
    OC4J however uses the file 'orion-ejb-jar.xml', and will generate this for you. Since you have already deployed your application, this will have already been generated. Go to the directory %J2EE_HOME%/application-deployments. There will be a directory corresponding to your application name. In there is a directory corresponding to your ejb module name, and in there is the 'orion-ejb-jar.xml' file. If you open this file, you will see the entity deployment tags, an example of one is below. Edit the tag so that the table attribute points to the table in your database (note that there is a data-source attribute as well, so make sure that this points to the correct schema). The example below points the EJB Emp to the table EMP:
    <entity-deployment name="Emp" max-instances="10" location="Emp" wrapper="EmpHome_EntityHomeWrapper3" table="EMP" data-source="jdbc/rob9iDS" locking-mode="optimistic" update-changed-fields-only="true" min-instances-per-pk="0" max-instances-per-pk="50" disable-wrapper-cache="true">
    Hope this helps,
    Rob

  • Help needed in Mapping servlet in web.xml file

    Hi Can anyone please tell me what am doing wrong here. I can't see this servlet in my url. Please help.
    <servlet>
    <servlet-name>HelloWorld</servlet-name>
         <servlet-class>HelloWorld</servlet-class>
    </servlet>
    <servlet-mapping>
    <servlet-name>HelloWorld</servlet-name>
    <url-pattern>/com/jci/fi/application/eems/HelloWorld/*</url-pattern>
    </servlet-mapping>
    package com.jci.fi.application.eems;
    import java.io.*;
    import javax.servlet.*;
    import javax.servlet.http.*;
    public class HelloWorld extends HttpServlet {
    public void doGet(HttpServletRequest request, HttpServletResponse response)
    throws IOException, ServletException
    response.setContentType("text/html");
    PrintWriter out = response.getWriter();
    out.println("<html>");
    out.println("<body>");
    out.println("<head>");
    out.println("<title>Naveen Wants to goto TollyWood......!</title>");
    out.println("</head>");
    out.println("<body>");
    out.println("<h1>Hello World!</h1>");
    out.println("</body>");
    out.println("</html>");

    your web.xml shoould be something like this...
    <servlet>
    <servlet-name>HelloWorld</servlet-name>
    <servlet-class>com.jci.fi.application.eems.HelloWorld</servlet-class>
    </servlet>
    <servlet-mapping>
    <servlet-name>HelloWorld</servlet-name>
    <url-pattern>HelloWorld</url-pattern>
    </servlet-mapping>
    Hope this helps..
    CK

  • Virtual directory path from weblogic.xml file in servlet

    HI All,
    I want to get the virtual directory path in servlet, which is specified in the weblogic.xml file
    Regards,
    Abhi

    Sure, put it in your WEB-INF directory. That should be in your servlet's classpath when it runs. UseInputStream is = this.getClass().getResourceAsStream("/config.xml");to read it.

  • How to display XML file in browser using Servlets?

    Hi My XML file is like this:
    <?xml version="1.0" encoding="UTF-8"?>
    <ns0:purchase_order_MT xmlns:ns0="http://filetohttp.com">
    <ORDER_HEADER_IN>
    <purchase_date/>
    <purchase_group/>
    <purchase_org/>
    </ORDER_HEADER_IN>
    <vendor/>
    <ORDER_ITEMS_IN>
    <item>
    <storage_location/>
    <plant/>
    <quantity/>
    <material/>
    </item>
    </ORDER_ITEMS_IN>
    <ORDER_SCHEDULES_IN>
    <item>
    <delivery_date/>
    </item>
    </ORDER_SCHEDULES_IN>
    </ns0:purchase_order_MT>
    My Servlet receives this XML file and it should display the above XML as it is, How to do that?
    Please help me I have posted same question in many forums but i got no proper reply.
    Thanks a lot

    If you only need to display it, consider your HttpServletRequest just as byte stream. :) Just read from ServletInputStream and write it back into ServletOutputStream of HttpServletResponse.
    If you want to process incoming XML, please give some details first, what exactly you want to do. :)

  • How to display XML file as is it is in my browser using servlets???

