【发布时间】:2015-08-27 08:24:06
【问题描述】:
我正在尝试创建一个Web Service,它将温度从C 转换为F。我遵循了一个基本教程:
http://crunchify.com/how-to-build-restful-service-with-java-using-jax-rs-and-jersey/
我遵循了帖子建议的相同步骤,并且正在使用apache Tomcat v7 运行该项目。但是,输出并没有按照教程所说的方式显示。当我运行我的项目时,我收到一个错误,上面写着HTTP Status 404 - /RestJerseyExample/ ; The resource is not available。我对此很陌生,我不明白我哪里可能出错了。我正在处理Windows 7。建议非常感谢!
这是我的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: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_3_0.xsd"
version="3.0">
<display-name>CrunchifyRESTJerseyExample</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/crunchify/*</url-pattern>
</servlet-mapping>
另外,这是我的pom.xml:
<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/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>RESTJerseyExample</groupId>
<artifactId>RESTJerseyExample</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<build>
<sourceDirectory>src</sourceDirectory>
<plugins>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.1</version>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
<plugin>
<artifactId>maven-war-plugin</artifactId>
<version>2.4</version>
<configuration>
<warSourceDirectory>WebContent</warSourceDirectory>
<failOnMissingWebXml>false</failOnMissingWebXml>
</configuration>
</plugin>
</plugins>
</build>
<dependencies>
<dependency>
<groupId>asm</groupId>
<artifactId>asm</artifactId>
<version>3.3.1</version>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-bundle</artifactId>
<version>1.19</version>
</dependency>
<dependency>
<groupId>org.json</groupId>
<artifactId>json</artifactId>
<version>20140107</version>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-server</artifactId>
<version>1.19</version>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-core</artifactId>
<version>1.19</version>
</dependency>
</dependencies>
</project>
【问题讨论】:
-
请提供web.xml等配置和资源文件
-
请检查我的编辑
-
您是否尝试通过“/RestJerseyExample/”上下文访问您的服务器?我认为应该是“/CrunchifyRESTJerseyExample/crunchify/{your_resource_context}”
-
教程中提到了第一个。然而,我的班级名称是第二个..
标签: eclipse web-services tomcat7 restful-url