I am getting HTTP Status 404 when running simple RESTful web service in Java using Jersey. I am following the tutorial REST with Java (JAX-RS) using Jersey. I have copied all the jars that I downloaded from Jersey download site to WEB-INF/lib folder of my project(please see the screenshot for jars).
When I run the application from eclipse development environment Eclipse Console shows that Tomcat was started successfully. My web-app is deployed and I can see index.html coming up. But hitting http://localhost:8080/com.kj.rest.jersey.first/ gives Http Status 404.
My Environment:
Spring Tool Suite as my eclipse dev environment
Jersey 2.22.2 jars
Apache Tomcat v8.0
Please note I am not using Maven in my project and I also looked at other similar questions here but none of them solved my issue.
What am I missing, where should I look for the issue, which logs?
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<!-- Register resources and providers under com.vogella.jersey.first package. -->
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.*******</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
The mistake that I was making was to not specify the service name(specified by #Path annotation) in the URL. After doing that it worked.
So essentially the URL to hit should be http://localhost:8080/com.kj.rest.jersey.first/rest/path_from_rest_class and I missed the path_from_rest_class earlier.
I am trying to deploy a RESTful web service in JBOSS7.1.1 and it is giving me the error below:
ERROR [org.apache.catalina.core.ContainerBase.[jboss.web].[default-host].[/anthut]] (MSC service thread 1-1) Servlet /anthut threw load() exception: java.lang.ClassCastException: com.sun.jersey.spi.spring.container.servlet.SpringServlet cannot be cast to javax.servlet.Servlet.
The version of jersey that I am using in my project is 1.18.3.
A section of my web.xml is given below
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>abdfserve</servlet-name>
<servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
please what could be responsible for this, I have being on this for some days now...I need help please..
This can happen when you have multiple versions of the servlet API in your classpath.
The servlet API is always provided by the web application server, you must not include it in your own web application bundles.
You need it as a compile-time dependency during development, but it should not be deployed. If you use Maven, make sure it has "provided" scope.
I am trying to build a simple hello world application for two days using Jersey + Google app engine. For simple AppEngine project I followed these tutorials and both works just fine
https://developers.google.com/appengine/docs/java/gettingstarted/creating
https://developers.google.com/appengine/docs/java/webtoolsplatform
But now I am trying to add Jersey and following this tutorial http://www.vogella.com/articles/REST/article.html.
But server keeps giving me
java.lang.ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer
when I add these lines in web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID" version="2.5">
<display-name>TestServer</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.test.myproject</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
I have downloaded Jersey JAX-RS 2.1 RI bundle from here and have added all jar files in WEB-INF/lib folder as described in tutorial. And even after two days nothing is working. I have searched several times on Google and apparently people who are using Maven have solved it somehow but I am not using Maven neither did the guy who wrote that tutorial.
Just to check if even com.sun.jersey.spi.container.servlet.ServletContainer exists in imported Jersey jars I tried to just write this fully qualified name in Java and let the intellisense finish names but I couldn't get any intellisense after com.sun.je so my last guess is that there have been some package rearrangement in latest Jersey build and jersey is no longer inside com.sun. I am exhausted and I would appreciate any kind of help.
You have downloaded Jersey 2 (which RI of JAX-RS 2). The tutorial you're referring to uses Jersey 1. Download Jersey 1.17.1 from (here), should be sufficient for you.
Jersey 1 uses com.sun.jersey, and Jersey 2 uses org.glassfish.jersey hence the exception.
Also note that also init-param starting with com.sun.jersey won't be recognized by Jersey 2.
Edit
Registering Resources and Providers in Jersey 2 contains additional info on how to register classes/instances in Jersey 2.
If you are using jersey 2.x then you need different configuration in web.xml as servlet class is change in it. you can update your web.xml with following configuration.
<servlet>
<servlet-name>myrest</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>your.package.path</param-value>
</init-param>
<init-param>
<param-name>unit:WidgetPU</param-name>
<param-value>persistence/widget</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>myrest</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
Add this in pom
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-server</artifactId>
<version>1.17.1</version>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-core</artifactId>
<version>1.17.1</version>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-servlet</artifactId>
<version>1.17.1</version>
</dependency>
It's an eclipse setup issue, not a Jersey issue.
From this thread ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer
Right click your eclipse project Properties -> Deployment Assembly -> Add -> Java Build Path Entries -> Gradle Dependencies -> Finish.
So Eclipse wasn't using the Gradle dependencies when Apache was starting .
