I've been struggling with this web xml file for Tomcat for a while.
<context-param>
<param-name>name</param-name>
<param-value>Bob</param-value>
</context-param>
<servlet>
<servlet-name>test</servlet-name>
<servlet-class>TestServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>test</servlet-name>
<url-pattern>/servlet/*</url-pattern>
</servlet-mapping>
I know that this file is being read because when i test using
public class TestServlet extends HttpServlet{
public void doGet(HttpServletRequest req, HttpServletResponse res){
res.setContentType("text/html");
String name = getServletContext().getInitParameter("name");
PrintWriter out = null;
try{
out = res.getWriter();
}catch(Exception e){}
out.println("<html><head></head><body><img src=\"/twitter.png\"><p>Hi my name is " + name + "</p></body></html>");
}
}
I am able to read the name I put into the context-param. My question for you guys is how do I create a URL mapping such that I do not have to go through /servlet/ to access my servlets in the URL? When I try to make a url-pattern such as
/test/*, i cannot access the servlet even if i say website/test/TestServlet. I get a 404 Not Found error from the browser
Put the servlet class in a package and do not rely on InvokerServlet. It's disabled since Tomcat 5.5 and removed in Tomcat 7.0.
Please do not read 10~15 year old tutorials/books. Technology changes every year.
See also:
How to invoke a servlet without mapping in web.xml?
Related
In our app for all the notification we trigger through mail.
All the templates have non sso link
>/Userlogin?param1=param2value¶m2=param2value">Link to access app
I need to modify this link in all templates to
>/Userloginsso?param1=param2value¶m2=param2value">Link to access app
Since there are many templates and takes lot of manual effort, is there any way we can redirect the request of Userlogin to Userloginsso. Any configuration that we can do in web.xml ?
Considering you have a mapping for Userlogin in web.xml as below:
<web-app>
<servlet>
<servlet-name>Userlogin</servlet-name>
<servlet-path>com.something.Userlogin</servlet-path>
</servlet>
<servlet-mapping>
<servlet-name>Userlogin</servlet-name>
<url-pattern>/Userlogin</url-pattern>
</servlet-mapping>
</web-app>
Modify existing mapping to :
<web-app>
<servlet>
<servlet-name>Userloginsso</servlet-name>
<servlet-path>com.something.Userloginsso</servlet-path>
</servlet>
<servlet-mapping>
<servlet-name>Userloginsso</servlet-name>
<url-pattern>/Userlogin</url-pattern>
</servlet-mapping>
</web-app>
Now all calls to Userlogin will be redirected to Userloginsso servlet.
You could do a simple redirect in your UserLogin servlet with the following:
public void doGet (HttpServletRequest request, HttpServletResponse response) throws IOException {
String param1 = request.getParameter ("param1");
String param2 = request.getParameter ("param2");
// other parameters
// Build the new url: if too much parameters, prefer using a StringBuilder over String concatenation for better performances
String baseUrl = request.getContextPath () + "/Userloginsso?param1=" + param1 + "¶m2=" + param2;
String encodedUrl = response.encodeRedirectURL (baseUrl);
response.sendRedirect (encodedUrl);
}
If I understand your question correctly you could use a filter example here get the url and foward it somplace else in your app. Or and url rewrite library such us this one
If you still want a servlet you could use a ProxyServlet. There are already many good implementations.
Examples:
Complex proxy servlet with all features
Simple proxy servlet, limited features
This question already has an answer here:
Using special auto start servlet to initialize on startup and share application data
(1 answer)
Closed 1 year ago.
I would like to load java class file that contain db related function. How can I load that java file while starting the tomcat server
you can use servlet for that as below, define into web.xml
<servlet>
<servlet-name>YourServletName</servlet-name>
<servlet-class>com.abc.xyz.YourServletClassName</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
in YourServletClassName.java file you can write your code.
Hope it helps you.
The answer by Psabuwala is correct but not complete.
Code that will run on startup should be placed in the init method of the servlet.
Web.xml:
...
<servlet>
<servlet-name>mainServlet</servlet-name>
<servlet-class>example.com.MainServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
...
MainServlet.java:
public class MainServlet extends HttpServlet
{
public void init() throws ServletException
{
DataLoader dataLoader = new DataLoader();
dataLoader.load();
}
...
}
I wanted to create a servlet in liferay that is listening to a URL such as
http://localhost:8080/my-servlet
I tried to add it to a portlet but the I have the URL
http://localhost:8080/my-portlet/my-servlet
I tried to add my servlet description to the web.xml of ext-web, but no luck.