    I have a question like my servlet receives an XML file in its request object. I just need display the same XML file as it is in my browser. How to do that? Please help me out. I have searched so many blogs, but i did not get right solution.
    My XML file file will be like this:
    <ns0:Http_Message_Type_Demo xmlns:ns0="http://abcdemo.com">
    <Name>A</Name>
    <RollNo>123</RollNo>
    <Address>a2</Address>
    </ns0:Http_Message_Type_Demo>
    Thnks

    You mean you want to see the XML source?
    You need to replace the characters '<' and '&' with corresponding entities '&lt;' and '&amp;'. You can use replaceAll, but do the ampersands first.
    Then I suggest you probably want to put them in a <PRE> block.

  • Loading XML file for parsing

    Folks,
    I have placed one xxx.XML file under /WEB-INF/ of my application server and i have configured a servlet in web.xml file to execute during server startup like below
    <servlet>
    <servlet-name>MonitorInitServlet</servlet-name>
    <servlet-class>com.xxx.xxx..services.MonitorInitServlet</servlet-class>
    <init-param>
    <param-name>configFile</param-name>
    <param-value>/WEB-INF/portalMonitorConfig.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
    </servlet>
    And the above servlet loads portalMonitorConfig.xml file and i have below code
    InputStream inputStream = xxx.class.getResourceAsStream("/WEB-INF/portalMonitorConfirg.xml");
    private static Document parseXmlFile(InputStream xml) throws SAXException, IOException,
    ParserConfigurationException
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    factory.setValidating(false);
    return factory.newDocumentBuilder().parse(xml);
    So when i do the above , i get
    java.lang.IllegalArgumentException :InputStream cannot be null,
    I know its sending a null InputStream, By the way i need to know what is the issue in the above code when i create InputStream object
    InputStream inputStream = xxx.class.getResourceAsStream("/WEB-INF/portalMonitorConfig.xml");
    Any help with the code is really appreciated

    when serializing the DOM, use this: transformer.setOutputProperty(OutputKeys.DOCTYPE_SYSTEM, scrReg.dtd");
    Merry Chrasmas too ;-)

  • XML File Within XML File

    Hi everyone,
    Plz help me out.....
    Currently, I am working on flex and Remote Object Accessing.
    My problem is that, in my application I have one Servlet which
    writes .xml file in browser and I use that .xml file to display
    data in Datagrid. Everything was woking fine. But then I have to
    change my .xml file in such a way that one of it's node is a .xml
    file. like this...
    <?xml version="1.0" encoding="UTF-8" ?>
    - <root>
    - <detail>
    <server>SOA </server>
    <resp_time>250</resp_time>
    - <request>
    <?xml version="1.0" encoding="UTF-8" ?>
    - <envelope>
    <operation>addition</operation>
    </envelope>
    </request>
    </detail>
    </root>
    Here request is one of the node in .xml file. now when i am
    displaying data in this file into datagrid, datagrid show correct
    data for <server> and <resp_time> tag. But it shows
    "[object Object]" for request tag.
    Can any one tell me how can I show that request node into my
    datagrid ,so that it shows me ..
    <request>
    <?xml version="1.0" encoding="UTF-8" ?>
    - <envelope>
    <operation>addition</operation>
    </envelope>
    </request>
    instead of [object Object] ???? Please Hepl me

    Tracy had replied me that you can use labelFunction. like
    this...
    private function lfRequest(oItem:Object,
    dgc:DataGridColumn):String
    return oItem.request.toXMLString()
    I have used resultHandler like this ....
    [Bindable]
    var xmlFile:String = null;
    private function resultHandler(event:ResultEvent):void
    trace("resultHandler is called...");
    var myO:Object = new Object();
    myO = event.result;
    trace("after new XML()");
    xmlFile=myO.request.toXMLString();
    trace("after toString()");
    and i m trying to display that ' xmlFile:String' into
    datagrid, but it gives me this runtime-error:
    Error #1009: Cannot access a property or method of a null
    object reference.
    then i tryed this...
    xmlFile=myO.request.root.detail.toXMLString(); then it gives
    me this error message :
    Error #1009: Cannot access a property or method of a null
    object reference.
    plz help me...

