Want to edit web.xml page to give name of jsp file

I want to organise some files in web application so thus want to edit the web.xml page.
What is the tag to give a name to a JSP page?
Below is the web.xml page that I have tried but it does not work
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
 PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN"
 "http://java.sun.com/j2ee/dtds/web-app_2_2.dtd">
<web-app>
 <servlet>
 <servlet-name>processpage</servlet-name>
 <servlet-class>processSystemlogin.jsp</servlet-class>
 </servlet>
 <servlet-mapping>
 <servlet-name>processpage</servlet-name>
 <url-pattern>/processpage</url-pattern>
 </servlet-mapping>
</web-app>

The element <servlet-class> is reserved for servlets and refers to classes in /WEB-INF/lib or /WEB-INF/classes of your webapp. You do not need to name your jsp(s) in this manner. Just keep them one level above /WEB-INF .
I want to organise some files in web application so
thus want to edit the web.xml page.
What is the tag to give a name to a JSP page?
Below is the web.xml page that I have tried but it
does not work
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web
Web Application 2.2//EN"
"http://java.sun.com/j2ee/dtds/web-app_2_2.dtd">
<web-app>
 <servlet>
 <servlet-name>processpage</servlet-name>
 <servlet-class>processSystemlogin.jsp</servlet-class>
 </servlet>
 <servlet-mapping>
 <servlet-name>processpage</servlet-name>
 <url-pattern>/processpage</url-pattern>
 </servlet-mapping>
</web-app>

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First, I sent an email to the author of PhotoME to inform him of the serious issues his addon caused with Firefox latest versions.
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And another question
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Hi,
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 <auth-constraint>
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 </security-constraint>As you can see, I exchanged *.do with *.go for testing. I simply defined all actions I want to be used without SSL with a .go instead of a .do in the struts.config.
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C:import and web.xml

Im using <c:import url="http://path_to_html" />. I want to put this value in the web.xml using context param. The problem is I don't know how my c:import tag will refer to that context param inside web.xml. Can anyone give a solution? Thanks.

web.xml
<context-param>
 <param-name>abc</param-name>
 <param-value>http://path_to_html</param-value>
</context-param>
JSP
<c:import url="<%= application.getInitParameter("abc") %>"></c:import>More info for c:import
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Web.xml parsing error

Hello,
I am a newbie to servlet programming. I am using Tomcat4.1.24. When I
am trying out some examples, I got a bunch of parsing errors. After double-checking my web.xml, I still do not see why it caused so many parsing errors. I am using NotePad to edit web.xml, does that affect this? These errors are always about the last line and the last column.
Very weird. Please help.
web.xml
<?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>HelloWorld</servlet-name>
 <servlet-class>HelloWorld</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern>*.myservlet</url-pattern>
</servlet-mapping>
<servlet>
 <servlet-name>Primes</servlet-name>
 <servlet-class>coreservlets.Primes</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Primes</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
[ERROR] Digester - -Parse Error at line 29 column 11: The content of element typ
e "web-app" must match "(icon?,display-name?,description?,distributable?,context
-param*,filter*,filter-mapping*,listener*,servlet*,servlet-mapping*,session-conf
ig?,mime-mapping*,welcome-file-list?,error-page*,taglib*,resource-env-ref*,resou
rce-ref*,security-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,e
jb-local-ref*)". <org.xml.sax.SAXParseException: The content of element type "we
b-app" must match "(icon?,display-name?,description?,distributable?,context-para
m*,filter*,filter-mapping*,listener*,servlet*,servlet-mapping*,session-config?,m
ime-mapping*,welcome-file-list?,error-page*,taglib*,resource-env-ref*,resource-r
ef*,security-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-lo
cal-ref*)".>

Hi.
The answer is rather simple. The file is not valid against the XML v.1.0 standard. That's because all <servlet> elements must come first before all <servlet-mapping> elements. If you change the order of the Primes <servlet> element, the file is valid and Tomcat should not throw an exception.
This is the correct web deployment descriptor:
<?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>HelloWorld</servlet-name>
<servlet-class>HelloWorld</servlet-class>
</servlet>
<servlet>
<servlet-name>Primes</servlet-name>
<servlet-class>coreservlets.Primes</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern>*.myservlet</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>Primes</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>

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...

Web.xml - security constraint question

I want to make a kind of inverted security constraint. I want to ALLOW access to one specific page over HTTP. All other pages should be HTTPS.
Is that possible???

Use the <security-role-ref> for the Faces Servlet to map the LDAP roles to the logical role names used by the managed bean to determine if links may be rendered.
Bean code:
this.isAdmin = context.getExternalContext().isUserInRole("admin");web.xml:
 <security-role>
 <role-name>Local Admin Group</role-name>
 </security-role>
 <security-role>
 <role-name>DBA Group</role-name>
 </security-role>
 
 <servlet>
 <servlet-name>Faces Servlet</servlet-name>
 <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
 <security-role-ref>
 <role-name>admin</role-name>
 <role-link>DBA Group</role-link>
 </security-role-ref>
 <security-role-ref>
 <role-name>admin</role-name>
 <role-link>Local Admin Group</role-link>
 </security-role-ref>
 </servlet>

