我有一个关于我想通过我的应用程序进行的休息通话的问题。我正在使用Tomcat 7和泽西岛。
这就是我的pom.xml的样子:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.example</groupId>
<artifactId>JavaWeb</artifactId>
<packaging>war</packaging>
<version>0.0.1-SNAPSHOT</version>
<name>JavaWeb Maven Webapp</name>
<url>http://maven.apache.org</url>
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>3.8.1</version>
<scope>test</scope>
</dependency>
<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>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-json</artifactId>
<version>1.17.1</version>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>eclipselink</artifactId>
<version>2.5.1</version>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>javax.persistence</artifactId>
<version>2.0.0</version>
<scope>compile</scope>
</dependency>
</dependencies>
<build>
<finalName>JavaWeb</finalName>
<plugins>
<!-- Eclipse project -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-eclipse-plugin</artifactId>
<version>2.9</version>
<configuration>
<!-- Always download and attach dependencies source code -->
<downloadSources>true</downloadSources>
<downloadJavadocs>false</downloadJavadocs>
<!-- Avoid type mvn eclipse:eclipse -Dwtpversion=2.0 -->
<wtpversion>2.0</wtpversion>
</configuration>
</plugin>
<!-- Set JDK Compiler Level -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>2.3.2</version>
<configuration>
<source>${jdk.version}</source>
<target>${jdk.version}</target>
</configuration>
</plugin>
<!-- For Maven Tomcat Plugin -->
<plugin>
<groupId>org.apache.tomcat.maven</groupId>
<artifactId>tomcat7-maven-plugin</artifactId>
<version>2.2</version>
<configuration>
<path>/JavaWeb</path>
</configuration>
</plugin>
</plugins>
</build>
</project>
我的web.xml:
<!DOCTYPE web-app PUBLIC
"-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app>
<display-name>Archetype Created Web Application</display-name>
<!-- .. -->
<servlet>
<servlet-name>jersey</servlet-name>
<servlet-class>
com.sun.jersey.spi.container.servlet.ServletContainer
</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>demo.rest</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
这是我的班级,我有我的服务。我重写了它以匹配你好的例子,看看我做错了什么,但它也没有这样工作:
@Path("/hello")
public class VehicleRest {
// 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> ";
}
}
我试图在浏览器中像这样访问它:http://localhost:8081/JavaWeb/rest/hello
当我运行服务器时,在我的路径中声明了我的路径,因此我认为这不是问题所在。也许Jersey配置存在问题。我也查了解其他问题和答案,但没有任何帮助。如果我再次进行配置并再次下载泽西罐子会有帮助吗?它只是在我尝试访问该URI时找不到404。
提前致谢!