In a website I have some links in a navbar like this
<li>PAGE</li>
and I load the linked page with JAVA in JSP page, then I submit data through a form by GET, after this my URL become
www.sitename.com/Servlet
then I'm not able to load the page by
?page=pagename
because I have the Servlet name in the URL.
How I can hide or delete the servlet name?
Use a servlet-mapping in your web.xml:
<servlet-mapping>
<servlet-name>Servlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
it's easier if all those JSPs are in a common path. E.g. /app/*.
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>com.example.FriendlyURLServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/app/*</url-pattern>
</servlet-mapping>
with
request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
This assumes the JSPs to be in /WEB-INF folder so that they cannot be requested directly. This will show /WEB-INF/search.jsp on http://example.com/app/search.
Related
I have a Tomcat server, serving two .war files. The first one is mapped to context /api, the second one to root /. The latter contains a single-page AngularJS app. It has the following web.xml config:
<servlet>
<servlet-name>webapp</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>
<servlet-mapping>
<servlet-name>webapp</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<error-page>
<error-code>404</error-code>
<location>/</location>
</error-page>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
When I go to https://my.url/, the index page of the single page app is properly served. So far, so good.
The problem is, when I deeplink to a page in my single page app, for example https://my.url/some/resource, Tomcat will give a 404. Because of the error-page config, it will still return the index page, but still with status 404. So, it kind of works, but not nicely.
Can I get Tomcat to return the index page with a proper 200 status code for all deep links? Of course, calls to /api should still resolve to the other deployed .war. I want to avoid duplicating the AngularJS url definitions in Tomcat, so it should just return the index page for any request that doesn't start with /api/.
For deeplinks you need to use wildcards on your webapp url pattern. Something like:
<url-pattern>/*</url-pattern>
Not sure would it not catch the /api/ calls then. Some more advanced url pattern might be required.
I want to develop a application and deploy in WebSphere where the requirement is:
if there are any request like http://appserver1:9080/ - it will reach to a landing jsp page
for example http://appserver1:9080/index.jsp
Is it at all possible to redirect to a page even if I don't mention
the resource name?
If you want to redirect from server's root, this is not about your java code or project configuration, it is about server configuration.
Look here for WebSphere configuration.
For JEE projects, on web.xml, you can define as;
<web-app>
....
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
</welcome-file-list>
</web-app>
So
http://localhost:8080/myproject
will load index.jsp
Source for details
From what you are describing, the redirection can be pretty much handled by servlets mapping. Read here:
https://docs.oracle.com/cd/E13222_01/wls/docs92/webapp/configureservlet.html
You can be able to intercept URL requests and process:
// Servlet definition on your web.xml
<servlet>
<servlet-name>ServletHandler</servlet-name>
<servlet-class>com.servlets.ServletHandler</servlet-class>
</servlet>
// This maps all requests to the above defined servlet for processing:
<servlet-mapping>
<servlet-name>ServletHandler</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
I hope this helps
You can define welcome files in web.xml
<web-app>
...
<welcome-file-list>
<welcome-file>index.jsp/welcome-file>
</welcome-file-list>
...
</web-app>
But according to the spec index.html, index.htm and index.jsp are welcome files by default. So you probably don't need to configure anything when the file is called index.jsp.
I am working with a tomcat app with multiple servlets.
I want to be able to initialize these and do dependency injection on server start up.
I understand that I will have to declare a org.springframework.web.servlet.DispatcherServlet .
But I am not sure how should my web.xml look like. Currently it looks like following:
<servlet>
<servlet-name>AddAccount</servlet-name>
<servlet-class>com.addressbook.servlets.AddAccount</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>AddAccount</servlet-name>
<url-pattern>/AddAccount</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>Login</servlet-name>
<servlet-class>com.addressbook.servlets.Login</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Login</servlet-name>
<url-pattern>/Login</url-pattern>
</servlet-mapping>
Currently, any request to add an account is sent directly to /AddAccount and for login the request is sent to /Login.
With DispatcherServler, how should my new web.xml and request structure look like? Do I have to make a new servlet that implements DispatcherServlet and forward each request to this new servlet which then forwards to correct Servlet?
You dont need multiple servlets for several kind of actions in your app. The DispatcherServlet is a front controller that handles all requests and dispatches them to your controllers. As Sotirios suggested, have a look at the Spring MVC manual first. Only the Dispatcher servlet is neccessary.
I have App engine application, with that servlet:
<servlet>
<servlet-name>Authorization</servlet-name>
<servlet-class>x.y.z.Authorization</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Authorization</servlet-name>
<url-pattern>/pattern</url-pattern>
</servlet-mapping>
everything works.
servlet invokes with this pattern http ://localhost/pattern
now I want to create AutorizationServlet default page. I need to invoke that servlet with this pattern:
http ://localhost
if I write <url-pattern></url-pattern> I have AppEngineConfigException:
Try this :
<url-pattern>/</url-pattern>
I am writing an application in JSP, and I need to remove the ".jsp" extension from the URL. For example, I need:
http://example.com/search.jsp?q=stackoverflow
To be:
http://example.com/search?q=stackoverflow
I know that this can be done using the ".htaccess" file, but I need some other way. I have tried the following code:
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>*.</url-pattern>
</servlet-mapping>
However, this did not work. Does anyone have some suggestions for ways to accomplish this? Thanks in advance for any help.
With a servlet mapping you need to specify each JSP individually like follows:
<servlet>
<servlet-name>search</servlet-name>
<jsp-file>/search.jsp</jsp-file>
</servlet>
<servlet-mapping>
<servlet-name>search</servlet-name>
<url-pattern>/search</url-pattern>
</servlet-mapping>
It's easier if all those JSPs are in a common path. E.g. /app/*.
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>com.example.FriendlyURLServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/app/*</url-pattern>
</servlet-mapping>
with
request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
This assumes the JSPs to be in /WEB-INF folder so that they cannot be requested directly. This will show /WEB-INF/search.jsp on http://example.com/app/search.
Alternatively, you can use Tuckey's URLRewriteFilter. It's much similar to Apache HTTPD's mod_rewrite.