  • XML  file and DataGrid

    Hi everyone,
    Plz help me out.....
    Currently, I am working on flex and Remote Object Accessing.
    My problem is that, in my application I have one Servlet which
    writes .xml file in browser and I use that .xml file to display
    data in Datagrid. Everything was woking fine. But then I have to
    change my .xml file in such a way that one of it's node is a .xml
    file. like this...
    <?xml version="1.0" encoding="UTF-8" ?>
    - <root>
    - <detail>
    <server>SOA </server>
    <resp_time>250</resp_time>
    - <request>
    <?xml version="1.0" encoding="UTF-8" ?>
    - <envelope>
    <operation>addition</operation>
    </envelope>
    </request>
    </detail>
    </root>
    Here request is one of the node in .xml file. now when i am
    displaying data in this file into datagrid, datagrid show correct
    data for <server> and <resp_time> tag. But it shows
    "[object Object]" for request tag.
    Can any one tell me how can I show that request node into my
    datagrid ,so that it shows me ..
    <request>
    <?xml version="1.0" encoding="UTF-8" ?>
    - <envelope>
    <operation>addition</operation>
    </envelope>
    </request>
    instead of [object Object] ???? Please Hepl me

    First, you are really posting in the wrong forum. This one is
    for Flex Builder 2, an IDE/code editor. You question is about the
    language Flex, and such questions should go in the "General
    Discussion" forum.
    Try using a labelFunction instead of dataField.
    private function lfRequest(oItem:Object,
    dgc:DataGridColumn):String
    return oItem.request.toXMLString()
    Tracy

  • Sending xml file from client to servlet

    Hi,
    I am writing the server component of an applcation, such that it will receive xml files from the clients(standalone application similar to javaSwing stuff but it's coded in C#), and the servlet will have to extract the data from the xml file and update the mySql database. it will also fulfill the client's request for xmlFiles (and extract data from DB, format to xml file and send back to client)
    I'm new to implementing the servlet receiving files from clients so would need some help.
    I've got 3 questions to ask:
    1) How does the servlet receive/returns the xml file from the client as a series of httpPost request/response. Do i send a File or the file's contents as a String to/from the client?
    2) Is it also a must to use socket for the file transfers? I have read in other posts about sockets as well as HttpURLConnection but i don't quite understand.
    3) When I send a file back to the client(client is standalone application written in C# whereas server is coded in java), what do i specify for the HttpResponse.setContentType() in my servlet? (i'm returning the xml file to client)
    Would really appreciate for any help rendered. If you have any useful links, would appreciate them too. Thanks a lot.
    Karen

    I've got 3 questions to ask:
    1) How does the servlet receive/returns the xml file
    from the client as a series of httpPost
    request/response. Do i send a File or the file's
    contents as a String to/from the client?The server will listen on some port for requests. The client has to open a socket to this server to send the file as string to the server.
    see http://java.sun.com/docs/books/tutorial/networking/index.html
    >
    2) Is it also a must to use socket for the file
    transfers? I have read in other posts about sockets as
    well as HttpURLConnection but i don't quite
    understand.You use HttpURLConnection to make a request using the http protocol, instead of opening a socket and then writing the html headers yourself.
    3) When I send a file back to the client(client is
    standalone application written in C# whereas server is
    coded in java), what do i specify for the
    HttpResponse.setContentType() in my servlet? (i'm
    returning the xml file to client)Its up to your receiving program how to interpret this though, so you probably dont need this.