I also faced a similar issue. Resolved the problem by going through the step step tutorial from the below link.
http://examples.javacodegeeks.com/enterprise-java/rest/jersey/jersey-hello-world-example/
The main thing to notice is that the jersey libraries should be placed correctly in TOMCAT WEB-INF/lib folder. It is done automatically by the Eclipse settings mentioned in the above link. It will create a WAR file with the dependent JAR Files. Else, you will run into problems with ClassNotFound Exception.
apache-tomcat-7.0.56-windows-x64\apache
-tomcat-7.0.56\webapps\JerseyJSONExample\WEB-INF\lib
"11/23/2014 12:06 AM 130,458 jersey-client-1.9.jar
11/23/2014 12:06 AM 458,739 jersey-core-1.9.jar
11/23/2014 12:06 AM 147,952 jersey-json-1.9.jar
11/23/2014 12:06 AM 713,089 jersey-server-1.9.jar"
4 File(s) 1,450,238 bytes
The second tutorial explains about how to create a Webservice which produces and consumes JSON output.
http://examples.javacodegeeks.com/enterprise-java/rest/jersey/json-example-with-jersey-jackson/
Both the links gave a good picture on how things work and save a lot of time.
try this :
org.glassfish.jersey.servlet.ServletContainer
on servlet-class
I had the same problem as you though I have followed a different guide: http://www.mkyong.com/webservices/jax-rs/jersey-hello-world-example/
The strange part is that, in this guide I have used, I should not have any problem with compatibility between versions (1.x against 2.x) because following the guide you use the jersey 1.8.x on pom.xmland in the web.xmlyou refer to a class (com.sun.jersey.spi.container.servlet.ServletContainer) as said before of 1.x version. So as I can infer this should be working.
My guess is because I'm using JDK 1.7 this class does not exist anymore.
After, I tried to resolve with the answers before mine, did not helped, I have made changes on the pom.xmland on the web.xml the error changed to: java.lang.ClassNotFoundException: org.glassfish.jersey.servlet.ServletContainer
Which supposedly should be exist!
As result of this error, I found a "new" solution: http://marek.potociar.net/2013/06/13/jax-rs-2-0-and-jersey-2-0-released/
With Maven (archetypes), generate a jersey project, likes this:
mvn archetype:generate -DarchetypeGroupId=org.glassfish.jersey.archetypes -DarchetypeArtifactId=jersey-quickstart-webapp -DarchetypeVersion=2.0
And it worked for me! :)
We get this error because of build path issue. You should add "Server Runtime" libraries in Build Path.
"java.lang.ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer"
Please follow below steps to resolve class not found exception.
Right click on project --> Build Path --> Java Build Path --> Add Library --> Server Runtime --> Apache Tomcat v7.0
I encountered the same error today although I was using Jersey 1.x, and had the right jars in my classpath. For those who'd like to follow the vogella tutorial to the letter, and use the 1.x jars, you'd need to add the jersey libraries to WEB-INF/lib folder. This will certainly resolve the problem.
you need to add jersey-bundle-1.17.1.jar to lib of project
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<!-- <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> -->
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<!-- <param-name>jersey.config.server.provider.packages</param-name> -->
<param-value>package.package.test</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
You must replace in your web.xml:
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.test.myproject</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
for this:
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.test.myproject</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
this is Jersey 2.x uses org.glassfish.jersey packages instead of com.sun.jersey (which is used by Jersey 1.x) and hence the exception. Note that also init-param starting with com.sun.jersey won't be recognized by Jersey 2.x once you migrate to JAX-RS 2.0 and Jersey 2.x
if at any moment you use maven, your pom.xml would be this:
<dependency>
<groupId>org.glassfish.jersey.core</groupId>
<artifactId>jersey-server</artifactId>
<version>2.X</version>
</dependency>
replace 2.X for your desire version, e.g. 2.15
A simple workaround is , check whether you have dependencies or libs in deployment assembly of eclipse.probably if you are using tomcat , the server might not have identified the libs we are using . in that case specify it explicitly in deployment assembly.
Coming back to the original problem - java.lang.ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer
As rightly said above, in JAX 2.x version, the ServletContainer class has been moved to the package - org.glassfish.jersey.servlet.ServletContainer. The related jar is jersey-container-servlet-core.jar which comes bundled within the jaxrs-ri-2.2.1.zip
JAX RS can be worked out without mvn by manually copying all jars contained within zip file jaxrs-ri-2.2.1.zip (i have used this version, would work with any 2.x version) to WEB-INF/lib folder. Copying libs to right folder makes them available at runtime.
This is required if you are using eclipse to build and deploy your project.
In pom.xml file we need to add
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-core</artifactId>
<version>1.8</version>
</dependency>
The same error and wasted 2+ hours debugging and trying all options. I was not using the Maven/POM, so I could not leverage that solution given by few.
Finally the following resolved it: Adding the jars directly to the tomcat/lib (NOT WEB-INF\lib) folder and restarting the tomcat.
If anyone is trying to build a hello world application using Jersey, I think one of the easiest ways is to follow Jersey documentation.
https://jersey.github.io/download.html
If you are already using maven, it'd take only a few minutes to see the result.
