Rest Web services returning a 404

2020-08-14 10:00发布

站内文章 / Java
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问题:

This is my first time using Eclipse, and is causing me to rage a lot.

I installed Tomcat 6.0, downloaded the Jersey libraries, and I followed the tutorials from : http://www.vogella.com/articles/REST/article.html#first_client

I created the Project Name as RestExample, and within that I have a package de.jay.jersey.first and within that I have a class HelloWorldResource, and here is what it looks like:

package de.jay.jersey.first;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/hello")
public class HelloWorldResource {
// This method is called if TEXT_PLAIN is request
@GET
@Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
    return "Hello Jersey";
}

// This method is called if XML is request
@GET
@Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
    return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}

// This method is called if HTML is request
@GET
@Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
    return "<html> " + "<title>" + "Hello Jersey" + "</title>"
            + "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
}
}

and my web.xml looks like 

<?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>RestExample</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>de.jay.jersey.first</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>

ANd I am trying to use curl as:

curl http://localhost:8081/RestExample/rest/hello

Apache Tomcat/6.0.35 - Error report

HTTP Status 404 - /RestExample/rest/Hello

type Status re port

message /RestExample/rest/hello

de scription The requested resource (/RestExample/rest/hello) is not available.

Apache Tomcat/6.0.35

The question is what should I change in the web.xml so that I can access that resource?

I tried RestExample/de.jay.jersey.first/rest/hello, and it still did not work. TOmcat is running without errors.

回答1:

I tried it with Tomcat 7.0 and it works fine:

package de.jay.jersey.first;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/hello")
public class HelloWorldResource {
// This method is called if TEXT_PLAIN is request
    @GET
    @Produces(MediaType.TEXT_PLAIN)
    public String sayPlainTextHello() {
        return "Hello Jersey";
    }

// This method is called if XML is request
    @GET
    @Produces(MediaType.TEXT_XML)
    public String sayXMLHello() {
        return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
    }

// This method is called if HTML is request
    @GET
    @Produces(MediaType.TEXT_HTML)
    public String sayHtmlHello() {
        return "<html> " + "<title>" + "Hello Jersey" + "</title>"
                + "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
    }
}

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<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>RestExample</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>de.jay.jersey.first</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>

Browsed to http://localhost:8084/RestExample/rest/hello and it works ok



回答2:

It took me a lot of time to figure out why it wasn't working for me inspite of looking all over the web for solution. The mistake I was making was that package names were not up to date to the new jersey api. Here's what updated package names should look like (Web.xml):

<?xml version="1.0" encoding="UTF-8"?>
<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>RestExample</display-name>
    <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>de.jay.jersey.first</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>

Notice that <servlet-class> and <param-name> are different(updated) from vogella tutorial. It may not be the answer to this particular question but might help someone. I found it from here.



回答3:

Please add all the given Jars in your project

Project (Right Click)>Properties>Java Build Path>Libraries>Add JARs/Add External JARs

  1. asm-3.1.jar
  2. jersey-bundle-1.14.jar
  3. jersey-client.jar
  4. jersey-core.1.17.1.jar
  5. jersey-server-1.17.jar
  6. jersey-servlet-1.17.jar


回答4:

i looked for a solution to the same problem for hours too.

this solved my problem:

if you use a Maven-Project (for example with archetype maven-archetype-webapp) and the class HelloWorldResource is implemented in the folder src/main/resources this class doesn't get compiled (for example then running "mvn clean package" or "run on server" in eclipse)

Implement HelloWorldResource in folder src/main/java instead and no more 404 occures.. (if you create Maven-Project with maven-archetype-webapp the folder needs to be manually created)



回答5:

Check if your path have this bar '/' example: @Path('/path') in some cases this problem is only the missing bar!



回答6:

If you are using Jersey 2.XX user ServletAdaptor instead. Like this, 

    <servlet-class>com.sun.jersey.server.impl.container.servlet.ServletAdaptor</servlet-class>
    <!-- <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> -->


回答7:

like the aticle says:

// Get the Todos
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.TEXT_XML).get(String.class));
    // Get XML for application
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.APPLICATION_JSON).get(String.class));
    // Get JSON for application
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.APPLICATION_XML).get(String.class));

you try to specify the method path which you want to call



标签： java eclipse web-services rest tomcat
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