Is there any way to add such a servlet ?
If you want to access Liferay service API, you may consider using PortalDelegateServlet : adding the following to your web.xml:
<servlet>
<servlet-name>myServlet</servlet-name>
<servlet-class>com.liferay.portal.kernel.servlet.PortalDelegateServlet</servlet-class>
<init-param>
<param-name>servlet-class</param-name>
<param-value>org.example.MyServlet</param-value>
</init-param>
<init-param>
<param-name>sub-context</param-name>
<param-value>myservlet</param-value>
</init-param>
</servlet>
will make your servelt accessible through
http://example.org/delegate/myservlet
in your servlet class, you then do things like extract the logged-in user and check permissions:
package org.example;
public class MyServlet extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
User user = PortalUtil.getUser(request);
PermissionChecker permissionChecker = PermissionCheckerFactoryUtil.create(user);
...
Liferay is also "Servlet"-Application - but a very-very big one. And Liferay need some servlet container like tomcat, jetty, jboss etc.
However, you can simple create servlet project and deploy it direct to servlet container where liferay is running.
edit: and put to web.xml by servlet-mapping a direct access like "/*".
http://127.0.0.1:8888/socialnetwork/contactsService
That's the current URL for one of my servlets. My question is, how do I change it?
In my web.xml file, altering
<servlet-mapping>
<servlet-name>contactsServiceServlet</servlet-name>
<url-pattern>/socialnetwork/contactsService</url-pattern>
</servlet-mapping>
to
<servlet-mapping>
<servlet-name>contactsServiceServlet</servlet-name>
<url-pattern>/a/contactsService</url-pattern>
</servlet-mapping>
Makes absolutely NO difference to the URL it requests when I make an RPC-call to the servlet.
Once you have done the above you need to change where you invoke (Which is described in the Annotation below) as in...
// The RemoteServiceRelativePath annotation automatically calls setServiceEntryPoint()
#RemoteServiceRelativePath("email")
public interface MyEmailService extends RemoteService {
void emptyMyInbox(String username, String password);
}
See http://google-web-toolkit.googlecode.com/svn/javadoc/1.6/com/google/gwt/user/client/rpc/RemoteServiceRelativePath.html
This question already has answers here:
Change default homepage in root path to servlet with doGet
(2 answers)
Closed 7 years ago.
I want my SampleServlet to be called first whenever my java web application is accessed in this manner :
http://server:8080/appname/
Is there any way to implement this?
Use a Servlet filter to call your Servlet.
If you want to make a servlet your homepage then this worked for me on http://feelitlive.com/
<welcome-file-list>
<welcome-file>homepage</welcome-file>
</welcome-file-list>
...
<servlet>
<description>Shows stuff on the homepage</description>
<display-name>Homepage Servlet</display-name>
<servlet-name>HomepageServlet</servlet-name>
<servlet-class>com.cantorva.gigcalendar.servlets.HomepageServlet</servlet-class>
</servlet>
...
<servlet-mapping>
<servlet-name>HomepageServlet</servlet-name>
<url-pattern>/homepage</url-pattern>
</servlet-mapping>
That means that that users arriving at your application via the URL you specified will be welcomed by your servlet. It also creates an alias for the homepage at "/homepage" but you don't have to use that.
If you want to run some code on start-up then asalamon74's answer looks right.
Not sure what you mean but you need to map your servlet to "/"
<servlet-mapping>
<servlet-name>SampleServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
Not sure what is your aim, but web application initialization can be achieved by ServletContextListener:
public class AppListener implements ServletContextListener {
public void contextInitialized(ServletContextEvent event) {
// place your code here
}
public void contextDestroyed(ServletContextEvent event) {
}
}
and later in web.xml:
<web-app>
<listener>
<listener-class>
package.AppListener
</listener-class>
</listener>
...
</web-app>
If you want to run code on start-up indeed asalamon74's answer should be fine. If you have a legacy situation and you must use a servlet, the parameter load-on-startup can do the trick for you:
<servlet>
<servlet-name>SampleServlet</servlet-name>
<display-name>SampleServlet</display-name>
<description>Sample Servlet</description>
<servlet-class>...</servlet-class>
<init-param>...</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
The load-on-startup tag specifies that the servlet should be loaded automatically when the web application is started; the number value just gives a loading order to those loading on startup. If no value is specified, the servlet will be loaded when the container decides it needs to be loaded - typically on it's first access.