I have the following problem:
In my web.xml I define how to serve pictures like so:
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>*.css</url-pattern>
<url-pattern>*.png</url-pattern>
<url-pattern>*.jpg</url-pattern>
</servlet-mapping>
On Tomcat this works fine but on IBM Websphere 8.5.5.9 I get the following error:
"SRVE0303E: Servlet name for the servlet mapping *.css could not be found."
On another site I already found a solution for this issue (https://www.ibm.com/developerworks/community/forums/html/topic?id=5f4420ba-0754-43fe-8c87-91acc588d9fc) so I also created the ibm-web-ext.xml exactly the same as in their solution but the error still persists.
Does anyone know what I could do differently?
I found the answer to my problem, thanks to #MigratedPigeon because he got me thinking about the class of my default servlet.
A tomcat server has a default servlet, the class for tomcats default servlet is
org.apache.catalina.servlets.DefaultServlet
Websphere on the other hand does not have a default servlet, thats why I get the error "Servlet Name could not be found".
As in the answer I linked in the original question, static file serving can be activated by websphere by using the web-ext.xml file but that still did not solve the issue of my web.xml file having a "default" servlet.
In our application we use spring, so in the end I replaced the default servlet in web.xml with springs dispatcher servlet and now my web.xml file is valid for both, tomcat and websphere.
you should also mention this in your web.xml
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>your servlet class</servlet-class>
</servlet>
Servlet mapping is done based on the value mentioned in the "" tags and respective servlet will be called.
Related
We run an application written with "jspx" (Java Server Page with XML I guess), it runs on web-logic but the web-logic is down currently.
I wish to know if I could copy the files and put it under Apache tomcat.
I have actually tried that but I got some errors which makes me feel Apache tomcat is strictly for "jsp" and not "jspx".
If my assuption is right then what else can I use to compile a ".jspx" program aside from weblogic?
As far as I know, jspx are simply jsp files with well-formed XML instead of "just any html".
Try editing your tomcat/conf/web.xml and add another mapping for the jsp-servlet:
<!-- Existing mapping -->
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>*.jsp</url-pattern>
</servlet-mapping>
<!-- New mapping -->
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>*.jspx</url-pattern>
</servlet-mapping>
But... for Tomcat 8 this is already in place. Can you share the error messages you got?
I have deployed Myapplication.war in tomcat webapps directory.
now i have, index.jsp in $CATALINAHOME/webapps/Myapplication and process.class in $CATALINAHOME/webapps/Myapplication/WEB-INF/classes.
When index.jsp post some variables to process, http://x.x.x.x:8080/Myapplication/process
Im getting below err,
type Status report
message /Myapplication/process
description The requested resource is not available..
if i convert the process file from java class to jsp, im able to post from index.jsp to process.jsp.
How can i achieve this? any other settings i need to do here?
Thanks in advance
You'll have to declare your servlet in web.xml otherwise Tomcat won't know which class to associate with which path:
<servlet>
<servlet-name>processServlet</servlet-name>
<servlet-class>process</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>processServlet</servlet-name>
<url-pattern>/process</url-pattern>
</servlet-mapping>
See also the tomcat sample deployment descriptor
A couple of minor niggles:
As per Java class naming, classes should start with an uppercase character, so it would be Process and Process.java.
Usually it makes more sense to put java classes into a package.
I have a WAR file with a web application that has been deployed to a weblogic stream.
The JSP part works fine, but it can't find the servlets. Possible due to the lack of mapping in my web.xml file.
I was working fine on Tomcat 6, but can't seem to find using weblogic.
I used annotation #WebServlet("/actionOne") but this doens't seem to work.
I am a little confused about how to map these correctly via the web.xml file.
the servlets are .java files and located at WEB-INF/classes/com/foo/bar/
So far I have added the following the web.xml file but the servlet-mapping section has me confused.
<servlet>
<servlet-name>actionOne</servlet-name>
<servlet-class>com.foo.bar.actionOne</servlet-class>
</servlet>
<servlet>
<servlet-name>actionTwo</servlet-name>
<servlet-class>com.foo.bar</servlet-class>
</servlet>
Hopefully the above is correct, the next section I'm not sure how to use and would appreciate some help.
<servlet-mapping>
<servlet-name>actionOne</servlet-name>
<url-pattern>/actionOne</url-pattern>
</servlet-mapping>
The servlets are being called from the jsp via a Form action="actionOne"
My mapping was correct, it seems the issue was related to a different version of servlet.api in the weblogic modules folder. 2.5 instead of 3.0. This resolved the issue.
I'm trying to write a web application using SpringMVC. Normally I'd just map some made-up file extension to Spring's front controller and live happily, but this time I'm going for REST-like URLs, with no file-name extensions.
Mapping everything under my context path to the front controller (let's call it "app") means I should take care of static files also, something I'd rather not do (why reinvent yet another weel?), so some combination with tomcat's default servlet (let's call it "tomcat") appears to be the way to go.
I got the thing to work doing something like
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>tomcat</servlet-name>
<url-pattern>*.ext</url-pattern>
</servlet-mapping>
and repeating the latter for each one of the file extensions of my static content. I'm just wondering why the following setups, which to me are equivalent to the one above, don't work.
