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问题描述

这是我第一次使用Eclipse，并且引起我的愤怒。

我安装了Tomcat 6.0，下载了泽西图书馆，并按照以下教程：

我创建了项目名称作为RestExample，在其中我有一个包de.jay.jersey.first，在我有一个类HelloWorldResource，这里是它的样子：

  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 {
 //如果TEXT_PLAIN请求，则调用此方法
 @GET 
 @Produces （MediaType.TEXT_PLAIN）
 public String sayPlainTextHello（）{
 returnHello Jersey; 
} 
 
 //如果XML请求，则调用此方法
 @GET 
 @Produces（MediaType.TEXT_XML）
 public String sayXMLHello（）{ 
 return<？xml version = \1.0\？> +< hello> Hello Jersey+< / hello>; 
} 
 
 //如果HTML请求
 @GET 
 @Produces（MediaType.TEXT_HTML）
 public String sayHtmlHello（）{ 
 return< html>+< title> +Hello Jersey+< / title> 
 +< body>< h1> +Hello Jersey+< / body>< / h1> +< / html>; 
} 
}

我的web.xml看起来像

 <？xml version =1.0encoding =UTF-8？> 
< web-app xmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexmlns =http://java.sun.com/xml/ns/javaeexmlns ：web =http://java.sun.com/xml/ns/javaee/web-app_2_5.xsdxsi：schemaLocation =http://java.sun.com/xml/ns/javaee http：// java.sun.com/xml/ns/javaee/web-app_2_5.xsdid =WebApp_IDversion =2.5> 
< display-name> RestExample< / display-name> 
< servlet> 
< servlet-name> Jersey REST服务< / 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服务< / servlet-name> 
< url-pattern> / rest / *< / url-pattern> 
< / servlet-mapping> 
< / web-app>

我正在尝试使用卷曲：

Apache Tomcat / 6.0.35 - 错误报告

HTTP状态404 - / RestExample / rest / Hello

类型状态
端口

消息 / RestExample / rest / hello

de
scription 请求的资源/ RestExample / rest / hello）是
不可用。

Apache Tomcat / 6.0.35

问题是我应该在网络上改变什么.xml，以便我可以访问该资源？

我尝试了RestExample / de.jay.jersey.first / rest / hello，它仍然没有起作用。 TOMcat正在运行没有错误。

解决方案

我尝试使用Tomcat 7.0，它工作正常：

  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 {
 //如果TEXT_PLAIN请求，则调用此方法
 @GET 
 @Produces （MediaType.TEXT_PLAIN）
 public String sayPlainTextHello（）{
 returnHello Jersey; 
} 
 
 //如果XML请求，则调用此方法
 @GET 
 @Produces（MediaType.TEXT_XML）
 public String sayXMLHello（）{ 
 return<？xml version = \1.0\？> +< hello> Hello Jersey+< / hello>; 
} 
 
 //如果HTML请求
 @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 =1.0encoding =UTF-8？> 
< web-app version =3.0xmlns =http://java.sun.com/xml/ns/javaeexmlns：xsi =http://www.w3.org/2001/ XMLSchema-instancexsi：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服务< / 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服务< / servlet-name> 
< url-pattern> / rest / *< / url-pattern> 
< / servlet-mapping> 
< / web-app>

浏览到，它工作正常

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 report

message /RestExample/rest/hello

description 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.

解决方案

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

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