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How to set a condition for different servlets in web.xml

I am new in Servlets and web.xml content. I would like to set a condition/dependency on two different servlets.
Consider I have below two servlets:
<servlet>
<servlet-name>servlet1</servlet-name>
<servlet-class>com.my.app.Servlet1</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>servlet2</servlet-name>
<servlet-class>com.my.app.Servlet2</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
And I would like to invoke servlet2 only in case servlet1 encounters an error or fails. As to be a scenario: Assume servlet1 is my main authenticator and the user login failed due to missing parameter in URL in servlet1. In that case I would like to try the second authenticator servlet2 that contains different set of checks...
Can I achieve this by modifying web.xml only?
If not, what would be an alternative solution to accomplish this need?
Note: I do have the source code of servlet1 but not the servlet2 (only built jar)...
Thanks in advance.

You could do this using reuqestDispatcher and forward the request to the other servlet2 like following in Servlet 1s doPost method.
//Here URL is path to servlet 2
RequestDispatcher rd = request.getRequestDispatcher(“url”);
rd.forward(request,response);

url-pattern /* serves jsp files and static assets as seperate requests

I have a simple servlet which renders different content based on different geo locations
for example:
example.com/us
example.com/de
example.com/uk
example.com/..
so actually the servlet should supports all geo locations as url-pattern. So thats why I defined the rule below in my web.xml file:
<servlet-mapping>
servlet-name>MyServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
this is how the servlet look like:
public class MyServlet extends HttpServlet{
String showPage = "/pages/show.jsp";
public void doGet(HttpServletRequest request, HttpServletResponse response){
System.out.println("servlet initiated... ~> "+request.getRequestURI());
...
request.getRequestDispatcher(showPage).forward(request, response);
}
}
As soon as I try to forward the page, the servlet starts again and again; it tries to also handle the request /page/show.jsp.
So I get such an ouput when I try to access example.com/us:
servlet initiated... ~> /us
servlet initiated... ~> /page/show.jsp
Exception...
servlet initiated... ~> /page/show.jsp
Exception...
servlet initiated... ~> /page/show.jsp
...
It throws an exception because my servlet is not expecting such a url-pattern /page/show.jsp but it triggers because I have define /* url-pattern in my web.xml file.
Any idea? how can I skip or exclude the unwanted requests like the one above? Thanks.

I could not find a specific reference but I believe that "/*" root url pattern conflicts with the default mapping in the the global web.xml which could be contributing to your problem.
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
<init-param>
<param-name>debug</param-name>
<param-value>0</param-value>
</init-param>
<init-param>
<param-name>listings</param-name>
<param-value>false</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<!-- The mapping for the default servlet -->
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
try adding explicit jsp servlet tag and servlet mapping to avoid processing the jsp through the default mapping which is created with the "/" root url pattern. I'm not sure that this will solve your problem.
<servlet>
<servlet-name>showpage</servlet-name>
<jsp-file>/pages/show.jsp</jsp-file>
</servlet>
<servlet-mapping>
<servlet-name>showpage</servlet-name>
<url-pattern>/show</url-pattern>
</servlet-mapping>
Another more explicit solution would be to create a url pattern that differentiates all of the country geo requests.
<servlet-mapping>
<servlet-name>MyServlet</servlet-name>
<url-pattern>/geo/*</url-pattern>
</servlet-mapping>
So example.com/geo/us etc will map explicitly
I attempted the same global setting and had a similar problem where it recursively attempted the request and threw an exception over and over until I stopped the Tomcat server. The problem was resolved when used more specific url pattern.

Bart answer gave me the clue. Actually there are many ways to handle this, one is to use Filters and the other one is to define all the static routes in web.xml and instead of the /* I have used /.
So to resolve the issue I change the url-pattern as below:
<!-- static assets -->
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>*.js</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>*.css</url-pattern>
</servlet-mapping>
<!-- my servlet -->
<servlet-mapping>
servlet-name>MyServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
You also don't need to define the pattern for org.apache.catalina.servlets.DefaultServlet its already there by default.
So After above changes everything works fine.

When resolving pages/show.jsp the container will match on the longest path first then look at the file extension (.jsp) afterwards. So pages/show.jsp matches /* and therefore the container loads the Servlet again and doesn't initiate the JSP engine servlet.
Are all your jsps in a pages directory? Maybe you can add that mapping to your web.xml
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>/pages/*</url-pattern>
</servlet-mapping>

