I am completely new to Jersey and am having trouble with getting it to work.
I first built a project in Eclipse using the Maven archetype for Jersey 2, jersey-quickstart-webapp (or something along those lines).
I then converted that project to a Dynamic Web Application so that I could run it on my Tomcat server on Eclipse.
After doing all of that, I cannot get the 'myresource' URL to work.
'http://localhost:8080/Testing/webapi/myresource' shows nothing.
Web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<!-- This web.xml file is not required when using Servlet 3.0 container,
see implementation details http://jersey.java.net/nonav/documentation/latest/jax-rs.html -->
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>Jersey Web Application</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.personal.Testing.Testing</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/webapi/*</url-pattern>
</servlet-mapping>
</web-app>
Resource class:
package com.personal.Testing.Testing;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
/**
* Root resource (exposed at "myresource" path)
*/
#Path("myresource")
public class MyResource {
/**
* Method handling HTTP GET requests. The returned object will be sent
* to the client as "text/plain" media type.
*
* #return String that will be returned as a text/plain response.
*/
#GET
#Produces(MediaType.TEXT_PLAIN)
public String getIt() {
return "Got it!";
}
}
From a tutorial (https://howtodoinjava.com/jersey/jersey2-hello-world-example/#webxml)
"First, run maven to create you war file.
Then hot deploy war file to your tomcat server..."
Then try hitting your URL. Should work just fine.
It seems you may be doing a simple run for the application you created in Eclipse. In that case, the application will not be deployed to you Web Server (Tomcat in your case).
You need to deploy the application as a war in the server for the URL to work.
Related
I am trying to get working a simple JAX RS example, but I am failing to do so.
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:jsp="http://java.sun.com/xml/ns/javaee/jsp" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>PLAYGROUND</display-name>
<servlet-mapping>
<servlet-name>playground.Rest</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
</web-app>
Rest.java
package playground;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.core.Application;
public class Rest extends Application {
#GET
#Path("hello")
public String helloworld() {
return "Hello World!";
}
}
Accessing http://localhost/{warcontext}/api/hello with the browser (GET) gives me 404 error status
It is probably something very silly but I can't figure out.
Using:
JBoss EAP 6.1.0 (Java EE 6)
web.xml
<servlet>
<servlet-name>RestServlet</servlet-name>
<servlet-class>javax.ws.rs.core.Application</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>RestServlet</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
Rest.java
public class Rest {
#GET
#Path("hello")
public String helloworld() {
return "Hello World!";
}
}
If you need, you can add
YourApplication.java
package playground;
import javax.ws.rs.core.Application;
...
public class YourApplication extends Application {
#Override
public Set<Class<?>> getClasses()
{
Set<Class<?>> yourResources = new HashSet<Class<?>>();
yourResources.add(InvoiceResource.class);
return yourResources;
}
}
Then
web.xml
<servlet>
<servlet-name>RestServlet</servlet-name>
<servlet-class>playground.YourApplication</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>RestServlet</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
If you add #ApplicationPath("/api") for YouApplication, then web.xml shouldn't have servlet-mapping.
You need to extend javax.ws.rs.core.Application (it can remain empty) and annotate it with #ApplicationPath("/ide"), then create a JAX-RS resource, ie, a class with an #Path("/hello") annotation. In this class, you'll just need to have your JAX-RS Resource Method annotated with #GET.
#ApplicationPath("/ide")
public class Rest extends Application { }
#Path("/hello")
public class HelloResource {
#GET
#Path("hello")
public String helloworld() {
return "Hello World!";
}
}
You can also take a look at this example: https://github.com/resteasy/Resteasy/tree/master/jaxrs/examples/oreilly-workbook/ex03_1
You need to declare servlet-class in web.xml (servlet name is not the same):
<servlet>
<servlet-name>Playground REST services</servlet-name>
<servlet-class>your.package.playground.Rest</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Playground REST services</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
Use the full name of your class (including package path name) in servlet-class.
I also had issues getting this to work, so I'm posting a more complete how-to for posterity:
If you're using a standard Tomcat install (or some other servlet container), AFAIK you can't avoid explicitly telling it what servlets to start in the web.xml file*. Since you have to use web.xml anyway, the simplest way to get restful web services working is to forget extending javax.ws.rs.core.Application entirely and just specify the context path there. You can still use standard jax-rs annotations to declare the actual web services.
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0"
>
<servlet>
<servlet-name>rest-test</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.domain.mypackage</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name> rest-test</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Two noteworthy points:
You will need to bundle a REST implementation in your WAR file, since servlet containers don't usually contain one. Since Jersey is the reference implementation for JAX-RS, that's the one I'm using in the servlet-class element above. You can replace this with Apache CXF implementation if you want.
The init-param element tells Jersey which of your packages to search for Java files with web service annotations. Edit this to point to your web services. Note that if you opt to use apache CXF instead of Jersey, the stuff needed in any init-param elements will be different. Someone who knows CXF please post what they would be.
If you're using Maven, just add a dependency to jersey-servlet in the dependencies section of your pom.xml file:
<dependencies>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-servlet</artifactId>
<version>1.18.2</version>
</dependency>
...