Access parameter from web.xml

<servlet>
<servlet-name>serverText</servlet-name>
<servlet-class>abc</servlet-class>
<init-param>
<param-name>serverParam</param-name>
<param-value>parameter</param-value>
</init-param>
</servlet>
Hi, above is what i defined in my web.xml.
i want to know how to get the serverParam from my jsp file.
i tried to use String param = pageContext.getServletConfig().getInitParameter("serverParam");
but i got null.
please help me.
thanks

hi,
well all jsps are compiled to servlets before running anyway, well if you using a jsp you can still do the
String param = getInitParameter("serverParam");
you could do this in the jspInit method
public void jspInit()
String param = config.getInitParameter("serverParam");
// config is the implicit ServletConfig object
The getInitParameter method is actually defined in the javax.servlet.ServletConfig class, the ServetConfig object is available as an implicit object by name config in a jsp page with page-level scope). well hope this was useful.
Cheerz

Replacing web.xml values with a deployment plan. Not working on WL 10.3.2

Hi All-
I am trying to replace a value from web.xml with one in a weblogic deployment plan. I deployed the application on WL 10.3.2 and used the deployment plan below. No matter what I do it doesn't seem to want to take the value.
web.xml:
<?xml version = '1.0' encoding = 'windows-1252'?>
<!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>
<context-param>
<param-name>TokenGroup</param-name>
<param-value>oldvalue</param-value>
</context-param>
</web-app>
Deployment Plan:
<?xml version='1.0' encoding='UTF-8'?>
<deployment-plan xmlns="http://xmlns.oracle.com/weblogic/deployment-plan"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.oracle.com/weblogic/deployment-plan http://xmlns.oracle.com/weblogic/deployment-plan/1.0/deployment-plan.xsd"
global-variables="false">
<application-name>APPName</application-name>
<variable-definition>
<variable>
<name>TokenGroup</name>
<value>Key123456</value>
</variable>
</variable-definition>
<module-override>
<module-name>APPName.war</module-name>
<module-type>war</module-type>
<module-descriptor external="false">
<root-element>web-app</root-element>
<uri>WEB-INF/web.xml</uri>
<variable-assignment>
<name>TokenGroup</name>
<xpath>/web-app/context-param/[param-name="TokenGroup"]/param-value</xpath>
<operation>replace</operation>
</variable-assignment>
</module-descriptor>
</module-override>
</deployment-plan>
I've also tried other formats for the xpath such as:
/web-app/context-param/{param-name="TokenGroup"}/param-value
/web-app/context-param[param-name="TokenGroup"]/param-value
Any ideas why this doesnt work?
Thanks,
Joe

"web.xml"
<?xml version='1.0' encoding='WINDOWS-1252'?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<context-param>
<param-name>name</param-name>
<param-value>Joe</param-value>
</context-param>
<context-param>
<param-name>password</param-name>
<param-value>password</param-value>
</context-param>
<servlet>
<servlet-name>ParameterServlet</servlet-name>
<servlet-class>ParameterServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ParameterServlet</servlet-name>
<url-pattern>/ParameterServlet</url-pattern>
</servlet-mapping>
</web-app>
ParameterServlet
import java.io.IOException;
import javax.servlet.ServletConfig;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class ParameterServlet extends HttpServlet {
private String dbName = "";
private String dbPassword = "";
public void init(ServletConfig config) throws ServletException {
super.init(config);
ServletContext context = getServletContext();
dbName = context.getInitParameter("name");
dbPassword = context.getInitParameter("password");
public void doGet(HttpServletRequest req, HttpServletResponse res)
throws IOException {
ServletOutputStream out = res.getOutputStream();
res.setContentType("text/html");
out.println("<html><head><title>Basic Servlet</title></head>");
out.println("<body>Database username is " + dbName);
out.println(" Database password is " + dbPassword + "");
out.println("</body></html>");
"Plan.xml"
<?xml version='1.0' encoding='UTF-8'?>
<deployment-plan xmlns="http://xmlns.oracle.com/weblogic/deployment-plan" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.oracle.com/weblogic/deployment-plan http://xmlns.oracle.com/weblogic/deployment-plan/1.0/deployment-plan.xsd" global-variables="false">
<application-name>virtual</application-name>
<variable-definition>
<variable>
 <name>name</name>
 <value>akshay1</value>
 </variable>
 <variable>
 <name>password</name>
 <value>koul1</value>
 </variable>
</variable-definition>
<module-override>
<module-name>virtual</module-name>
<module-type>war</module-type>
<module-descriptor external="false">
<root-element>web-app</root-element>
<uri>WEB-INF/web.xml</uri>
 <variable-assignment>
 <name>name</name>
 <xpath>/web-app/context-param/[param-name="name"]/param-value</xpath>
 <operation>replace</operation>
 </variable-assignment>
 <variable-assignment>
 <name>password</name>
 <xpath>/web-app/context-param/[param-name="password"]/param-value</xpath>
 <operation>replace</operation>
 </variable-assignment>
</module-descriptor>
</module-override>
<config-root>C:\DOCUME~1\user\LOCALS~1\Temp\user\.\config\deployments\virtual\plan</config-root>
</deployment-plan>
It works for me... I think your "web.xml" is causing the issue.
Try replacing
<?xml version = '1.0' encoding = 'windows-1252'?>
<!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>
with
<?xml version='1.0' encoding='WINDOWS-1252'?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
Hope this helps.
-Akshay

Want to edit web.xml page to give name of jsp file

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