So after installing UrlRewriteFilter, I set up a rule and a corresponding velocity template. And when I go to the test page, the velocity script is shown as raw code instead of being compiled.
example of the code for the rule:
<rule>
<from>/test/([0-9]+)</from>
<to>/downloads/test.vm?Id=$1</to>
</rule>
example of the urlrewritefilter
<filter>
<filter-name>UrlRewriteFilter</filter-name>
<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
<init-param>
<param-name>logLevel</param-name>
<param-value>DEBUG</param-value>
</init-param>
</filter>
any ideas why the velocity template wouldn't render even though the rule executed correctly? All other vm pages render correctly (when accessed directly) and the rewrite works perfectly with jsp.
This happens when the test.vm is served by the default servlet for static content.
I suspect you changed the mapping for velocity somehow. If you are using Velocity Tools, you should have a mapping like this,
<servlet>
<servlet-name>velocity</servlet-name>
<servlet-class>
org.apache.velocity.tools.view.VelocityViewServlet
</servlet-class>
</servlet>
<!-- Map all *.vm files to Velocity -->
<servlet-mapping>
<servlet-name>velocity</servlet-name>
<url-pattern>*.vm</url-pattern>
</servlet-mapping>
Another possibility is that other filter might interfere with UrlRewrite filter. It would be helpful if you can post your web.xml.
Related
Currently, we have "root" (/) mapped to a static index.html page, but we want to upgrade to a jsp to have dynamic content. Trying to figure out how to do this. We have content that is mapped to the default content server (e.g. /css), so we don't want to change too much.
We tried:
Changing the .html to .jsp. This resulted in a blank page.
Changing the .html to .jsp and then moving the file into the WEB-INF directory. This resulted in a 404.
Trying to subclass the DefaultServlet class that content servlet is currently mapped to. This through a 500, with a class assertion error (it checked to see if it was the same class).
Adding another servlet to that url, but it overwrote the current one.
I've searched StackOverflow, but still haven't found an answer that works.
Thanks!
If I understand your question correctly, this is trivial using Spring MVC:
<mvc:default-servlet-handler/>
And in web.xml:
<servlet>
<servlet-name>spring-mvc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<async-supported>true</async-supported>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:/META-INF/spring/your-applicationContext-web.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-mvc</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
Is this what you have tried already?
Just set up a controller method mapped to / that returns a view name, which is your jsp file. And make sure your view resolver is set up correctly. Any of the spring mac tutorial hellos world programs out there will show how.
How can we access web application context url from inside the CSS/JS file on java web server?
We can map URL's (for background image for example) relatively with url('../img/bg.gif') etc. but this will not work with web application with mappings like:
/shop/
/shop/show/3/
/shop/payment/
because browser will search for this file relative to current "virtual" directory.
Also, we can't universally map image URL global like url('/images/bg.gif'), because we forces deployment in top-level directory like example.com/ (not example.com/myproject/).
How to avoid changing CSS/JS(ajax) URL's when changing application context URL?
It is possible to access aplication context in CSS file in easy way like accessing contextPath in default servlet wich serves those static files?
You could do something like this (assuming you're using Tomcat, if not, look up the doc to find the correct servlet name to add the mapping):
<servlet>
<servlet-name>jsp</servlet-name>
<servlet-class>org.apache.jasper.servlet.JspServlet</servlet-class>
<init-param>
<param-name>fork</param-name>
<param-value>false</param-value>
</init-param>
<init-param>
<param-name>xpoweredBy</param-name>
<param-value>false</param-value>
</init-param>
<load-on-startup>3</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>*.css</url-pattern>
</servlet-mapping>
Then you can add jsp el code in your css:
url('${pageContext.request.contextPath}/images/bg.gif')
This will make your CSS files serve MUCH slower, so take into account if you can do it with absolute pathing at the root. We've used this approach in an app that allowed skinning of the CSS, so that we could output custom color schemes.
I'm at the end of my rope on this one. I'm try to get a super simple webapp up and I can't seem to get tomcat to not 404 static files.
I'm using the gradle tomcat plugin with tomcat version 7.0.39
My html file is at hey-world/src/main/webapp/index.html
My web.xml looks like this:
<servlet>
<servlet-name>Resteasy</servlet-name>
<servlet-class>
org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>HeyWorldApplication</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>/static/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>Resteasy</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
So I thought this setup would map localhost:8080/hey-world/static/index.html to the file, but it 404's everytime. Is this a problem with some convention in the gradle tomcat plugin?
The URL-patterns used in web.xml/servlet-mapping is often a little simplistic. I think in your case, the /* pattern for Resteasy will work as a catch-all, so that no other mapping will really matter.
For debugging, I suggest you remove the Resteasy-servlet altogether, and see if you can serve static files from a custom URL with your mapping.
If that works, re-enable Resteasy, but on a different URL-pattern (eg. /rest/*).
If that works, well, then everything really works fine, it's just that the URL-mapping for /* blocks anything else from working.
The easiest solution would probably be to server static files as per default (no mapping), and serve rest-stuff from another URL.
Alternatively use two web apps. One with context root /static, one with context root /.
Assume it is a struts project..
the filter configuration is as follows,
<filter>
<filter-name>samplefilter</filter-name>
<filter-class>org.samplepack.SampleFilterXXX</filter-class>
</filter>
<filter-mapping>
<filter-name>samplefilter</filter-name>
<servlet-name>action</servlet-name>
</filter-mapping>
and the servlet mapping is as follows,
<servlet-mapping>
<servlet-name>action</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
I want to know whether the samplefilter will be executed for every action class in the struts project?
can you post the situations whereever i can apply this effectively ?
The simple answer is Yes, provided that all your struts actions are accessed via the pattern *.do. This is the default configuration for Struts, so I expect that this is the case.
However as the Struts mapping us configurable, it is possible for you to define a different mapping to access a Struts action, and therefore your filter will not be picked up. But, as mentioned above, this is uncommon practice, so I expect you will be okay.
I have two servlets defined in the web.xml file, namely the default2 and myservlet. The default2 servlet is used to map the static files like the javascript and css. The myservlet is used for getting dynamic content.
<servlet>
<servlet-name>default2</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>myservlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:my-servlet.xml
</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
The servlet mapping is defined as follows
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>default2</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
When i try to access any files under /resources, i get a 404. Any ideas why this config is not working or change this config to make it work.
Tomcat's default servlet before 6.0.30 actually serves a static resource identified by HttpServletRequest.getPathInfo(), so that /style.css will be returned when /resources/style.css is requested.
Tomcat's behavior has changed from version 6.0.30 onwards. So the original configuration from the question works in newer versions of Tomcat. See "50026: Add support for mapping the default servlet to URLs other than /. (timw)" in the changelog.
Jetty's default servlet uses a full path.
It should work fine. Are those files in real also located in the /resources folder?
Your web.xml looks correct (except I would change your <load-on-startup> constants).
Make sure that your /resources exists and is a publicly visible folder in your project path and not under /WEB-INF folder.
Try changing your url-pattern for myservlet to /, and optionally adding <mvc:default-servlet-handler /> (see here) to your Spring configuration.
Removed wrong portion of the answer as per #BalusC comment.
Set a break point in your servlet and perform a debug session. Look for the path that your servlet is picking up these files at. Make sure they match the location