</dependencies>
After this, declaring your web services is straight forward using the standard JAX-RS annotations in your Java classes:
package com.domain.mypackage;
import javax.ws.rs.Consumes;
import javax.ws.rs.Produces;
import javax.ws.rs.GET;
import javax.ws.rs.MatrixParam;
import javax.ws.rs.Path;
// It's good practice to include a version number in the path so you can have
// multiple versions deployed at once. That way consumers don't need to upgrade
// right away if things are working for them.
#Path("calc/1.0")
public class CalculatorV1_0 {
#GET
#Consumes("text/plain")
#Produces("text/plain")
#Path("addTwoNumbers")
public String add(#MatrixParam("firstNumber") int n1, #MatrixParam("secondNumber") int n2) {
return String.valueOf(n1 + n2);
}
}
This should be all you need. If your Tomcat install is running locally on port 8080 and you deploy your WAR file to the context myContext, going to
http://localhost:8080/myContext/rest/calc/1.0/addTwoNumbers;firstNumber=2;secondNumber=3
...should produce the expected result (5).
Cheers!
* Someone please correct me if you know of a way to a add the Jersey servlet to the context in Tomcat without using web.xml--maybe by using a context or life cycle listener?
I'm trying to create and deploy a RESTful Web Service using JAX-RS and deploy it to tomcat. I don't want to use any IDE.
In Tomcat I have the following directory structure inside webapps\
notifire\WEB-INF\
|
---> web.xml
|
---> \classes/Notifier.class
|
---> \lib\javax.ws.rs-api-2.0
my web.xml contains:
<servlet>
<servlet-name>Web Service Servlet</servlet-name>
<servlet-class>javax.ws.rs.core.Application</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Web Service Servlet</servlet-name>
<url-pattern>/webservice/*</url-pattern>
</servlet-mapping>
and the class file Notifier.class was compiled from the file Notifier.java.
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.UriInfo;
#Path("notifier")
public class Notifier {
#Context
private UriInfo context;
#GET
#Produces("text/html")
public String getHTML() {
return "<p></p>";
}
}
When I try to access the Web Service at http://localhost:8080/notifire/webservice/notifier I get the following error:
--type Exception report
--message Class javax.ws.rs.core.Application is not a Servlet
--description The server encountered an internal error that prevented it from fulfilling this request.
Any help is appreciated.
You have the wrong class for your Servlet. Not sure why you are not wanting to use an IDE, but there is a maven archetype that will layout your project structure for you using the appropriate classes that the Jersey developers have defined. My web.xml looks like this:
<servlet>
<servlet-name>Jersey Web Application</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.pluralsight</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/webapi/*</url-pattern>
</servlet-mapping>
I cover all of this in this course here.
I had trouble getting this to work as well. Since you have to use web.xml anyway, the simplest way to get restful web services working is to forget extending javax.ws.rs.core.Application entirely and just specify the context path there. You can still use standard jax-rs annotations to declare the actual web services.
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0"
>
<servlet>
<servlet-name>rest-test</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.domain.mypackage</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name> rest-test</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Two noteworthy points:
You will need to bundle a REST implementation in your WAR file, since servlet containers don't usually contain one. Since Jersey is the reference implementation for JAX-RS, that's the one I'm using in the servlet-class element above. You can replace this with Apache CXF implementation if you want.
The init-param element tells Jersey which of your packages to search for Java files with web service annotations. Edit this to point to your web services. Note that if you opt to use apache CXF instead of Jersey, the stuff needed in any init-param elements will be different. Someone who knows CXF please post what they would be.
If you're using Maven, just add a dependency to jersey-servlet in the dependencies section of your pom.xml file:
<dependencies>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-servlet</artifactId>
<version>1.18.2</version>
</dependency>
...
</dependencies>
After this, declaring your web services is straight forward using the standard JAX-RS annotations in your Java classes:
package com.domain.mypackage;
import javax.ws.rs.Consumes;
import javax.ws.rs.Produces;
import javax.ws.rs.GET;
import javax.ws.rs.MatrixParam;
import javax.ws.rs.Path;
// It's good practice to include a version number in the path so you can have
// multiple versions deployed at once. That way consumers don't need to upgrade
// right away if things are working for them.
#Path("calc/1.0")
public class CalculatorV1_0 {
#GET
#Consumes("text/plain")
#Produces("text/plain")
#Path("addTwoNumbers")
public String add(#MatrixParam("firstNumber") int n1, #MatrixParam("secondNumber") int n2) {
return String.valueOf(n1 + n2);
}
}
This should be all you need. If your Tomcat install is running locally on port 8080 and you deploy your WAR file to the context myContext, going to
http://localhost:8080/myContext/rest/calc/1.0/addTwoNumbers;firstNumber=2;secondNumber=3
...should produce the expected result (5).
Cheers!
* Someone please correct me if you know of a way to a add the Jersey servlet to the context in Tomcat without using web.xml--maybe by using a context or life cycle listener?
I wrote a very simple example using Jersey.
I downloaded the latest jar files from the jersey website into lib folder in WEB-INF.
My class and web.xml are below.
