What's going wrong here?
The ResourceConfig instance does not contain any root resource classes.
Dec 10, 2010 10:21:24 AM com.sun.jersey.spi.spring.container.servlet.SpringServlet initiate
SEVERE: Exception occurred when intialization
com.sun.jersey.api.container.ContainerException: The ResourceConfig instance does not contain any root resource classes.
at com.sun.jersey.server.impl.application.RootResourceUriRules.<init>(RootResourceUriRules.java:103)
at com.sun.jersey.server.impl.application.WebApplicationImpl._initiate(WebApplicationImpl.java:1182)
at com.sun.jersey.server.impl.application.WebApplicationImpl.access$600(WebApplicationImpl.java:161)
at com.sun.jersey.server.impl.application.WebApplicationImpl$12.f(WebApplicationImpl.java:698)
at com.sun.jersey.server.impl.application.WebApplicationImpl$12.f(WebApplicationImpl.java:695)
at com.sun.jersey.spi.inject.Errors.processWithErrors(Errors.java:197)
at com.sun.jersey.server.impl.application.WebApplicationImpl.initiate(WebApplicationImpl.java:695)
at com.sun.jersey.spi.spring.container.servlet.SpringServlet.initiate(SpringServlet.java:117)
Filter:
<filter>
<filter-name>JerseyFilter</filter-name>
<filter-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</filter-class>
<init-param>
<param-name>com.sun.jersey.config.feature.Redirect</param-name>
<param-value>true</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.JSPTemplatesBasePath</param-name>
<param-value>/views/</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.WebPageContentRegex</param-name>
<param-value>/(images|css|jsp)/.*</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>JerseyFilter</filter-name>
<url-pattern>/myresource/*</url-pattern>
</filter-mapping>
Code:
#Path ("/admin")
public class AdminUiResource {
#GET
#Produces ("text/html")
#Path ("/singup")
public Viewable getSignUp () {
return new Viewable("/public/signup", "Test");
}
}
Have you tried adding
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>my.package.name</param-value>
</init-param>
to your SpringServlet definition? Obviously replace my.package.name with the package that AdminUiResource is in and make sure it is in the classpath.
I am new to Jersey - I had the same issue, But when I removed the "/" and just used the #path("admin") it worked.
#Path("admin")
public class AdminUiResource { ... }
YOU NEED TO ADD YOUR PACKAGE NAME AT
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>your.package.name</param-value>
</init-param>
ALSO ONE SILLY THING I HAVE NOTICED,
I Need to refresh my project after MAVEN BUILD else it show me same error.Please comment If you know reason why we need to refresh project?
This means, it couldn't find any class which can be executed as jersey RESTful web service.
Check:
Whether 'com.sun.jersey.config.property.packages' is missing in your
web.xml.
Whether value for 'com.sun.jersey.config.property.packages'
param is missing or invalid (the mentioned package doesn't exists). It should be a package where you have put your POJO classes which runs as jersey services.
Whether there exists at least one POJO class, which has a method annotated with #Path attribute.
Your resource package should contain at least one pojo which is either annotated with #Path or have at least one method annotated with #Path or a request method designator, such as #GET, #PUT, #POST, or #DELETE. Resource methods are methods of a resource class annotated with a request method designator. This resolved my issue...
I ran across this problem with JBOSS EAP 6.1. I was able to deploy my code through eclipse to the JBOSS server but once I attempted to deploy the file as a WAR file to JBOSS I started getting this error.
The solution was configuring the web.xml to work properly with JBOSS by allowing the two to work together.
The following two lines were commented out in web.xml to allow JBOSS to do it's own configurations
<!--
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.your.package</param-value>
</init-param> -->
And then add the following context params after
<context-param>
<param-name>resteasy.scan</param-name>
<param-value>false</param-value>
</context-param>
<context-param>
<param-name>resteasy.scan.resources</param-name>
<param-value>false</param-value>
</context-param>
<context-param>
<param-name>resteasy.scan.providers</param-name>
<param-value>false</param-value>
</context-param>
Basically I corrected it like below and everything worked fine.