<!-- failed attempt #1 -->
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>tomcat</servlet-name>
<url-pattern>*.ext</url-pattern>
</servlet-mapping>
<!-- failed attempt #2 -->
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>tomcat</servlet-name>
<url-pattern>/some-static-content-folder/*</url-pattern>
</servlet-mapping>
Can anyone shed some light?
I think I may know what is going on.
In your working web.xml you have set your servlet to be the default servlet (/ by itself is the default servlet called if there are no other matches), it will answer any request that doesn't match another mapping.
In Failed 1 your /* mapping does appear to be a valid path mapping. With the /* mapping in web.xml it answers all requests except other path mappings. According to the specification extension mappings are implicit mappings that are overwritten by explicit mappings. That's why the extension mapping failed. Everything was explicitly mapped to app.
In Failed 2, App is responsible for everything, except content that matches the static content mapping. To show what is happening in the quick test I set up. Here is an example. /some-static-content-folder/ contains test.png
Trying to access test.png I tried:
/some-static-content-folder/test.png
and the file was not found. However trying
/some-static-content-folder/some-static-content-folder/test.png
it comes up. So it seems that the Tomcat default servlet (6.0.16 at least) drops the servlet mapping and will try to find the file by using the remaining path. According to this post Servlet for serving static content Jetty gives the behavior you and I were expecting.
Is there some reason you can't do something like map a root directory for your rest calls. Something like app mapped to /rest_root/* than you are responsible for anything that goes on in the rest_root folder, but anywhere else should be handled by Tomcat, unless you make another explicit mapping. I suggest setting your rest servlet to a path mapping, because it declares the intent better. Using / or /* don't seem appropriate, since you have to map out the exceptions. Using SO as an example, my rest mappings would be something like
/users/* for the user servlet
/posts/* for the posts servlet
Mapping order
Explicit (Path mappings)
Implicit (Extension mappings)
Default (/)
Please correct anything that I got wrong.
For reference, the "failed attempt #2" is perfectly correct in version of Tomcat >= to 6.0.29.
It was the result of a Tomcat bug that get fixed in version 6.0.29:
https://issues.apache.org/bugzilla/show_bug.cgi?id=50026
<!-- Correct for Tomcat >= 6.0.29 or other Servlet containers -->
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>/some-static-content-folder/*</url-pattern>
</servlet-mapping>
I've never tried to map a servlet like this, but I would argue that /* does technically both start with / and end with /*, even though the same character is used for both matches.
I have a JavaEE 1.4 web application running on WebSphere Application Server 6.0. In web.xml, there is a servlet configured to intercept all server requests:
<servlet-mapping>
<servlet-name>name</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
This works fine until I try to request something ending with *.jsp. In this case, server tries to find JSP with this name and fails with the error:
java.io.FileNotFoundException: JSPG0036E: Failed to find resource /cfvct/search_criteria.jsp
at com.ibm.ws.jsp.webcontainerext.JSPExtensionProcessor.findWrapper(JSPExtensionProcessor.java:279)
at com.ibm.ws.jsp.webcontainerext.JSPExtensionProcessor.handleRequest(JSPExtensionProcessor.java:261)
at com.ibm.ws.webcontainer.webapp.WebApp.handleRequest(WebApp.java:3226)
at com.ibm.ws.webcontainer.webapp.WebGroup.handleRequest(WebGroup.java:253)
at com.ibm.ws.webcontainer.VirtualHost.handleRequest(VirtualHost.java:229)
at com.ibm.ws.webcontainer.WebContainer.handleRequest(WebContainer.java:1970)
at com.ibm.ws.webcontainer.channel.WCChannelLink.ready(WCChannelLink.java:120)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.handleDiscrimination(HttpInboundLink.java:434)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.handleNewInformation(HttpInboundLink.java:373)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.ready(HttpInboundLink.java:253)
at com.ibm.ws.tcp.channel.impl.NewConnectionInitialReadCallback.sendToDiscriminaters(NewConnectionInitialReadCallback.java:207)
at com.ibm.ws.tcp.channel.impl.NewConnectionInitialReadCallback.complete(NewConnectionInitialReadCallback.java:109)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.requestComplete(WorkQueueManager.java:566)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.attemptIO(WorkQueueManager.java:619)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.workerRun(WorkQueueManager.java:952)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager$Worker.run(WorkQueueManager.java:1039)
at com.ibm.ws.util.ThreadPool$Worker.run(ThreadPool.java:1475)
I need to have this request processed by the servlet, but seems server uses some JSPExtensionProcessor to process all paths ending with .jsp. Is there any way to change this behaviour?
Yes, you'll need to map your servlet to *.jsp in order to get *.jsp support redirected to your servlet.
<servlet-mapping>
<servlet-name>name</servlet-name>
<url-pattern>*.jsp</url-pattern>
</servlet-mapping>
It is normally a bad idea to have jsps accessible directly, however. Placing them in WEB-INF in some directory, then mapping an appropriate url (.do, .action, etc) to a servlet that then redirects internally to that JSP is the better practice.
So instead of typing thisUrl.jsp, the user would type thisUrl.do or thisUrl.action, and it would then get hit by the servlet to redirect to thisUrl.jsp.