When I provide URL localhost:8080/SimpleJersey/rest/test I get 404 error (not found).
However when I use Maven, it works.
I use Eclipse Kepler, Glassfish 4 server and Java 7.
What am I doing wrong in the non-Maven version?
Thanks.
Class:
package com.simplejersey;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
#Path("/test")
public class MyResources
{
#GET
#Produces("text/plain")
public String getIt()
{
return "Hello there!";
}
}
Web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
<display-name>SimpleJersey</display-name>
<servlet>
<servlet-name>jersey-servlet</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.simplejersey</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-servlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
I found the solution in this post by Michal Gajdos: Jersey REST Web Service, Tomcat, Eclipse and 404's
The issue is (quoting from the abovementioned post):
Jersey 2.0 does not recognize init-param with name com.sun.jersey.config.property.packages (web.xml). Try to change it to jersey.config.server.provider.packages as described in ServerProperties.PROVIDER_PACKAGES (link)."
Be carefeul when you copy web.xml from websites that show older solutions or versions (like I did). Jersey is being updated too...
I am testing a very simple REST server with Jersey and Servlet 3.0 implementation on Tomcat 7.0. I have programmed a simple PoJo:
package toplevel;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
#Path("/pojo")
public class PoJo {
#GET
#Produces("text/plain")
public String hello() {
return "Hello, World";
}
}
I have put the following in the WEB-INF/web.xml file (running on Servlet 3.0):
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
<display-name>RestTest</display-name>
<servlet>
<servlet-name>toplevel.PoJo</servlet-name>
</servlet>
<servlet-mapping>
<servlet-name>toplevel.PoJo</servlet-name>
<url-pattern>/pojo</url-pattern>
</servlet-mapping>
</web-app>
When I deploy, I get a HTTP Status 500 response. This seems to me that the webserver is recognizing that something should be served from /pojo, but that the corresponding class PoJo is not found. The jersey specific jars (version 1.17) are in the WEB-INF/lib dir:
activation-1.1.1.jar jersey-client-1.17.jar junit-4.9.jar
asm-3.3.1.jar jersey-core-1.17.jar persistence-api-1.0.2.jar
jaxb-api-2.2.4.jar jersey-server-1.17.jar stax-api-1.0-2.jar
jaxb-impl-2.2.4-1.jar jsr311-api-1.1.1.jar
Does anyone recognize this ?
You need to tell Jersey where to find your REST resource. Your web.xml should look something like this:
<servlet>
<servlet-name>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>toplevel</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Service</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Adding jersey-servlet:1.17.jar took care of that problem for me.
Within SpringSource Tool Suite I created a standard google app engine project. I added Jersey for REST support. The development server starts up fine, but when I try to GET a URL (e.g. http://localhost:8888/sibibjersey/api) I'm simply getting a 404. I suppose this is a simple configuration issue, but the solutions seems to hide away from me....
Here the main files:
web.xml looks like this.
<?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" version="2.5">
<init-param>
<param-name>com.sun.jersey.config.feature.DisableWADL</param-name>
<param-value>true</param-value>
</init-param>
<servlet>
<servlet-name>Jersey Web Application</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.sibib.main</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
I tried variations of the url-pattern like /* and /rest/*, but none seemed to work.
The only Java class in com.sibib.main is InfoResource.java:
package com.sibib.main;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
#Path("/api")
public class InfoResource {
#GET
public String info() {
return "Hello Jersey on Google App Engine";
}
}
I tried adding #Path to the info function, but no effect. When I start the server and navigate e.g. to http://localhost:8888/sibibjersey/api I'm simply getting a 404.
Loading http://localhost:8888 loads the index.html in the war folder.
These are the lib referenced in the project:
activation-1.1.1.jar
appengine-api-1.0-sdk-1.6.1.jar
appengine-api-labs-1.6.1.jar
appengine-jsr107cache-1.6.1.jar
asm-3.3.1.jar
datanucleus-appengine-1.0.10.final.jar
datanucleus-core-1.1.5.jar
datanucleus-jpa-1.1.5.jar
geronimo-jpa_3.0_spec-1.1.1.jar
geronimo-jta_1.1_spec-1.1.1.jar
google_sql.jar
jackson-core-asl-1.9.2.jar
jackson-jaxrs-1.9.2.jar
jackson-mapper-asl-1.9.2.jar
jaxb-api-2.2.4.jar
jaxb-impl-2.2.4-1.jar
jdo2-api-2.3-eb.jar
jersey-bundle-1.11.jar
jersey-client-1.11.jar
jersey-core-1.11.jar
jersey-json-1.11.jar
jersey-server-1.11.jar
jettison-1.1.jar
jsr107cache-1.1.jar
persistence-api-1.0.2.jar
stax-api-1.0-2.jar
Any hint is greatly appreciated!
Thanks!
Can you try removing sibibjersey from the context path, just try with http: //localhost:8888/api
In my local, I have not seen Google App Engine's application having project path along with http: //localhost:8888
I followed http://tugdualgrall.blogspot.com/2010/02/create-and-deploy-jax-rs-rest-service.html and it worked for me without any issues.