Per the servlet specification (2.5 here, but it should be the smae for newer version) :
The container will try to find an exact match of the path of the request to the
path of the servlet. A successful match selects the servlet.
The container will recursively try to match the longest path-prefix. This is done
by stepping down the path tree a directory at a time, using the ’/’ character as
a path separator. The longest match determines the servlet selected.
If the last segment in the URL path contains an extension (e.g. .jsp), the servlet
container will try to match a servlet that handles requests for the extension.
An extension is defined as the part of the last segment after the last ’.’ character.
If neither of the previous three rules result in a servlet match, the container will
attempt to serve content appropriate for the resource requested. If a "default"
servlet is defined for the application, it will be used.
(source : servlet-spec-2.5)
So for your case, it should be sufficient to add a more precise mapping (as said by Si Kelly) :
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>/pages/show.jsp</url-pattern>
</servlet-mapping>

How to transmit parameter to redirected jsp that this parameter is accesable into this jsp

I have this controller's method:
#RequestMapping(value="/path", method = RequestMethod.GET )
public String path(Model model, RedirectAttributes redirectAttributes) {
redirectAttributes.addFlashAttribute("attr", "valueFromPath");
return "redirect:jspPage.jsp";//this page located in webApp folder
}
jspPage.jsp:
...
<h1>${attr}</h1>
...
In my case this row is empty, but I want that valueFromPath will shown on this jsp.
How I can make it?
web.xml:
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
...
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>

I'm going to make an assumption here. Because of your URL mapping of
<url-pattern>/*</url-pattern>
The DispatcherServlet is not handling the request for jspPage.jsp. That jobs falls on the default Servlet. As such, the DispatcherServlet cannot perform the logic that adds the flash attributes from the HttpSession attributes back into the request attributes.
You need to make sure that the request for jspPage.jsp is handled by the DispatcherServlet. Either by moving the file to some other location or changing your url-pattern to / and providing a handler.

Servlet mapping: url-pattern for URLs with trailing slash

I have a problem related to the servlet mapping. I have the following in web.xml:
<servlet>
<servlet-name>HelloWorldServlet</servlet-name>
<servlet-class>test.HelloWorldServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorldServlet</servlet-name>
<url-pattern>/HelloWorld</url-pattern>
</servlet-mapping>
If I access to http://localhost:<port>/MyApp/HelloWorld the servlet HelloWorldServlet is called.
I also want my servelet to respond to http://localhost:<port>/MyApp/HelloWorld/. How can I achieve this effect? I'm developing with NetBeans but it does not allow me to put a pattern ended with /.

After you've added your wildcard on your <url-pattern>
<url-pattern>/HelloWorld/*</url-pattern>
You can get the extra path associated with the URL by using HttpServletRequest.getPathInfo().
E.g.
http://localhost:<port>/MyApp/HelloWorld/one/
The result will be
/one/
From the JavaDoc:
Returns any extra path information
associated with the URL the client
sent when it made this request. The
extra path information follows the
servlet path but precedes the query
string and will start with a "/"
character.

Use a wildcard. You can redirect all the traffic going to a specific URL to the same servlet. For example, you can add the following:
<servlet-mapping>
<servlet-name>HelloWorldServlet</servlet-name>
<url-pattern>/HelloWorld/*</url-pattern>
</servlet-mapping>
This will redirect the URL with a slash to your original servlet.
One thought - this would redirect anything to this URL pattern to the servlet. If you want to have other URL's past this URL, you should create a servlet that will redirect to the correct URL (by looking at the URL specified). Alternatively, you could use a framework that provides mapping for you.

Call a servlet on click of hyperlink

Is there a way to call a Java Servlet on click of hyperlink without using JavaScript?

Make the hyperlink have a URL that you have a servlet mapping defined for in the web.xml file.
The servlet-mapping element defines a mapping between a servlet and a URL pattern. The example below maps the servlet named myservlet to any URL that starts with /foo:
<servlet>
<servlet-name>myservlet</servlet-name>
<servlet-class>com.stackoverflow.examples.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/foo/*</url-pattern>
</servlet-mapping>
For this example, a hyperlink such as Click Me would invoke the servlet.

you declare your servlet in web.xml by setting its name, class and url-pattern
(let's say your url-pattern is /myServlet)
write mylink
override the doGet(..) method of the servlet to do whatever you want

Think that you've defined a servlet "callme" and web.xml has been configured for this servlet. Use the following syntax to call it using hyperlink
web.xml
<servlet>
<description>callme Functions</description>
<display-name>callme</display-name>
<servlet-name>callme</servlet-name> <servlet-class>com.test.Projects.callme</servlet-
class>
</servlet>
<servlet-mapping>
<servlet-name>callme</servlet-name>
<url-pattern>/callme</url-pattern>
</servlet-mapping>
in JSP:
Call the servlet

What exactly do you mean with "call a Java Servlet? The most normal (i.e. without any JavaScript magic) browser behaviour for clicking on a link is to send a HTTP request to fetch the document at the URL specified in the link and display it - and Servlets exist to respond to HTTP requests.
So you don't have to do anything special at all. Just have a regular HTML link and make sure that the servlet you want to "call" corresponds to that link's URL. Of course the next question is what that Servlet returns and what you want the browser to do with it.