  • How to set two different welcome-file to different servlets in web.xml

    Hi friends,
    Can some one help with web.xml file
    I have to different servlets in applications.
    I want to set two different welcome files for my two different
    servlets in web.xml
    Please, can some one give me sample code to achieve this.
    Here web.xml code:
    <?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>reports</servlet-name>
              <servlet-class>com.caremark.ivr.servlet.IvrReportsServlet</servlet-class>
         </servlet>
         <servlet-mapping>
              <servlet-name>reports</servlet-name>
              <url-pattern>/reports</url-pattern>
         </servlet-mapping>
         <session-config>
              <session-timeout>30</session-timeout>
         </session-config>
         <welcome-file-list>
              <welcome-file>reportsindex.jsp</welcome-file>
         </welcome-file-list>
         <servlet>
              <servlet-name>calldetails</servlet-name>
              <servlet-class>com.caremark.ivr.servlet.IvrCallDetailsServlet</servlet-class>
         </servlet>
         <servlet-mapping>
              <servlet-name>calldetails</servlet-name>
              <url-pattern>/calldetails</url-pattern>
         </servlet-mapping>
    <welcome-file-list>
              <welcome-file>calldetailsindex.jsp</welcome-file>
         </welcome-file-list>
    </web-app>
    I need this fix immediately. Your help will be really appreciated. Thanks in advance.

    First of all, my understanding is that you need to have a Servlet 2.4 complaint server to do this...Given that, you're supposed to be able to simply specify the Name of the servlet as the welcome-file to get it to work. For example:
    Your servlet section:
    <servlet>
      <servlet-name>Controller</servlet-name>
      <servlet-class>com.company.ControllerServlet</servlet-class>
    </servlet>Now you would add the following to your welcome file list:
    <welcome-file-list>
      <welcome-file>Controller</welcome-file>
    </welcome-file-list>Note that you use the servlet-name value from the servlet definition, NOT the url-mapping from the servlet-mapping section.
    Please note that I cannot verify this as I am running a 2.3 server at home. When I get to work Friday I can test on a 2.4 machine. However this shouldn't take you more than a minute to test.
    Note that I did try this on the 2.3 compliant server and it did NOT work (as I expected)...
    HTH.

  • XSQL-003: Failed to find 'XSQLConfig.xml' file in server CLASSPATH.

    Hi,
    I'm trying to make the Tomcat 1.3.2 to work with xsql servlet, and have followed the instructions on how to do it, but i get the : XSQL-003: Failed to find 'XSQLConfig.xml' file in server CLASSPATH. - error.
    What am I doing wrong?
    null

    Forget it...
    I had pasted in:
    set CLASSPATH=%CLASSPATH%;directory_where_XSQLConfig.xml_resides
    istead of:
    set CLASSPATH=%CLASSPATH%;C:\xsql\lib
    works now....
    null

Maybe you are looking for

  • Importing, converting and editing home video mostly

    Guys - I am new to Apple. Please be patient !! I basically need help . I shoot mostly family videos with my Sony cam (.mts format). I need basic conversion, editing and subtitling etc, but since my cam is HD, I would like to keep the quality. I have

  • Macbook Air Virus?

    my macbook air just froze on me while viewing a photo gallary on the internet. I turned it off, and restarted it. However, all I see is a blank gray screen. I tried Command + R, which prompts a message of Apple searching on Internet, but in less that

  • Server product support

    We recently upgraded our image server to Windows 2008 R2.  Our old server had Adobe Acrobat pro 6 and we had no trouble opening diffent type of image files - specifically TIFF files in Adobe.  I have been told that there are issues with Adobe X and W

  • How to use adress-independent e-mail?

    Hi, Guru! I made this steps: 1) Create BP with e-mail in adress-independent communication 2) Create target group with this BP 3) I export to chanel this target group in segmentation After this steps i have one problem: SAP CRM7 using e-mail in addres

  • Music app keeps crashing

    When I press the music app it just reverts back to the home page. I have turned of ipad etc... and it still does it I have tried to circumvent the app by searching for a song and it still crashes any ideas???