这是我第一次使用Eclipse,这使我大为恼火。
我安装了Tomcat 6.0,下载了Jersey库,然后按照以下教程进行操作:http : //www.vogella.com/articles/REST/article.html#first_client
我将项目名称创建为RestExample,其中有一个程序包de.jay.jersey.first,其中有一个类HelloWorldResource,如下所示:
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> "; } }
和我的web.xml看起来像
<?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>
我正在尝试使用curl作为:
curl http:// localhost:8081 / RestExample / rest / hello
Apache Tomcat / 6.0.35-错误报告
类型 状态报告
消息 / RestExample / rest / hello
说明 所请求的资源(/ RestExample / rest / hello)不可用。
Apache Tomcat / 6.0.35
问题是我应该在web.xml中进行哪些更改,以便可以访问该资源?
我尝试过RestExample / de.jay.jersey.first / rest / hello,但仍然无法正常工作。TOmcat正在正常运行。
我在Tomcat 7.0上试过了,效果很好:
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>
浏览到http:// localhost:8084 / RestExample / rest / hello,它可以正常运行