<servlet>
<servlet-name >MyWebApplication</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.feature.Redirect</param-name>
<param-value>true</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.JSPTemplatesBasePath</param-name>
<param-value>/views/</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.WebPageContentRegex</param-name>
<param-value>/(images|css|jsp)/.*</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>MyWebApplication</servlet-name>
<url-pattern>/myapp/*</url-pattern>
</servlet-mapping>
I am getting this exception, because of a missing ResourseConfig in Web.xml.
Add:
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>/* Name of Package where your service class exists */</param-value>
</init-param>
Service class means: class which contains services like: #Path("/orders")
I had the same issue with trying to run the webapp from an eclipse project. As soon I copied the .class files to /WEB-INF/classes it worked perfectly.
I had the same issue, testing a bunch of different examples, and tried all the possible solutions. What finally got it working for me was when I added a #Path("") over the class line, I had left that out.
Had the same issue and found out it was a problem with the way I deployed my source code. As the error message says: "...does not contain any root resource classes". So it couldn't find any resource classes in the configured package. I just deployed the classes wrong - that's why it didn't pick it up.
I forgot to deploy my class files in the /WEB-INF/classes directory of the WAR - initially I just had it directly in the root of the WAR file. So when it looked for resource classes it didn't find them - because they existed in a different (wrong) location.
Same issue - web.xml looked like this:
<servlet>
<servlet-name>JerseyServlet</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.mystuff.web.JerseyApplication</param-value>
</init-param>
...
Providing a custom application overrides any XML configured auto detection of classes. You need to implement the right methods to write your own code to wire up the classes. See the javadocs.
Another possible cause of this error is that you have forgotten to add the libraries that are already in the /WEBINF/lib folder to the build path (e.g. when importing a .war-file and not checking the libraries when asked in the wizard). Just happened to me.
It happened to me when I deployed my main.jar, without checking the add directory entries box in the export jar menu in Eclipse.
Well, it's a little late to reply. I have faced the same problem and my Google searches were in vain. However, I managed to find what the problem was. There might be many reasons for getting this error but I got the error due to the following and I wanted to share this with my fellow developers.
I previously used Jersey 1.3 and I was getting this error. But when I upgraded the jars to the latest version of Jersey, this issue was resolved.
Another instance in which I got this error was when I was trying to deploy my service into JBoss by building a war file. I made the mistake of including the Java files in the .war instead of java classes.
I had to add a trailing forward slash to the end of #path
#Path ("/admin/")
Ok... For me work fine just only assigning the "servlet-class" to com.sum.jersey.spi.container.servlet.ServletContainer, I am using IDE (Eclipse Mars)
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/frontend/*</url-pattern>
</servlet-mapping>
but for some reason I had to reboot my computer in order to work in my localhost. If still not work? You have to add in your web.xml this code in between "servlet" tag.
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>the.package.name</param-value>
</init-param>
"the.package.name" is the package name where you have your classes. If you are using IDE, refresh the project and run again in Tomcat. still not work? reboot your computer and will work.
Another thing to check is a combination of previous entries
You can have in your web.xml file this:
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.acme.rest</param-value>
</init-param>
and you can have
<context-param>
<param-name>resteasy.scan</param-name>
<param-value>false</param-value>
</context-param>
<context-param>
<param-name>resteasy.scan.providers</param-name>
<param-value>false</param-value>
</context-param>
<context-param>
<param-name>resteasy.scan.resources</param-name>
<param-value>false</param-value>
</context-param>
but you cannot have both or you get this sort of error. The fix in this case would be to comment out one or the other (probably the first code snippet would be commented out)
yes adding the init param for com.sun.jersey.config.property.packages fixed this issue for me.
was merging a jersey rest services into maven based spring application and got this error.
I also got this kind of error, please take care of the configurations in xml.
I wrote
com.sun.jersey.comfig.property.packages
Instead of
com.sun.jersey.config.property.packages
After correction it's working.
that issue is because jersey can't find a dependecy package for your rest
service declarated
check your project package distribution and assert that is equals to your web.xml param value
Probably too late but this is how I resolved this error.