I used below.
mvn archetype:generate -DarchetypeGroupId=org.glassfish.jersey.archetypes -DarchetypeArtifactId=jersey-quickstart-webapp -DarchetypeVersion=2.26
It basically depends on which version jersey you are using. If you are using Jersey ver.1.X.X you need to add
Jersey 1 uses "com.sun.jersey", and Jersey 2 uses org.glassfish. on servlet class tag.
Also, note that also init-param starting with com.sun.jersey won't be recognized by Jersey 2.
And Add all the jar file into WEB-INF lib folder
In my case, it worked after adding the jersey-bundle jar in my tomcat lib.
I have set up my maven project using the m2e plugin in eclipse indigo, and transformed it to an eclipse dynamic web project using mvn eclipse:eclipse -Dwtpversion=1.5. I have managed to get the project up and running in tomcat7, except for my servlets, for which I cannot create the servlet mappings.
I have tried modifying the web.xml file but it throws a ClassNotFoundException. Directory Structure and web.xml :
(ROOT)
+src
+main
+resources
+DrawInitialMap.java
+webapp
(WebContent here)
<web-app>
<servlet>
<servlet-name>DrawInitialMap</servlet-name>
<servlet-class>(groupId).(artifactId).src.main.resources.DrawInitialMap</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>DrawInitialMap</servlet-name>
<url-pattern>/drawInitialMap.do</url-pattern>
</servlet-mapping>
(...)
</web-app>
While the #WebServlet annotation also fails to map the servlet :
#WebServlet(name="drawInitialMap", description="visualizes ttrp on html5 canvas", urlPatterns={"/drawInitialMap.do"})
Thank you in advance, and notify if you need any more of the code.
PS : Keep in mind that the servlet worked perfectly in Dynamic Web Project mode, without Maven
There are several issues.
You should stop using eclipse:eclipse. Instead, install WTP integration for M2E from Eclipse Marketplace
In Maven project, your DrawInitialMap should be in /src/main/classes folder. So, it will be compiled as per default Maven project conventions
The servlet-class element in web.xml requires full class name, i.e. no things like (groupId).(artifactId).src.main.resources.
I have web app, based on Spring 3.0.3, that I've been developing using Eclipse 3.4. While doing so I've been running the web app in Tomcat 6.0.18 from Eclipse. That is, I have Eclipse use the Tomcat installation meaning that Tomcat will, as need, modify files etc. (at least, that's my understanding of what it's doing).
My problem is specifying the values for the contextConfigLocations in the web.xml. When running from within Eclipse this worked fine:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:applicationContext.xml
classpath:applicationContext-security.xml
</param-value>
</context-param>
However, when I package the app into a war file (ROOT.war) and then added it to Tomcat's webapp directory and the try to start Tomcat, I get an error that neither of these applicationContext files can be found. But when I change it to below, Tomcat can find the files:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/config/applicationContext.xml
/WEB-INF/config/applicationContext-security.xml
</param-value>
</context-param>
I should note that applicationContext.xml includes other applicationContext files that also use the classpath: short hand. When running within Tomcat, I need to drop all use of classpath: in favor of relative paths to get Tomcat to see these files.
Great. Tomcat and Eclise are getting along nicely. But JUnit 4.7 is no longer happy. For whatever reason, files specified using #ContextConfiguration in a test class can't be found unless the classpath: short hand is used. Here is an example:
#ContextConfiguration(locations = {"classpath:applicationContext.xml", "classpath:applicationContext-security.xml"})
public class UserDaoTest extends AbstractTransactionalJUnit4SpringContextTests {
#Test
public void testCreateUser() {
}
So applicationContext.xml and applicationContext-security.xml are found without a problem; however, property files that are specified in applicationContext.xml are not found.
<bean id="appProperties" class="org.springframework.beans.factory.config.PropertiesFactoryBean">
<property name="singleton" value="true" />
<property name="ignoreResourceNotFound" value="true" />
<property name="locations">
<list>
<value>/WEB-INF/config/spring/base.spring-config.properties</value>
<value>/WEB-INF/config/spring/local.spring-config.properties</value>
</list>
</property>
</bean>
But if I specify the location of these files using the classpath: short hand, the property files are found. If I do this though, the files won't be found when running from a war file in Tomcat.
For now I've created a applicationContext-test.xml that is a cut-and-paste conglomeration of all of the other applicationContext files wherein I'm using the classpath: short hand. This seems hacky and error prone and I'm wondering what the issue might be across all of these technologies.
Feedback most welcome!
web.xml content should look like
<context-param>
<description>
Spring Context Configuration.
</description>
<param-name>contextConfigLocation</param-name>
<!-- spring loads all -->
<param-value>
classpath*:spring/*.xml,
classpath*:spring/persistence/*.xml,
classpath*:spring/webapp/*.xml</param-value>
</context-param>
see http://static.springsource.org/spring/docs/2.5.x/reference/resources.html#resources-app-ctx-wildcards-in-resource-paths for further reference
the junit config should follow the same convention with classpath*:
but beware spring might load .xml context files you don't want it to do