If this solution is not working,
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>/* Name of Package where your service class exists */</param-value>
</init-param>
In eclipse:
RightClick on your Project Or Select Project and press Alt + Enter On the left-hand side of the opened window find Java Build Path
Select Libraries from the right tab panel: If there is anything which is corrupted or showing cross mark on top of the jars, remove and add the same jar again
Apply and Close
Rebuild your project
In my case I have added the jars twice in build path after importing from war.
It worked fine after removing the extra jars which was showing error deployment descriptor error pages
adding
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>service.package.name</param-value>
</init-param>
Also came accross this problem, twice for different reasons. The first time I forgot to include
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>my.package.name</param-value>
</init-param>
as described in previous comments, and once I did that, it started working.
Yet... another day I started Eclipse, expecting to continue where I left off, and instead of having my program working, it showed the very same error once again. I started checking if I accidentally had made some changes and saved corrupted file, but could find no such error and the file looked exactly like examples I have, all in order. Since it worked the day before, after some initial searching, I thought, well, maybe it's a Eclipse, or Tomcat glitch or something, so let's just try to make some changes and see if it reacts. So, I did a space + backspace in web.xml file, just to fool Eclipse that the file is changed, and saved it then. The next step was restarting Tomcat server (from Eclipse IDE) and voila, it works again!
Maybe someone with broader experience could explain what the problem really was behind all of this?
Main cause of this Exception is:
You have not given the proper package name where you using the #Path or forgot to configure in web.xml / Configuration file(Rest API Class File package Name, Your Class Package Name)
Check this Configuration inside <init-param>
When I deploy my web application (Spring MVC) on WebLogic, the welcome page does not launch with the default URL http://my.site.com/myApp. I have my welcome page under home directory and I set as follows in web.xml.
<welcome-file-list>
<welcome-file>/home/index.html</welcome-file>
</welcome-file-list>
I am able to access the page using the full URL http://my.site.com/myApp/home/index.html.
Also, if I put index.html directly under the root and update web.xml as follows, the welcome page launches with the default URL
<welcome-file>/index.html</welcome-file>
What should I do to make the default URL launch the index.html under home directory ?
Here is the code in web.xml and applicationContext.xml.
web.xml:
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
applicationContext.xml:
<mvc:default-servlet-handler/>
I assume it has something to do with the leading slash. Have you tried simply home/index.html instead of /home/index.html.
This user asked the same question but had it work without the initial slash: can-i-set-tomcat-with-a-welcome-file-in-a-subfolder
Another discussion on the same topic:
I also had the same issue.
Make sure that the file you are trying to show is outside the
web-inf folder.
There are some configurations available in weblogic.xml file but none seems to work.I think it's a security feature or something else not sure exactly what.
In my case dispatcher servlet was capturing the empty request and showing no mapping found. May be you can create a controller mapping for the same and be done with it.
The funny thing is the same thing was working flawlessly in tomcat 7
I'm trying to map my servlet to server everything off of the root URL. It looks as follows:
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
My war file is register.war. When I run this in tomcat, I'm having to use www.myurl.com/register/. What I want to use however, is www.myurl.com/
Can someone tell me how I can achieve this without renaming my register.war to root.war?
You have to go to your Tomcat configuration and change the context path for that particular web application. You will find it in
/tomcat_installation/conf/context.xml
The various elements are detailed here. You're looking for docBase and path attributes.
I am using Spring MVC with such definition:
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>0</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
But the problem that spring always trying to find /favicon.ico by default and as a result I can't render any page. How can I disable such behaviour?
Thank you
If you are using Spring Security, then make sure you have omitted the favicon request (and any other static resources) from the security filter chain.
It has nothing to do with spring mvc but that is the default behaviour of the browser you are using. Also that should not break anything at all and your app should work as normal even if it does not have an ico. (Unless you specifically coded to make it fail in case of missing ico)